NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

Access premium articles, webinars, resources to make the best decisions for career, course, exams, scholarships, study abroad and much more with

Plan, Prepare & Make the Best Career Choices

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

Edited By Vishal kumar | Updated on Nov 17, 2023 08:45 AM IST

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- In this article, you will get NCERT solutions for Class 11 Maths Chapter 9 miscellaneous exercise. The miscellaneous exercise consists of mixed kinds of questions from all the topics covered in this chapter. You must have solved the previous exercise of this chapter. Now you can try to solve problems from the miscellaneous exercise chapter 9 Class 11.

This Story also Contains

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- Download Free PDF
NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise
Access Sequences And Series Class 11 Chapter 9:Miscellaneous Exercise
Question:1 Show that the sum of and terms of an A.P. is equal to twice the term.
More About NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-
Topic Covered in Class 11 Chapter 9 Maths Miscellaneous Solutions
Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-
Key Features of Class 11 Maths ch 9 Miscellaneous Exercise Solutions
NCERT Solutions of Class 11 Subject Wise
NCERT Solutions for Class 11 Maths
Subject Wise NCERT Exampler Solutions

As you have already familiar with the NCERT syllabus Class 11 Maths chapter 9 concepts, you can try to solve problems from miscellaneous exercise chapter 9 Class 11. These problems are a bit lengthy and complex as compared to the other exercise of this chapter but these problems will check your understanding of this chapter. You don't need to worry if you are not able to solve these problems by yourself at first. Class 11 Maths Chapter 9 miscellaneous exercise solutions are here to help you where you will get detailed miscellaneous solutions. The class 11 maths ch 9 miscellaneous exercise solutions Maths Miscellaneous Exercises are meticulously crafted by subject experts from Careers360. Each step is explained in detail, providing comprehensive and well-explained solutions. Furthermore, the availability of a PDF version of class 11 chapter 9 maths miscellaneous solutions allows students to access the solutions offline, providing flexibility and convenience to study according to their preferences. You can check NCERT Solutions link if you are looking for NCERT solutions for all the Classes at one place.

Also, see

** As per the CBSE Syllabus for the academic year 2023-24, it has been noted that miscellaneous exercise class 11 chapter 9 has been renumbered and is now recognized as Chapter 8.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise

Download PDF

Access Sequences And Series Class 11 Chapter 9:Miscellaneous Exercise

Question:1 Show that the sum of $( m+n)^{th}$ and $( m-n)^{th}$ terms of an A.P. is equal to twice the $m^{th}$ term.

Answer:

Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

$a_k=a+(k-1)d$

$\therefore a_{m+n}=a+(m+n-1)d$

$\therefore a_{m-n}=a+(m-n-1)d$

$a_m=a+(m-1)d$

$a_{m+n}+ a_{m-n}=a+(m+n-1)d+a+(m-n-1)d$

$=2a+(m+n-1+m-n-1)d$

$=2a+(2m-2)d$

$=2(a+(m-1)d)$

$=2.a_m$

Hence, the sum of $( m+n)^{th}$ and $( m-n)^{th}$ terms of an A.P. is equal to twice the $m^{th}$ term.

Question:2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer:

Let three numbers of AP are a-d, a, a+d.

According to given information ,

$a-d+a+a+d=24$

$3a=24$

$\Rightarrow a=8$

$(a-d)a(a+d)=440$

$\Rightarrow (8-d)8(8+d)=440$

$\Rightarrow (8-d)(8+d)=55$

$\Rightarrow (8^2-d^2)=55$

$\Rightarrow (64-d^2)=55$

$\Rightarrow d^2=64-55=9$

$\Rightarrow d=\pm 3$

When d=3, AP= 5,8,11 also if d=-3 ,AP =11,8,5.

Thus, three numbers are 5,8,11.

Question:3 Let the sum of n, 2n, 3n terms of an A.P. be $S_1 , S_2 , S_3$ , respectively, show that $S_3 = 3(S_2 - S_1)$

Answer:

Let a be first term and d be common difference of AP.

$S_n=\frac{n}{2}[2a+(n-1)d]=S_1..................................1$

$S_2n=\frac{2n}{2}[2a+(2n-1)d]=S_2..................................2$

$S_2_n=\frac{3n}{2}[2a+(3n-1)d]=S_3..................................3$

Subtract equation 1 from 2,

$S_2-S_1=\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[4a+4nd-2d-2a-nd+d]$

$=\frac{n}{2}[2a+3nd-d]$

$=\frac{n}{2}[2a+(3n-1)d]$

$\therefore 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1)d]=S_3$

Hence, the result is proved.

Question:4 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer:

Numbers divisible by 7 from 200 to 400 are $203,210,.............399$

This sequence is an A.P.

Here , first term =a =203

common difference = 7.

We know , $a_n = a+(n-1)d$

$399 = 203+(n-1)7$

$\Rightarrow \, \, 196 = (n-1)7$

$\Rightarrow \, \, 28 = (n-1)$

$\Rightarrow \, \, n=28+1=29$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{29}{2}[2(203)+(29-1)7]$

$= \frac{29}{2}[2(203)+28(7)]$

$= 29\times 301$

$= 8729$

The sum of numbers divisible by 7from 200 to 400 is 8729.

Question:5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer:

Numbers divisible by 2 from 1 to 100 are $2,4,6................100$

This sequence is an A.P.

Here , first term =a =2

common difference = 2.

We know , $a_n = a+(n-1)d$

$100= 2+(n-1)2$

$\Rightarrow \, \, 98 = (n-1)2$

$\Rightarrow \, \, 49 = (n-1)$

$\Rightarrow \, \, n=49+1=50$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{50}{2}[2(2)+(50-1)2]$

$= \frac{50}{2}[2(2)+49(2)]$

$= 25\times 102$

$= 2550$

Numbers divisible by 5 from 1 to 100 are $5,10,15................100$

This sequence is an A.P.

Here , first term =a =5

common difference = 5.

