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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

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

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

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 Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

NCERT Class 11 Biology Chapter 22 Notes Chemical Coordination And Integration- Download PDF Notes

Some basic concepts of Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 1 Notes - Download PDF

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Books for class 11 Hindi 2024 - Download Free PDF

NCERT Book for class 11 Biology 2024 - Download Free PDF

NCERT Books for class 11 Chemistry 2024 - Download Free Pdf

NCERT Books for class 11 Maths 2024 - Download Free PDF

NCERT Books for Class 11 Physics 2024 - Download PDF

NCERT Books for Class 11 English 2024 - Download Free PDF

NCERT Syllabus for Class 11 2024 - Download All Subjects Pdf Here

NCERT Syllabus for Class 11 English 2024 - Download PDF here

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

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 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

Diploma in Stenography - Duration, Eligibility, Top Colleges, Fees

Apr 18, 2024

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

Apr 17, 2024

Best Lab Technician Courses After 12th: Eligibility, Admission 2024, Colleges, Duration, Fees, Scope

Apr 16, 2024

Graphic Designing Courses after 12th - Eligibility & Top Institutes

Apr 03, 2024

Navyug School Admission 2024: Apply now | Admissions Open for 2024-25

Mar 26, 2024

NSSNET Result 2024, Check Navyug School Admission Test Here

Mar 22, 2024

NSSNET Admit Card 2024 (Out), Navyug School Entrance NTA Hall ticket

Mar 22, 2024

NSSNET Syllabus 2024, Download Navyug School Admission Test Syllabus

Mar 22, 2024

Diploma in Stenography - Duration, Eligibility, Top Colleges, Fees

Apr 18, 2024

Best Lab Technician Courses After 12th: Eligibility, Admission 2024, Colleges, Duration, Fees, Scope

Apr 16, 2024

6 min ⋆ Jan 13, 2023

Parenting

Getting Over The Pink and Blue Divide: Revising Gender Roles

Result

Mock Test

View All School Exams

Get answers from students and experts

Explore Career Options (By Industry)

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

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

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

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

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

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

Database Architect

If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi.

3 Jobs Available

Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive.

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available

Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available

Data Analyst

3 Jobs Available

Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available

5 Jobs Available

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

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available

Welding Engineer

5 Jobs Available

QA Manager

4 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

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

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

JEE Test Series

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

Ranking

Products & Resources

NCERT Solutions

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

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

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

Understand Your Attachment Style And Learn How You Can Reform Your Relationships

7 Tips To Convey Your Struggles To Your Loved Ones

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.