# NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

NCERT solutions for class 11 maths chapter 9 Sequences and Series: Sequence means the progression of numbers in a definite order and series means the sum of the objects of the sequence. In the previous classes, you have studied about arithmetic progression(A.P). In this chapter, we will discuss more arithmetic progression(A.P) and geometric progression(G.P). In this article, you will get NCERT solutions for class 11 maths chapter 9 sequences and series. Important topics like arithmetic progression(A.P), geometric progression(G.P), arithmetic means(A.M), geometric mean(G.M), the relationship between A.M. and G.M, sum to n terms of special series, sum to n terms of squares and cubes of natural numbers are covered in this chapter. You will get questions related to these topics in the solutions of NCERT for class 11 maths chapter 9 sequences and series. In this chapter, there are two types of sequence.

• Finite sequence( A sequence containing a finite number of terms)
• Infinite sequence( A sequence has a first term but doesn't have last term or a sequence which is not finite)

Check all NCERT solutions from class 6 to 12 to understand the concepts in a much easy way. There are four exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

Miscellaneous Exercise

## Topics of NCERT Grade 11 Maths Chapter- 9 Sequences and Series

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.5 Geometric Progression (G.P.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n terms of Special Series

## Question:1  Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

$a _n = n ( n +2)$

Given : $a _n = n ( n +2)$

$a _1 = 1 ( 1 +2)=3$

$a _2 = 2 ( 2 +2)=8$

$a _3 = 3 ( 3 +2)=15$

$a _4 = 4 ( 4 +2)=24$

$a _5 = 5 ( 5 +2)=35$

Therefore, the required number of terms =3, 8, 15, 24, 35

$a _n = \frac{n }{n+1}$

Given : $a _n = \frac{n }{n+1}$

$a _1 = \frac{1}{1+1}=\frac{1}{2}$

$a _2 = \frac{2}{2+1}=\frac{2}{3}$

$a _3 = \frac{3}{3+1}=\frac{3}{4}$

$a _4 = \frac{4}{4+1}=\frac{4}{5}$

$a _5 = \frac{5}{5+1}=\frac{5}{6}$

Therefore, the required number of terms $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$

$a _ n = 2 ^n$

Given : $a _ n = 2 ^n$

$a _ 1 = 2 ^1=2$

$a _ 2 = 2 ^2=4$

$a _ 3 = 2 ^3=8$

$a _ 4 = 2 ^4=16$

$a _ 5 = 2 ^5=32$

Therefore, required number of terms $=2,4,8,16,32.$

$a _n = \frac{2n-3 }{6}$

Given : $a _n = \frac{2n-3 }{6}$

$a _1 = \frac{2\times 1-3 }{6}=\frac{-1}{6}$

$a _2 = \frac{2\times 2-3 }{6}=\frac{1}{6}$

$a _3 = \frac{2\times 3-3 }{6}=\frac{3}{6}=\frac{1}{2}$

$a _4 = \frac{2\times 4-3 }{6}=\frac{5}{6}$

$a _5 = \frac{2\times 5-3 }{6}=\frac{7}{6}$

Therefore, the required number of terms $=\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

$a _ n = ( -1) ^{n-1} 5 ^{n+1}$

Given : $a _ n = ( -1) ^{n-1} 5 ^{n+1}$

$a _ 1 = ( -1) ^{1-1} 5 ^{1+1}=(-1)^{0}.5^2=25$

$a _ 2 = ( -1) ^{2-1} 5 ^{2+1}=(-1)^{1}.5^3=-125$

$a _ 3 = ( -1) ^{3-1} 5 ^{3+1}=(-1)^{2}.5^4= 625$

$a _ 4 = ( -1) ^{4-1} 5 ^{4+1}=(-1)^{3}.5^5= -3125$

$a _ 5 = ( -1) ^{5-1} 5 ^{5+1}=(-1)^{4}.5^6= 15625$

Therefore, the required number of terms $=25,-125,625,-3125,15625$

$a _n = n \frac{n^2 + 5}{4}$

Given : $a _n = n \frac{n^2 + 5}{4}$

$a _1 = 1. \frac{1^2 + 5}{4}=\frac{6}{4}=\frac{3}{2}$

$a _2 = 2. \frac{2^2 + 5}{4}=\frac{18}{4}=\frac{9}{2}$

$a _3 = 3. \frac{3^2 + 5}{4}=\frac{42}{4}=\frac{21}{2}$

$a _4 = 4. \frac{4^2 + 5}{4}=\frac{84}{4}=21$

$a _5 = 5. \frac{5^2 + 5}{4}=\frac{150}{4}=\frac{75}{2}$

Therefore, the required number of terms $=\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$