We know , $a_n = a+(n-1)d$

$100= 5+(n-1)5$

$\Rightarrow \, \, 95 = (n-1)5$

$\Rightarrow \, \, 19 = (n-1)$

$\Rightarrow \, \, n=19+1=20$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{20}{2}[2(5)+(20-1)5]$

$= \frac{20}{2}[2(5)+19(5)]$

$= 10\times 105=1050$

Numbers divisible by both 2 and 5 from 1 to 100 are $10,20,30................100$

This sequence is an A.P.

Here , first term =a =10

common difference = 10

We know , $a_n = a+(n-1)d$

$100= 10+(n-1)10$

$\Rightarrow \, \, 90 = (n-1)10$

$\Rightarrow \, \, 9 = (n-1)$

$\Rightarrow \, \, n=9+1=10$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{10}{2}[2(10)+(10-1)10]$

$= \frac{10}{2}[2(10)+9(10)]$

$= 5\times 110=550$

$\therefore Required \, \, sum=2550+1050-550=3050$

Thus, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.

Question:6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

Numbers divisible by 4, yield remainder as 1 from 10 to 100 are $13,17,..................97$

This sequence is an A.P.

Here , first term =a =13

common difference = 4.

We know , $a_n = a+(n-1)d$

$97 = 13+(n-1)4$

$\Rightarrow \, \, 84 = (n-1)4$

$\Rightarrow \, \, 21 = (n-1)$

$\Rightarrow \, \, n=21+1=22$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{22}{2}[2(13)+(22-1)4]$

$= \frac{22}{2}[2(13)+21(4)]$

$= 11\times 110$

$=1210$

The sum of numbers divisible by 4 yield 1 as remainder from 10 to 100 is 1210.

Question:7 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y $\epsilon$ N such that f(1) = 3 and

$\sum_{x=1}^{n} f(x) = 120$ , find the value of n.

Answer:

Given : f (x +y) = f(x) f(y) for all x, y $\epsilon$ N such that f(1) = 3

$f(1) = 3$

Taking $x=y=1$ , we have

$f(1+1)=f(2)=f(1)*f(1)=3*3=9$

$f(1+1+1)=f(1+2)=f(1)*f(2)=3*9=27$

$f(1+1+1+1)=f(1+3)=f(1)*f(3)=3*27=81$

$f(1),f(2),f(3),f(4).....................$ is $3,9,27,81,..............................$ forms a GP with first term=3 and common ratio = 3.

$\sum_{x=1}^{n} f(x) = 120=S_n$

$S_n=\frac{a(1-r^n)}{1-r}$

$120=\frac{3(1-3^n)}{1-3}$

$40=\frac{(1-3^n)}{-2}$

$-80=(1-3^n)$

$-80-1=(-3^n)$

$-81=(-3^n)$

$3^n=81$

Therefore, $n=4$

Thus, value of n is 4.

Question:8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Answer:

Let the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2

$S_n=\frac{a(1-r^n)}{1-r}$

$315=\frac{5(1-2^n)}{1-2}$

$63=\frac{(1-2^n)}{-1}$

$-63=(1-2^n)$

$-63-1=(-2^n)$

$-64=(-2^n)$

$2^n=64$

Therefore, $n=6$

Thus, the value of n is 6.

Last term of GP=6th term $=a.r^{n-1}=5.2^5=5*32=160$

The last term of GP =160

Question:9 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Answer:

Given: The first term of a G.P. is 1. The sum of the third term and fifth term is 90.

$a=1$

$a_3=a.r^2=r^2$ $a_5=a.r^4=r^4$

$\therefore \, \, r^2+r^4=90$

$\therefore \, \, r^4+r^2-90=0$

$\Rightarrow \, \, r^2=\frac{-1\pm \sqrt{1+360}}{2}$

$r^2=\frac{-1\pm \sqrt{361}}{2}$ $r^2=-10 \, or \, 9$

$r=\pm 3$

Thus, the common ratio of GP is $\pm 3$ .

Question:10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer:

Let three terms of GP be $a,ar,ar^2.$

Then, we have $a+ar+ar^2=56$

$a(1+r+r^2)=56$ ...............................................1

$a-1,ar-7,ar^2-21$ from an AP.

$\therefore ar-7-(a-1)=ar^2-21-(ar-7)$

$ar-7-a+1=ar^2-21-ar+7$

$ar-6-a=ar^2-14-ar$

$\Rightarrow ar^2-2ar+a=8$

$\Rightarrow ar^2-ar-ar+a=8$
$\Rightarrow a(r^2-2r+1)=8$

$\Rightarrow a(r^2-1)^2=8$ ....................................................................2

From equation 1 and 2, we get

$\Rightarrow 7(r^2-2r+1)=1+r+r^2$

$\Rightarrow 7r^2-14r+7-1-r-r^2=0$

$\Rightarrow 6r^2-15r+6=0$

$\Rightarrow 2r^2-5r+2=0$

$\Rightarrow 2r^2-4r-r+2=0$

$\Rightarrow 2r(r-2)-1(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r=2,r=\frac{1}{2}$

If r=2, GP = 8,16,32

If r=0.2, GP= 32,16,8.

Thus, the numbers required are 8,16,32.

Question:11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Answer:

Let GP be $A_1,A_2,A_3,...................A_2_n$

Number of terms = 2n

According to the given condition,

$(A_1,A_2,A_3,...................A_2_n)=5(A_1,A_3,...................A_{2_n-1})$

$\Rightarrow (A_1,A_2,A_3,...................A_2_n)-5(A_1,A_3,...................A_{2n-1})=0$

$\Rightarrow (A_2,A_4,A_6,...................A_2_n)=4(A_1,A_3,...................A_{2n-1})$

Let the be GP as $a,ar,ar^2,..................$

$\Rightarrow \frac{ar(r^n-1)}{r-1}=\frac{4.a(r^{n}-1)}{r-1}$

$\Rightarrow ar=4a$

$\Rightarrow r=4$

Thus, the common ratio is 4.

Question:12 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Answer:

Given : first term =a=11

Let AP be $11,11+d,11+2d,11+3d,...........................................11+(n-1)d$

Given: The sum of the first four terms of an A.P. is 56.

