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

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 :

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

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

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

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

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

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

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

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

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

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

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

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.