$a _ n = 4 n - 3 ; a _{17} , a _{24}$

$a _ n = 4 n - 3$

Put $n=17,$

$a _ 1_7 = 4 (17) - 3=68-3=65$

Put  n=24,

$a _ 2_4 = 4 (24) - 3=96-3=93$

Hence, we have $a_1_7=65\, \, and\, \, a _ 2_4 =93$

$a _n = \frac{n^2 }{2^n } ; a_7$

Given :  $a _n = \frac{n^2 }{2^n }$

Put n=7,

$a _7 = \frac{7^2 }{2^7 } =\frac{49}{128}$

Heence, we have $a _7 =\frac{49}{128}$

$a _ n = ( -1) ^{n-1} n ^ 3 , a _9$

Given : $a _ n = ( -1) ^{n-1} n ^ 3$

Put n =9,

$a _ 9 = ( -1) ^{9-1} 9 ^ 3= (1).(729)=729$

The value of $a _ 9 =729$

$a _n = \frac{n ( n-2)}{ n+3 }; a _{20}$

Given : $a _n = \frac{n ( n-2)}{ n+3 }$

Put n=20,

$a _2_0 = \frac{20 ( 20-2)}{ 20+3 }=\frac{360}{23}$

Hence, value of $a _2_0=\frac{360}{23}$

$a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$

Given : $a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$

$a_2 = 3a_{2 - 1} + 2=3a_1+2=3(3)+2=11$

$a_3 = 3a_{3 - 1} + 2=3a_2+2=3(11)+2=35$

$a_4 = 3a_{4 - 1} + 2=3a_3+2=3(35)+2=107$

$a_5 = 3a_{5 - 1} + 2=3a_4+2=3(107)+2=323$

Hence, five terms of series are $3,11,35,107,323$

Series $=3+11+35+107+323+...............$

$a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$

Given : $a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$

$a _ 2 = \frac{a_{2-1}}{2} =\frac{a_1}{2}=\frac{-1}{2}$

$a _ 3 = \frac{a_{3-1}}{3} =\frac{a_2}{3}=\frac{-1}{6}$

$a _ 4 = \frac{a_{4-1}}{4} =\frac{a_3}{4}=\frac{-1}{24}$

$a _ 5 = \frac{a_{5-1}}{5} =\frac{a_4}{5}=\frac{-1}{120}$

Hence, five terms of series are $-1,\frac{-1}{2},\frac{-1}{-6},\frac{-1}{24},\frac {-1}{120}$

Series

$=-1+\frac{-1}{2}+\frac{-1}{-6}+\frac{-1}{24}+\frac {-1}{120}.........................$

Given : $a_1 = a_2 = 2, a_n = a_{n - 1}-1, n > 2$

$a_3 = a_{3 - 1}-1=a_2-1=2-1=1$

$a_4 = a_{4 - 1}-1=a_3-1=1-1=0$

$a_5 = a_{5 - 1}-1=a_4-1=0-1=-1$

Hence, five terms of series are $2,2,1,0,-1$

Series $=2+2+1+0+(-1)+..................$

Find$\frac{a _{n+1}}{a_n}$, for n = 1, 2, 3, 4, 5

Given : The Fibonacci sequence is defined by $1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2$

$a _3 = a _{3-1} + a _{3-2} =a_2+a_1=1+1=2$

$a _4 = a _{4-1} + a _{4-2} =a_3+a_2=2+1=3$

$a _5 = a _{5-1} + a _{5-2} =a_4+a_3=3+2=5$

$a _6 = a _{6-1} + a _{6-2} =a_5+a_4=5+3=8$

$For \,\,n=1,\frac{a _{n+1}}{a_n}=\frac {a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$

$For \,\, n=2,\frac{a _{n+1}}{a_n}=\frac {a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$

$For \,\, n=3,\frac{a _{n+1}}{a_n}=\frac {a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$

$For \,\, n=4,\frac{a _{n+1}}{a_n}=\frac {a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$

$For \,\, n=5,\frac{a _{n+1}}{a_n}=\frac {a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$

Solutions of NCERT for class 11 maths chapter 9 sequences and series-Exercise: 9.2

Odd integers from 1 to 2001 are $1,3,5,7...........2001.$

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know , $a_n = a+(n-1)d$

$2001 = 1+(n-1)2$

$\Rightarrow \, \, 2000 = (n-1)2$

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

$\Rightarrow \, \, n=1000+1=1001$

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

$= \frac{1001}{2}[2(1)+(1001-1)2]$

$= \frac{1001}{2}[2002]$

$= 1001\times 1001$

$= 1002001$

The , sum of odd integers from 1 to 2001 is 1002001.