$11+11+d+11+2d+11+3d=56$

$\Rightarrow 44+6d=56$

$\Rightarrow 6d=56-44=12$

$\Rightarrow 6d=12$

$\Rightarrow d=2$

Also, The sum of the last four terms is 112.

$11+(n-4)d+11+(n-3)d+11+(n-2)d+11+(n-1)d=112$ $\Rightarrow 44+(n-4)2+(n-3)2+(n-2)2+(n-1)2=112$

$\Rightarrow 44+2n-8+2n-6+2n-4+2n-2=112$

$\Rightarrow 44+8n-20=112$

$\Rightarrow 24+8n=112$

$\Rightarrow 8n=112-24$

$\Rightarrow 8n=88$

$\Rightarrow n=11$

Thus, the number of terms of AP is 11.

Question:13 If $\frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x\neq 0 )$ then show that a, b, c and d are in G.P.

Answer:

Given :

$\frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} (x\neq 0 )$

Taking ,

$\frac{a+ bx}{a-bx} = \frac{b+cx}{b-cx}$

$\Rightarrow (a+ bx) (b-cx) = (b+cx) (a-bx)$

$\Rightarrow ab+ b^2x-bcx^2-acx) = ba-b^2x+acx-bcx^2$

$\Rightarrow 2 b^2x = 2acx$

$\Rightarrow b^2 = ac$

$\Rightarrow \frac{b}{a}=\frac{c}{b}..................1$

Taking,

$\frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}$

$\Rightarrow (b+cx)(c-dx)=(c+dx)(b-cx)$

$\Rightarrow bc-bdx+c^2x-cdx^2=bc-c^2x+bdx-cdx^2$

$\Rightarrow 2bdx=2c^2x$

$\Rightarrow bd=c^2$

$\Rightarrow \frac{d}{c}=\frac{c}{b}..............................2$

From equation 1 and 2 , we have

$\Rightarrow \frac{d}{c}=\frac{c}{b}=\frac{b}{a}$

Thus, a,b,c,d are in GP.

Question:14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^2 R^n = S ^n$

Answer:

Ler there be a GP $=a,ar,ar^2,ar^3,....................$

According to given information,

$S=\frac{a(r^n-1)}{r-1}$

$P=a^n \times r^{(1+2+...................n-1)}$

$P=a^n \times r^{\frac{n(n-1)}{2}}$

$R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+..................\frac{1}{ar^{n-1}}$

$R=\frac{r^{n-1}+r^{n-2}+r^{n-3}+..............r+1}{a.r^{n-1}}$

$R=\frac{1}{a.r^{n-1}}\times \frac{1(r^n-1)}{r-1}$

$R= \frac{(r^n-1)}{a.r^{n-1}.(r-1)}$

To prove : $P^2 R^n = S ^n$

LHS : $P^2 R^n$

$= a^{2n}.r^{n(n-1)}\frac{(r^n-1)^n}{a^n.r^{n(n-1)}.(r-1)^n}$

$= a^{n} \frac{(r^n-1)^n}{(r-1)^n}$

$= \left ( \frac{a(r^n-1)}{(r-1)} \right )^{n}$

$=S^n=RHS$

Hence proved

Question:15 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

$(q - r )a + (r - p )b + (p - q )c = 0$

Answer:

Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively.

To prove : $(q - r )a + (r - p )b + (p - q )c = 0$

Let the first term of AP be 't' and common difference be d

$a_p=t+(p-1)d=a...............................1$

$a_q=t+(q-1)d=b...............................2$

$a_r=t+(r-1)d=c...............................3$

Subtracting equation 2 from 1, we get

$(p-1-q+1)d=a-b$

$\Rightarrow (p-q)d=a-b$

$\Rightarrow d=\frac{a-b}{p-q}....................................4$

Subtracting equation 3 from 2, we get

$(q-1-r+1)d=b-c$

$\Rightarrow (q-r)d=b-c$

$\Rightarrow d=\frac{b-c}{q-r}....................................5$

Equating values of d, from equation 4 and 5, we have

$d=\frac{a-b}{p-q}=\frac{b-c}{q-r}.$

$\Rightarrow \frac{a-b}{p-q}=\frac{b-c}{q-r}.$

$\Rightarrow (a-b)(q-r)=(b-c)(p-q)$

$\Rightarrow aq-ar-bq+br=bp-bq-cp+cq$

$\Rightarrow aq-ar+br=bp-cp+cq$

$\Rightarrow aq-ar+br-bp+cp-cq=0$

$\Rightarrow a(q-r)+b(r-p)+c(p-q)=0$

Hence proved.

Question:16 If $a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b})$ are in A.P., prove that a, b, c are in A.P.

Answer:

Given: $a (\frac{1}{b}+\frac{1}{c}) , b ( \frac{1}{c}+\frac{1}{a}) , c ( \frac{1}{a}+ \frac{1}{b})$ are in A.P.

$\therefore \, \, \, b ( \frac{1}{c}+\frac{1}{a})-a (\frac{1}{b}+\frac{1}{c}) = c ( \frac{1}{a}+ \frac{1}{b})- b ( \frac{1}{c}+\frac{1}{a})$

$\therefore \, \, \, ( \frac{b(a+c)}{ac})- (\frac{a(c+b)}{bc}) = ( \frac{c(b+a)}{ab})- ( \frac{b(a+c)}{ac})$

$\therefore \, \, \, ( \frac{ab^2+b^2c-a^2c-a^2b}{abc}) = ( \frac{c^2b+c^2a-b^2a-b^2c}{abc})$

$\Rightarrow \, \, \, ab^2+b^2c-a^2c-a^2b= c^2b+c^2a-b^2a-b^2c$

$\Rightarrow \, \, \, ab(b-a)+c(b^2-a^2)= a(c^2-b^2)+bc(c-b)$

$\Rightarrow \, \, \, ab(b-a)+c(b-a)(b+a)= a(c-b)(c+b)+bc(c-b)$

$\Rightarrow \, \, \, (b-a)(ab+c(b+a))= (c-b)(a(c+b)+bc)$

$\Rightarrow \, \, \, (b-a)(ab+cb+ac)= (c-b)(ac+ab+bc)$

$\Rightarrow \, \, \, (b-a)= (c-b)$

Thus, a,b,c are in AP.

