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.

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

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

$\text{Double subscripts: use braces to clarify}$

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

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

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

$\Rightarrow 9=(n-1)$

$\Rightarrow n=9+1=10$

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

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

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

$=5\times 110=550$

$\therefore Requiredsum=2550+1050-550=3050$

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 $ϵ$ 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 $ϵ$ N such that f(1) = 3

$f(1)=3$

Taking $x=y=1$ , we have

$f(1+1)=f(2)=f(1)\ast f(1)=3\ast 3=9$

$f(1+1+1)=f(1+2)=f(1)\ast f(2)=3\ast 9=27$

$f(1+1+1+1)=f(1+3)=f(1)\ast f(3)=3\ast 27=81$

$f(1),f(2),f(3),f(4).....................$ 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$=a.r^{n-1}={5.2}^5=5\ast 32=160$

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=-10or9$

$r=\pm 3$

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 $a,ar,a{r}^2.$

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

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

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

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

$ar-7-a+1=a{r}^2-21-ar+7$

$ar-6-a=a{r}^2-14-ar$

$\Rightarrow a{r}^2-2ar+a=8$

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

$\Rightarrow a(r^2-1{)}^2=8$....................................................................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 $\text{Double subscripts: use braces to clarify}$

Number of terms = 2n

According to the given condition,

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

$\text{Double subscripts: use braces to clarify}$

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

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

$\Rightarrow ar=4a$

$\Rightarrow r=4$

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

Answer:

Given :

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

Taking ,

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

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

$\Rightarrow ab+b^{2}x-bc{x}^2-acx)=ba-b^{2}x+acx-bc{x}^2$

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

$\Rightarrow b^2=ac$

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

Taking,

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

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

$\Rightarrow bc-bdx+c^{2}x-cd{x}^2=bc-c^{2}x+bdx-cd{x}^2$

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

$\Rightarrow bd=c^2$

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

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 $=a,ar,a{r}^2,a{r}^3,....................$

According to given information,

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

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

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

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

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

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

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

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.

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

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

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

$\Rightarrow a{b}^2+b^{2}c-a^{2}c-a^{2}b=c^{2}b+c^{2}a-b^{2}a-b^{2}c$

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

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

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

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

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

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: $(b^n+c^n{)}^2$

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

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

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

$=a^n{c}^n+b^n{c}^n+a^n{d}^n+b^n{d}^n$

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

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

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 $x^2-3x+p=0$ and c, d are roots of $x^2-12x+q=0$, 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 $x^2-3x+p=0$

Then, $a+b=3andab=p......................1$

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

$c+d=12andcd=q......................2$

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

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

From 1 and 2, we get

$x+xr=3$ and $x{r}^2+x{r}^3=12$