Numbers divisible by 5 from 100 to 1000 are $105,110,.............995$

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know , $a_n = a+(n-1)d$

$995 = 105+(n-1)5$

$\Rightarrow \, \, 890 = (n-1)5$

$\Rightarrow \, \, 178 = (n-1)$

$\Rightarrow \, \, n=178+1=179$

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

$= \frac{179}{2}[2(105)+(179-1)5]$

$= \frac{179}{2}[2(105)+178(5)]$

$= 179\times 550$

$= 98450$

The  sum of numbers divisible by 5 from 100 to 1000 is 98450.

First term =a=2

Let the series be $2,2+d,2+2d,2+3d,.......................$

Sum of first five terms $=10+10d$

Sum of next five terms $=10+35d$

Given : The sum of the first five terms is one-fourth of  the next five terms.

$10+10d=\frac{1}{4}(10+35d)$

$\Rightarrow \, \, 40+40d=10+35d$

$\Rightarrow \, \, 40-10=35d-40d$

$\Rightarrow \, \, 30=-5d$

$\Rightarrow \, \, d=-6$

To prove  : $a_2_0=-112$

L.H.S : $a_2_0=a+(20-1)d=2+(19)(-6)=2-114=-112=R.H.S$

Hence, 20th term is –112.

Given : A.P. = $-6 , -11/2 , -5...$

$a=-6$

$d=\frac{-11}{2}+6=\frac{1}{2}$

Given : sum = -25

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

$\Rightarrow \, \, -25=\frac{n}{2}[2(-6)+(n-1)\frac{1}{2}]$

$\Rightarrow \, \, \, \, -50= n[-12+(n-1)\frac{1}{2}]$

$\Rightarrow \, \, \, \, -50= -12n+ \frac{n^2}{2}-\frac{n}{2}$

$\Rightarrow \, \, \, \, -100= -24n+ n^2-n$

$\Rightarrow \, \, \, \, n^2-25n+100=0$

$\Rightarrow \, \, \, \, n^2-5n-20n+100=0$

$\Rightarrow \, \, \, \, n(n-5)-20(n-5)=0$

$\Rightarrow \, \, \, \, (n-5)(n-20)=0$

$\Rightarrow \, \, \, \, n=5\, \, or\, \, 20.$

Given : In an A.P., if pth term is 1/q  and qth term is 1/p

$a_p=a+(p-1)d=\frac{1}{q}.................(1)$

$a_q=a+(q-1)d=\frac{1}{p}.................(2)$

Subtracting (2) from (1), we get

$\Rightarrow \, \, a_p-a_q$

$\Rightarrow \, \, (p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}$

$\Rightarrow \, \, pd-d-qd+d=\frac{p-q}{pq}$

$\Rightarrow \, \, (p-q)d=\frac{p-q}{pq}$

$\Rightarrow \, \, d=\frac{1}{pq}$

Putting value of d in equation (1),we get

$a+(p-1)\frac{1}{pq} = \frac{1}{q}$

$\Rightarrow a+\frac{1}{q}-\frac{1}{pq} = \frac{1}{q}$

$\Rightarrow a= \frac{1}{pq}$

$\therefore \, \, S_p_q=\frac{pq}{2}[2.\frac{1}{pq}+(pq-1).\frac{1}{pq}]$

$\Rightarrow \, \, S_p_q=\frac{1}{2}[2+(pq-1)]$

$\Rightarrow \, \, S_p_q=\frac{1}{2}[pq+1]$

Hence,the sum of first pq  terms is 1/2  (pq +1), where $p \neq q$.

Given : A.P. 25, 22, 19, ….....