Question:17 If a, b, c, d are in G.P, prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.

Answer:

Given: a, b, c, d are in G.P.

To prove: $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.

Then we can write,

$b^2=ac...............................1$

$c^2=bd...............................2$

$ad=bc...............................3$

Let $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ be in GP

$(b^n + c^n)^2 =(a^n + b^n) (c^n + d^n)$

LHS: $(b^n + c^n)^2$

$(b^n + c^n)^2 =b^{2n}+c^{2n}+2b^nc^n$

$(b^n + c^n)^2 =(b^2)^n+(c^2)^n+2b^nc^n$

$=(ac)^n+(bd)^n+2b^nc^n$

$=a^nc^n+b^nc^n+a^nd^n+b^nd^n$

$=c^n(a^n+b^n)+d^n(a^n+b^n)$

$=(a^n+b^n)(c^n+d^n)=RHS$

Hence proved

Thus, $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in GP

Question:18 If a and b are the roots of $x^2 -3 x + p = 0$ and c, d are roots of $x^2 -12 x + q = 0$ , where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Answer:

Given: a and b are the roots of $x^2 -3 x + p = 0$

Then, $a+b=3\, \, \, \, and\, \, \, \, ab=p......................1$

Also, c, d are roots of $x^2 -12 x + q = 0$

$c+d=12\, \, \, \, and\, \, \, \, cd=q......................2$

Given: a, b, c, d form a G.P

Let, $a=x,b=xr,c=xr^2,d=xr^3$

From 1 and 2, we get

$x+xr=3$ and $xr^2+xr^3=12$

$\Rightarrow x(1+r)=3$ $xr^2(1+r)=12$

On dividing them,

$\frac{xr^2(1+r)}{x(1+r)}=\frac{12}{3}$

$\Rightarrow r^2=4$

$\Rightarrow r=\pm 2$

When , r=2 ,

$x=\frac{3}{1+2}=1$

When , r=-2,

$x=\frac{3}{1-2}=-3$

CASE (1) when r=2 and x=1,

$ab=x^2r=2\, \, \, and \, \, \, \, cd=x^2r^5=32$

$\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$

i.e. (q + p) : (q – p) = 17:15.

CASE (2) when r=-2 and x=-3,

$ab=x^2r=-18\, \, \, and \, \, \, \, cd=x^2r^5=-288$

$\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-305}{-270}=\frac{17}{15}$

i.e. (q + p) : (q – p) = 17:15.

Question:19 The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that $a: b = \left ( m + \sqrt {m^2 -n^2 }\right ) : \left ( m- \sqrt {m^2 - n^2} \right )$

Answer:

Let two numbers be a and b.

$AM=\frac{a+b}{2}\, \, \, and\, \, \, \, \, GM=\sqrt{ab}$

According to the given condition,

$\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$

$\Rightarrow \frac{(a+b)^2}{4ab}=\frac{m^2}{n^2}$

$\Rightarrow (a+b)^2=\frac{4ab.m^2}{n^2}$

$\Rightarrow (a+b)=\frac{2\sqrt{ab}.m}{n}$ ...................................................................1

$(a-b)^2=(a+b)^2-4ab$

We get,

$(a-b)^2=\left ( \frac{4abm^2}{n^2} \right )-4ab$

$(a-b)^2=\left ( \frac{4abm^2-4abn^2}{n^2} \right )$

$\Rightarrow (a-b)^2=\left ( \frac{4ab(m^2-n^2)}{n^2} \right )$

$\Rightarrow (a-b)=\left ( \frac{2\sqrt{ab}\sqrt{(m^2-n^2)}}{n} \right )$ .....................................................2

From 1 and 2, we get

$2a =\left ( \frac{2\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )$

$a =\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )$

Putting the value of a in equation 1, we have

$b=\left ( \frac{2.\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )$

$b=\left ( \frac{\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( \sqrt{(m^2-n^2)} \right )$

$=\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )$

$\therefore a:b=\frac{a}{b}=\frac{\left ( \frac{\sqrt{ab}}{n} \right )\left (m+ \sqrt{(m^2-n^2)} \right )}{\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )}$

$=\frac{\left (m+ \sqrt{(m^2-n^2)} \right )}{\left (m- \sqrt{(m^2-n^2)} \right )}$

$a:b=\left (m+ \sqrt{(m^2-n^2)} \right ) : \left (m- \sqrt{(m^2-n^2)} \right )$

Question:20 If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.

Answer:

Given: a, b, c are in A.P

$b-a=c-b..............................1$

Also, b, c, d are in G.P.

$c^2=bd..............................2$

Also, 1/c, 1/d, 1/e are in A.P

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}...........................3$

To prove: a, c, e are in G.P. i.e. $c^2=ae$

From 1, we get $2b=a+c$

$b=\frac{a+c}{2}$

From 2, we get

$d=\frac{c^2}{b}$

Putting values of b and d, we get

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\frac{2}{d}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2b}{c^2}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2(a+c)}{2c^2}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{(a+c)}{c^2}=\frac{e+c}{ce}$

$\Rightarrow \frac{(a+c)}{c}=\frac{e+c}{e}$

$\Rightarrow e(a+c)=c(e+c)$

$\Rightarrow ea+ec=ec+c^2$

$\Rightarrow ea=c^2$

Thus, a, c, e are in G.P.

Question:21(i) Find the sum of the following series up to n terms: $5 + 55+ 555 + ....$

Answer:

$5 + 55+ 555 + ....$ is not a GP.

It can be changed in GP by writing terms as

$S_n=5 + 55+ 555 + ....$ to n terms

$S_n=\frac{5}{9}[9+99+999+9999+................]$

$S_n=\frac{5}{9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+................]$

$S_n=\frac{5}{9}[(10+10^2+10^3+........)-(1+1+1.....................)]$

$S_n=\frac{5}{9}[\frac{10(10^n-1)}{10-1}-(n)]$

$S_n=\frac{5}{9}[\frac{10(10^n-1)}{9}-(n)]$

$S_n=\frac{50}{81}[(10^n-1)]-\frac{5n}{9}$

Thus, the sum is

$S_n=\frac{50}{81}[(10^n-1)]-\frac{5n}{9}$

Question:21(ii) Find the sum of the following series up to n terms: .6 +. 66 +. 666+…

Answer:

Sum of 0.6 +0. 66 + 0. 666+….................

It can be written as

$S_n=0.6+0.66+0.666+..........................$ to n terms

$S_n=6[0.1+0.11+0.111+0.1111+................]$

$S_n=\frac{6}{9}[0.9+0.99+0.999+0.9999+................]$

$S_n=\frac{6}{9}[(1-\frac{1}{10})+(1-\frac{1}{10^2})+(1-\frac{1}{10^3})+(1-\frac{1}{10^4})+................]$

$S_n=\frac{2}{3}[(1+1+1.....................n\, terms)-\frac{1}{10}(1+\frac{1}{10}+\frac{1}{10^2}+...................n \, terms)]$

$S_n=\frac{2}{3}[n-\frac{\frac{1}{10}(\frac{1}{10}^n-1)}{\frac{1}{10}-1}]$

$S_n=\frac{2n}{3}-\frac{2}{30}[\frac{10(1-10^-^n)}{9}]$

$S_n=\frac{2n}{3}-\frac{2}{27}(1-10^-^n)$

Question:22 Find the 20th term of the series $2 \times 4+4\times 6+\times 6\times 8+....+n$ terms.

Answer:

the series = $2 \times 4+4\times 6+\times 6\times 8+....+n$

$\therefore n^{th}\, term=a_n=2n(2n+2)=4n^2+4n$

$\therefore a_2_0=2(20)[2(20)+2]$

$=40[40+2]$

$=40[42]$

$=1680$

Thus, the 20th term of series is 1680

Question:23 Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…

Answer:

The series: 3+ 7 +13 +21 +31 +…..............

n th term = $n^2+n+1=a_n$

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k^{2}+k+1$

$=\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} k+\sum _{k=1}^{n} 1$

$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$

$=n\left ( \frac{(n+1)(2n+1)}{6}+\frac{n+1}{2}+1 \right )$

$=n\left ( \frac{(n+1)(2n+1)+3(n+1)+6}{6} \right )$

$=n\left ( \frac{2n^2+n+2n+1+3n+3+6}{6} \right )$

$=n\left ( \frac{2n^2+6n+10}{6} \right )$

$=n\left ( \frac{n^2+3n+5}{3} \right )$

Question:24 If $S_1 , S_2 , S_3$ are the sum of first n natural numbers, their squares and their cubes, respectively, show that $9 S ^2 _2 = S_3 ( 1+ 8 S_1)$

Answer:

To prove : $9 S ^2 _2 = S_3 ( 1+ 8 S_1)$

From given information,

$S_1=\frac{n(n+1)}{2}$

$S_3=\frac{n^2(n+1)^2}{4}$

Here , $RHS= S_3 ( 1+ 8 S_1)$

$\Rightarrow S_3 ( 1+ 8 S_1)=\frac{n^2(n+1)^2}{4}\left ( 1+8\frac{n(n+1)}{2} \right )$

$=\frac{n^2(n+1)^2}{4}\left ( 1+4.n(n+1) \right )$

$=\frac{n^2(n+1)^2}{4}\left ( 1+4n^2+4n \right )$

$=\frac{n^2(n+1)^2}{4}\left ( 2n+1 \right )^2$

$=\frac{n^2(n+1)^2(2n+1)^2}{4}.........................................................1$

Also, $RHS=9S_2^2$

$\Rightarrow 9S_2^2=\frac{9\left [ n(n+2)(2n+1) \right ]^2}{6^2}$

$=\frac{9\left [ n(n+2)(2n+1) \right ]^2}{36}$

$=\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}..........................................2$

From equation 1 and 2 , we have

$9S_2^2=S_3(1+8S_1)=\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}$

Hence proved .

Question:25 Find the sum of the following series up to n terms: $\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ...$

Answer:

n term of series :

$\frac{1^3}{1} + \frac{1^3+2^3}{1+3}+ \frac{1^3+2^3+3^3}{1+3+5}+ ........=\frac{1^3+2^3+3^3+..........n^3}{1+3+5+...........(2n-1)}$

$=\frac{\left [ \frac{n(n+1)}{2} \right ]^2}{1+3+5+............(2n-1)}$

Here, $1,3,5............(2n-1)$ are in AP with first term =a=1 , last term = 2n-1, number of terms =n

$1+3+5............(2n-1)=\frac{n}{2}\left [ 2(1)+(n-1)2 \right ]$

$=\frac{n}{2}\left [ 2+2n-2 \right ]=n^2$

$a_n=\frac{n^2(n+1)^2}{4n^2}$

$=\frac{(n+1)^2}{4}$

$=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}$

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{(k+1)^2}{4}$

$=\frac{1}{4}\sum _{k=1}^{n} k^2+\frac{1}{2}\sum _{k=1}^{n} k+\sum _{k=1}^{n}\frac{1}{4}$

$=\frac{1}{4}\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\frac{n(n+1)}{2}+\frac{n}{4}$

$=n\left ( \frac{(n+1)(2n+1)}{24}+\frac{n+1}{4}+\frac{1}{4} \right )$

$=n\left ( \frac{(n+1)(2n+1)+6(n+1)+6}{24} \right )$

$=n\left ( \frac{2n^2+n+2n+1+6n+6+6}{24} \right )$

$=n\left ( \frac{2n^2+9n+13}{24} \right )$

Question:26 Show that $\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}$

Answer:

To prove :

$\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}$

the nth term of numerator $=n(n+1)^2=n^3+2n^2+n$

nth term of the denominator $=n^2(n+1)=n^3+n^2$

$RHS:\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)}..........................1$

$=\frac{\sum _{k=1}^{n} a_k}{\sum _{k=1}^{n} a_k}=\frac{\sum _{k=1}^{n} k^{3}+2k^2+k}{\sum _{k=1}^{n} (k^3+k^2)}$

Numerator :

$S_n=\sum _{k=1}^{n} k^3+2\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} k$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{2.n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}$