$S_n=116$

a=25  , d = -3

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

$\Rightarrow \, \, 116=\frac{n}{2}[2(25)+(n-1)(-3)]$

$\Rightarrow \, \, 232=n[50-3n+3]$

$\Rightarrow \, \, 232=n[53-3n]$

$\Rightarrow \, \, 3n^2-53n+232=0$

$\Rightarrow \, \, 3n^2-24n-29n+232=0$

$\Rightarrow \, \, 3n(n-8)-29(n-8)=0$

$\Rightarrow \, \, (3n-29)(n-8)=0$

$\Rightarrow \, \, n=8\, \, or\, \, \, n=\frac{29}{3}$

n could not be $\frac{29}{3}$ so n=8.

Last term $=a_8=a+(n-1)d$

$=25+(8-1)(-3)$

$=25-21=4$

The, last term of A.P. is 4.

Given : $a_k=5k+1$

$\Rightarrow \, \, a+(k-1)d=5k+1$

$\Rightarrow \, \, a+kd-d=5k+1$

Comparing LHS and RHS , we have

$a-d=1$        and     $d=5$

Putting value of d,

$a=1+5=6$

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

$S_n=\frac{n}{2}[2(6)+(n-1)5]$

$S_n=\frac{n}{2}[12+5n-5]$

$S_n=\frac{n}{2}[7+5n]$

If the sum of n terms of an A.P. is $( pn + qn ^ 2 )$,

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

$\Rightarrow \, \, \frac{n}{2}[2a+(n-1)d]=pn+qn^2$

$\Rightarrow \, \, \frac{n}{2}[2a+nd-d]=pn+qn^2$

$\Rightarrow \, \, an+\frac{n^2}{2}d-\frac{nd}{2}=pn+qn^2$

Comparing coefficients of $n^2$ on both side , we get

$\frac{d}{2}=q$

$\Rightarrow \, \, d=2q$

The common difference of AP is 2q.

Given: The sums of n terms of two arithmetic progressions are in the ratio.$5n + 4 : 9n + 6$

There are two AP's with first terms =$a_1,a_2$    and common difference =  $d_1,d_2$

$\Rightarrow \, \, \frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{5n+4}{9n+6}$

$\Rightarrow \, \, \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}$

Substituting n=35,we get

$\Rightarrow \, \, \frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_2}=\frac{5(35)+4}{9(35)+6}$

$\Rightarrow \, \, \frac{2a_1+34 d_1}{2a_2+34d_2}=\frac{5(35)+4}{9(35)+6}$

$\Rightarrow \, \, \frac{a_1+17 d_1}{a_2+17d_2}=\frac{179}{321}$

$\Rightarrow \, \, \frac{18^t^h \, term \, of\, first \, AP}{18^t^h\, term\, of\, second\, AP}=\frac{179}{321}$

Thus, the ratio of the 18th term of AP's is $179:321$

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

Then,

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

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

Given : $S_p=S_q$

$\Rightarrow \frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$

$\Rightarrow p[2a+(p-1)d]=q[2a+(q-1)d]$

$\Rightarrow 2ap+p^2d-pd=2aq+q^2d-qd$

$\Rightarrow 2ap+p^2d-pd-2aq-q^2d+qd=0$

$\Rightarrow 2a(p-q)+d(p^2-p-q^2+q)=0$

$\Rightarrow 2a(p-q)+d((p-q)(p+q)-(p-q))=0$

$\Rightarrow 2a(p-q)+d[(p-q)(p+q-1)]=0$

$\Rightarrow (p-q)[2a+d(p+q-1)]=0$

$\Rightarrow 2a+d(p+q-1)=0$

$\Rightarrow d(p+q-1)=-2a$

$\Rightarrow d=\frac{-2a}{p+q-1}$

Now, $S_(_p_+_q_)=$ $\frac{p+q}{2}[2a+(p+q-1)d]$

$=\frac{p+q}{2}[2a+(p+q-1)\frac{-2a}{p+q-1}]$

$=\frac{p+q}{2}[2a+(-2a)]$

$=\frac{p+q}{2}[0]=0$

Thus, sum of p+q terms of AP is 0.

$\frac{a}{p} ( q-r ) + \frac{b}{q}( r-p ) + \frac{c}{r} ( p-q ) = 0$

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

Let $a_1$ and d be the first term and the common difference of AP, respectively.