$=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1)$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+8n+4+6}{6} \right )$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+11n+10}{6} \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+11n+10 \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+5n+10 \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+5(n+2) \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ((n+2)(3n+5) \right )$

$=\frac{n(n+1)(n+2)(3n+5)}{12}........................................................2$

Denominator :

$S_n=\sum _{k=1}^{n} k^3+\sum _{k=1}^{n} k^2$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{6}$

$=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2n+1}{3})$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+4n+2}{6} \right )$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+7n+2}{6} \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+7n+2 \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+n+2 \right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+1(n+2)\right )$

$=\left [ \frac{n(n+1)}{12} \right ] \left ( (n+2)(3n+1)\right )$

$=\left [ \frac{n(n+1)(n+2)(3n+1)}{12} \right ].....................................3$

From equation 1,2,3,we have

$\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)}$ $=\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$

$=\frac{3n+5}{3n+1}$

Hence, the above expression is proved.

Question:27 A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Answer:

Given : Farmer pays Rs 6000 cash.

Therefore , unpaid amount = 12000-6000=Rs. 6000

According to given condition, interest paid annually is

12% of 6000,12% of 5500,12% of 5000,......................12% of 500.

Thus, total interest to be paid

$=12\%of\, 6000+12\%of\, 5500+.............12\%of\, 500$

$=12\%of\, (6000+ 5500+.............+ 500)$

$=12\%of\, (500+ 1000+.............+ 6000)$

Here, $500, 1000,.............5500,6000$ is a AP with first term =a=500 and common difference =d = 500

We know that $a_n=a+(n-1)d$

$\Rightarrow 6000=500+(n-1)500$

$\Rightarrow 5500=(n-1)500$

$\Rightarrow 11=(n-1)$

$\Rightarrow n=11+1=12$

Sum of AP:

$S_1_2=\frac{12}{2}\left [ 2(500)+(12-1)500 \right ]$

$S_1_2=6\left [ 1000+5500 \right ]$

$=6\left [ 6500 \right ]$

$=39000$

Thus, interest to be paid :

$=12\%of\, (500+ 1000+.............+ 6000)$

$=12\%of\, ( 39000)$

$=Rs. 4680$

Thus, cost of tractor = Rs. 12000+ Rs. 4680 = Rs. 16680

Question:28 Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Answer:

Given: Shamshad Ali buys a scooter for Rs 22000.

Therefore , unpaid amount = 22000-4000=Rs. 18000

According to the given condition, interest paid annually is

10% of 18000,10% of 17000,10% of 16000,......................10% of 1000.

Thus, total interest to be paid

$=10\%of\, 18000+10\%of\, 17000+.............10\%of\, 1000$

$=10\%of\, (18000+ 17000+.............+ 1000)$

$=10\%of\, (1000+ 2000+.............+ 18000)$

Here, $1000, 2000,.............17000,18000$ is a AP with first term =a=1000 and common difference =d = 1000

We know that $a_n=a+(n-1)d$

$\Rightarrow 18000=1000+(n-1)1000$

$\Rightarrow 17000=(n-1)1000$

$\Rightarrow 17=(n-1)$

$\Rightarrow n=17+1=18$

Sum of AP:

$S_1_8=\frac{18}{2}\left [ 2(1000)+(18-1)1000 \right ]$

$=9\left [ 2000+17000 \right ]$

$=9\left [ 19000 \right ]$

$=171000$

Thus, interest to be paid :

$=10\%of\, (1000+ 2000+.............+ 18000)$

$=10\%of\, ( 171000)$

$=Rs. 17100$

Thus, cost of tractor = Rs. 22000+ Rs. 17100 = Rs. 39100

Question:29 A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

Answer:

The numbers of letters mailed forms a GP : $4,4^2,4^3,.............4^8$

first term = a=4

common ratio=r=4

number of terms = 8

We know that the sum of GP is

$S_n=\frac{a(r^n-1)}{r-1}$

$=\frac{4(4^8-1)}{4-1}$

$=\frac{4(65536-1)}{3}$

$=\frac{4(65535)}{3}$

$=87380$

costs to mail one letter are 50 paise.

Cost of mailing 87380 letters

$=Rs. \, 87380\times \frac{50}{100}$

$=Rs. \,43690$

Thus, the amount spent when the 8th set of the letter is mailed is Rs. 43690.

Question:30 A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Answer:

Given : A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

$=\frac{5}{100}\times 10000=Rs.500$

$\therefore$ Interest in fifteen year 10000+ 14 times Rs. 500

$\therefore$ Amount in 15 th year $=Rs. 10000+14\times 500$

$=Rs. 10000+7000$

$=Rs. 17000$

$\therefore$ Amount in 20 th year $=Rs. 10000+20\times 500$

$=Rs. 10000+10000$

$=Rs. 20000$

Question:31 A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Answer:

Cost of machine = Rs. 15625

Machine depreciate each year by 20%.

Therefore, its value every year is 80% of the original cost i.e. $\frac{4}{5}$ of the original cost.

$\therefore$ Value at the end of 5 years

$=15625\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}$

$=5120$

Thus, the value of the machine at the end of 5 years is Rs. 5120

Question:32 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on.

Answer:

Let x be the number of days in which 150 workers finish the work.

According to the given information, we have

$150x=150+146+142+............(x+8)terms$

Series $150x=150+146+142+............(x+8)terms$ is a AP

first term=a=150

common difference= -4

number of terms = x+8

$\Rightarrow 150x=\frac{x+8}{2}\left [ 2(150)+(x+8-1)(-4) \right ]$

$\Rightarrow 300x=x+8\left [ 300-4x-28 \right ]$

$\Rightarrow 300x= 300x-4x^2-28x+2400-32x+224$

$\Rightarrow x^2+15x-544=0$

$\Rightarrow x^2+32x-17x-544=0$

$\Rightarrow x(x+32)-17(x+32)=0$

$\Rightarrow (x+32)(x-17)=0$

$\Rightarrow x=-32,17$

Since x cannot be negative so x=17.

Thus, in 17 days 150 workers finish the work.

Thus, the required number of days = 17+8=25 days.

Topic Covered in Class 11 Chapter 9 Maths Miscellaneous Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 9 - Sequences and Series covers the following topics:

Introduction
Sequences
Series
Arithmetic Progression (A.P.)
Geometric Progression (G.P.)
Relationship Between A.M. and G.M.
Sum to n terms of Special Series

By engaging with the concepts presented in NCERT Solutions for Class 11 Maths, students can address any doubts related to these topics, establishing a strong foundation for their understanding of Class 12 Maths.

Also Read| Sequences And Series Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-

There are 32 questions in the miscellaneous exercise chapter 9 Class 11 which you can try to solve by yourself.
You can use Class 11 Maths chapter 9 miscellaneous exercise solutions when you are finding it difficult to solve miscellaneous problems.
Class 11 Maths chapter 9 miscellaneous solutions are designed by subject matter experts in a detailed manner that could be understood by an average student also.

Key Features of Class 11 Maths ch 9 Miscellaneous Exercise Solutions

Comprehensive Approach: The miscellaneous exercise class 11 chapter 9 offers a thorough explanation of topics, ensuring a well-rounded understanding of Sequences and Series.
Stepwise Clarity: Each problem in Class 11 maths miscellaneous exercise chapter 9 is solved with step-by-step solutions, facilitating a logical and clear progression of concepts for Class 11 students.
Accessible Language: The class 11 chapter 9 miscellaneous exercise solutions are presented in clear and concise language, making intricate mathematical concepts more accessible for students in Chapter 9 of Class 11 Maths.
Variety of Problems: The Miscellaneous Exercise in Class 11 Maths Chapter 9 provides a diverse range of problems, catering to different difficulty levels and scenarios, allowing for a comprehensive understanding.
Free PDF Access: Solutions for Chapter 9 of Class 11 Maths Miscellaneous Exercise are accessible in a free PDF format, providing students with easy access to study materials according to their convenience.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. What is the meaning of sequence ?

Sequence means an arrangement of numbers in a definite order according to some rule.

2. What is the meaning of finite sequence ?

The finite sequence is s sequence containing a finite number of terms.

3. What is the meaning of infinite sequence ?

The infinite sequence is s sequence containing an infinite number of terms.

4. What is the meaning of series ?

If the terms of a sequence are expressed as the sum of terms then it is called a series.

5. What is the sum of A.P. if the first term is 'a' and the last term of the A.P. is 'l' and the number of terms are 'n' ?

Sum of A.P. = n(a+l)/2

6. What is the arithmetic mean of two numbers 5, 9 ?

Arithmetic mean (A.M.) = (5+9)/2 = 7

Also Read

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 13 Class 11 - Limits and Derivatives

NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 11 - Conic Section

NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 11 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

NCERT Book for class 11 Biology 2023 - Download PDF

NCERT Books for Class 11 - Download All Subjects PDF here

NCERT Books for class 11 Hindi 2023 - Download PDF Here

NCERT Books for class 11 Chemistry 2023 - Download Pdf here

NCERT Books for class 11 Maths 2023 - Download Pdf

NCERT Books for Class 11 Physics 2023 (Topic wise) - Download PDF

NCERT Syllabus for Class 11 Maths 2023

NCERT Syllabus For Class 11 Chemistry 2023 - Download PDF here

NCERT Syllabus for Class 11 2023 (All Subjects) - Download Pdf Here

NCERT Syllabus for Class 11 Physics 2023 - Download Syllabus Pdf

NCERT Syllabus for Class 11 English 2023 - Download PDF here

NCERT Syllabus for Class 11 Biology 2023 - Download Unit Wise Syllabus

NCERT Notes for Class 11 Biology Chapter 9 Biomolecules

NCERT Class 11th Notes Subject wise - Download Free PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

Hydrocarbons Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 13 Notes - Download PDF

Organic Chemistry- Some Basic Principles and Techniques Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 12 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Articles

ITI Courses After 12th - Course Duration & Top Institutes

Nov 16, 2023

Lab Technician Courses After 12th - Check Eligibility, Duration, Top Colleges

Dec 05, 2023

10th Time Table 2024 - Download All Boards Class 10th Date Sheet Here

Nov 28, 2023

UP Madarsa Board Result 2024 for Molvi, Munshi, Alim, Kamil & Fazil - UPBME Result Here

Nov 15, 2023

Diwali Essay: About Deepavali, Short Paragraphs, 10 Lines On Diwali

Nov 09, 2023

Class 10 Sample Papers - Download UP Board, CBSE, HBSE, All Boards Class 10 Sample Papers

Nov 01, 2023

7 Smart Strategies for Acing Board Examinations

Oct 18, 2023

How Can Schools Inculcate Social Responsibility Among Students?

Sep 20, 2023

Lab Technician Courses After 12th - Check Eligibility, Duration, Top Colleges

Dec 05, 2023

ITI Courses After 12th - Course Duration & Top Institutes

Dec 05, 2023

10th Time Table 2024 - Download All Boards Class 10th Date Sheet Here

Nov 28, 2023

Diwali Essay: About Deepavali, Short Paragraphs, 10 Lines On Diwali

Nov 09, 2023

UPTAC Participating Institutes 2023 - Check here List of Colleges

Jul 27, 2023

Children's Day 2023: Why Is It Celebrated, How Schools Can Celebrate

Sep 12, 2023

How to join DRDO after 12th - A Complete Guide

Jul 03, 2023

Best Career Options For Girls - Check Eligibility, Salary, Qualification

Jun 21, 2023

Explore Premium

Mental Health

Exam Preparation: Get Over Self-Doubt With These Effective Measures

5 min ⋆ Dec 02, 2023

Mental Health

Gandhian Principles Can Aid Mental Well-Being

3 min ⋆ Oct 21, 2023

Mental Health

Every Student Should Build On These Must Have Skills

3 min ⋆ Sep 23, 2023

Parenting

Why Children Should Be Trained In Disaster Preparedness

5 min ⋆ Dec 03, 2023

Parenting

Is Playschool Mandatory For Your Child’s Development?

5 min ⋆ Nov 26, 2023

Parenting

Parents, Don’t Shy Away From Talking Good And Bad Touch

5 min ⋆ Nov 19, 2023

Upcoming School Exams

All India Sainik Schools Entrance Examination

Application Date:06 November,2023 - 15 December,2023

Exam Pattern

Answer Key

Result

National Institute of Open Schooling 12th Examination

Application Date:20 November,2023 - 19 December,2023

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Application Date:20 November,2023 - 19 December,2023

Application Process

Exam Pattern

Mock Test

West Bengal Board 12th Examination

Exam Date:30 November,2023 - 14 December,2023

Result

Exam Pattern

Admit Card

National Means Cum-Merit Scholarship

Application Date:03 December,2023 - 18 December,2023

Application Process

Exam Pattern

Mock Test

View All School Exams

Get answers from students and experts

Explore Career Options (By Industry)

Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available

Data Administrator

Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.

4 Jobs Available

Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available

Geotechnical engineer

The role of geotechnical engineer starts with reviewing the projects needed to define the required material properties. The work responsibilities are followed by a site investigation of rock, soil, fault distribution and bedrock properties on and below an area of interest. The investigation is aimed to improve the ground engineering design and determine their engineering properties that include how they will interact with, on or in a proposed construction.

The role of geotechnical engineer in mining includes designing and determining the type of foundations, earthworks, and or pavement subgrades required for the intended man-made structures to be made. Geotechnical engineering jobs are involved in earthen and concrete dam construction projects, working under a range of normal and extreme loading conditions.

3 Jobs Available

Cartographer

How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.

3 Jobs Available

Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available

GIS Expert

GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

3 Jobs Available

Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available

Bank Probationary Officer (PO)

A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.

3 Jobs Available

Operations Manager

3 Jobs Available

Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available

Finance Executive

A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.

3 Jobs Available

Investment Banker

An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.

3 Jobs Available

Bank Branch Manager

Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.

3 Jobs Available

Treasurer

Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.

3 Jobs Available

Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available

Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available

Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability.

2 Jobs Available

Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

2 Jobs Available

Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials.

2 Jobs Available

Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available

Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available

Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available

Team Lead

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available

Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available

Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available

Veterinary Doctor

A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

5 Jobs Available

Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.

4 Jobs Available

Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available

Surgical Technologist

When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.

Also Read: Career as Nurse

3 Jobs Available

Critical Care Specialist

A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.

2 Jobs Available

Ophthalmologist

Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.

2 Jobs Available

Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available

Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available

Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available

Talent Agent

The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

3 Jobs Available

Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available

Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available

Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available

Talent Director

Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.

2 Jobs Available

Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available

Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available

Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available

Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available

Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available

Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning).

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian.

2 Jobs Available

Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

Resource Links for Online MBA

3 Jobs Available

Product Manager

3 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available

Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available

Metrologist

You might be googling Metrologist meaning. Well, we have an easily understandable Metrologist definition for you. A metrologist is a professional who stays involved in measurement practices in varying industries including electrical and electronics. A Metrologist is responsible for developing processes and systems for measuring objects and repairing electrical instruments. He or she also involved in writing specifications of experimental electronic units.

2 Jobs Available

Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available

Process Engineer

As the name suggests, a Process Engineer stays involved in designing, overseeing, assessing and implementing processes to make products and provide services efficiently. Process Engineers are responsible for creating systems to enhance productivity and cut costs.

2 Jobs Available

Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

Product Manager

3 Jobs Available

ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available

.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available

Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available

DevOps Architect

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

2 Jobs Available

Cloud Solution Architect

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

2 Jobs Available

Applications for Admissions are open.

NEET 2024 Most scoring concepts

Apply

Just Study 32% of the NEET syllabus and Score upto 100% marks

ETS ® TOEFL ®

Apply

Thinking of Studying Abroad? Think the TOEFL® test & make your dreams come true

JEE Main high scoring chapters and topics

Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

NEET previous year papers with solutions

Apply

Solve NEET previous years question papers & check your preparedness

JEE Main Important Mathematics Formulas

Apply

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

JEE Main Important Physics formulas

Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

View All Application Forms

News and Notifications

37,484 chairs, 6,246 desks ordered for 800 schools of MCD: Official

22/09/2023, 02:01:05

CBI books Kendriya Vidyalaya principal for admitting 193 students on fake certificates

21/09/2023, 17:04:41

Aadhaar card not mandatory for EWS admission in private schools: Delhi High Court

21/09/2023, 11:34:59

25 Assam schools have 0 students; those lacking ‘sustainable' enrolment will be merged, says Ranoj Pegu

21/09/2023, 11:15:37

Tamil Nadu Rains: Schools closed in Vellore for Class 1 to 5 due to heavy rainfall

21/09/2023, 05:22:28

MCD to open 5 new schools soon, start work on 5 more by December: Officials

20/09/2023, 17:20:17

Odisha primary school teachers to hold demonstration outside assembly on September 26

20/09/2023, 16:17:14

Everything about Education

Latest updates, Exclusive Content, Webinars and more.

Subscribe to Premium

Download Careers360 App's

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

250M+
Students
30,000+
Colleges
500+
Exams
1500+
E-Books
12000+
Cetifications

We Appeared in

Explore Top Colleges Accepting Applications

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Ranking

Products & Resources

NCERT Solutions

Exams

Colleges

Top Countries

Student Visas

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise

Access Sequences And Series Class 11 Chapter 9:Miscellaneous Exercise

Question:1 Show that the sum of and terms of an A.P. is equal to twice the term.

More About NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-

Topic Covered in Class 11 Chapter 9 Maths Miscellaneous Solutions

Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-

Key Features of Class 11 Maths ch 9 Miscellaneous Exercise Solutions

NCERT Solutions of Class 11 Subject Wise

NCERT Solutions for Class 11 Maths

Subject Wise NCERT Exampler Solutions

Frequently Asked Question (FAQs)

Also Read

Articles

Explore Premium

Exam Preparation: Get Over Self-Doubt With These Effective Measures

Gandhian Principles Can Aid Mental Well-Being

Question:1 Show that the sum of $( m+n)^{th}$ and $( m-n)^{th}$ terms of an A.P. is equal to twice the $m^{th}$ term.