According to the given information, we have

$S_p=\frac{p}{2}[2a_1+(p-1)d]=a$

$\Rightarrow [2a_1+(p-1)d]=\frac{2a}{p}............(1)$

$S_q=\frac{q}{2}[2a_1+(q-1)d]=b$

$\Rightarrow [2a_1+(q-1)d]=\frac{2b}{q}............(2)$

$S_r=\frac{r}{2}[2a_1+(r-1)d]=c$

$\Rightarrow [2a_1+(r-1)d]=\frac{2c}{r}............(3)$

Subtracting equation (2) from (1), we have

$\Rightarrow (p-1)d-(q-1)d=\frac{2a}{p}-\frac{2b}{q}$

$\Rightarrow d(p-q-1+1)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d(p-q)=\frac{2(aq-bp)}{pq}$

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$

Subtracting equation (3) from (2), we have

$\Rightarrow (q-1)d-(r-1)d=\frac{2b}{q}-\frac{2c}{r}$

$\Rightarrow d(q-r-1+1)=\frac{2(br-cq)}{qr}$

$\Rightarrow d(q-r)=\frac{2(br-qc)}{qr}$

$\Rightarrow d=\frac{2(br-qc)}{qr(q-r)}$

Equating values of d, we have

$\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow \frac{2(aq-bp)}{pq(p-q)}$$=\frac{2(br-qc)}{qr(q-r)}$

$\Rightarrow \, \, (aq-bp)qr(q-r)=(br-qc)pq(p-q)$

$\Rightarrow \, \, (aq-bp)r(q-r)=(br-qc)p(p-q)$

$\Rightarrow \, \, (aqr-bpr)(q-r)=(bpr-pqc)(p-q)$

Dividing both sides from pqr, we get

$\Rightarrow \, \, (\frac{a}{p}-\frac{b}{q})(q-r)=(\frac{b}{q}-\frac{c}{r})(p-q)$

$\Rightarrow \, \, \frac{a}{p}(q-r)-\frac{b}{q}(q-r+p-q)+\frac{c}{r}(p-q)=0$

$\Rightarrow \, \, \frac{a}{p}(q-r)-\frac{b}{q}(p-r)+\frac{c}{r}(p-q)=0$

$\Rightarrow \, \, \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Hence, the given result is proved.

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is $m^2 : n^2$ .

To prove :  the ratio of mth and nth term is $( 2m-1) : ( 2n- 1 )$.

$\therefore \, \frac{sum\, of\, m\, \, terms}{sum\, of\, n\, \, terms }=\frac{m^2}{n^2}$

$\Rightarrow \, \, \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}$

$\Rightarrow \, \, \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$

Put $m=2m-1\, \, and\, \, n=2n-1$, we get

$\Rightarrow \, \, \frac{2a+(2m-2)d}{2a+(2n-2)d}=\frac{2m-1}{2n-1}$

$\Rightarrow \, \, \frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}.........1$

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{a+(m-1)d}{a+(n-1)d}$

From equation (1) ,we get

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{2m-1}{2n-1}$

Hence proved.

Given : If the sum of n terms of an A.P. is  $3 n^2 + 5 n$  and its $m^{th }$ term is 164

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

Sum of n terms = $3 n^2 + 5 n$

$\Rightarrow \, \, \frac{n}{2}[2a+(n-1)d]=3n^2+5n$

$\Rightarrow \, \, 2a+(n-1)d=6n+10$

$\Rightarrow \, \, 2a+nd-d=6n+10$

Comparing the coefficients of n on both side , we have

$\Rightarrow \, \, d=6$

Also , $2a-d=10$

$\Rightarrow \, \, 2a-6=10$

$\Rightarrow \, \, 2a=10+6$

$\Rightarrow \, \, 2a=16$

$\Rightarrow \, \, a=8$

m th term is 164.

$\Rightarrow \, \, a+(m-1)d=164$

$\Rightarrow \, \, 8+(m-1)6=164$

$\Rightarrow \, \, (m-1)6=156$

$\Rightarrow \, \, m-1=26$

$\Rightarrow \, \, m=26+1=27$

Hence, the value of m is 27.

Let five numbers be A,B,C,D,E.

Then  $AP=8,A,B,C,D,E,26$

Here we have,

$a=8,a_7=26,n=7$

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

$\Rightarrow \, \, 8+(7-1)d=26$

$\Rightarrow \, \, 6d=18$

$\Rightarrow \, \, d=\frac{18}{6}=3$

Thus, we have $A=a+d=8+3=11$

$B=a+2d=8+(2)3=8+6=14$

$C=a+3d=8+(3)3=8+9=17$

$D=a+4d=8+(4)3=8+12=20$

$E=a+5d=8+(5)3=8+15=23$

Thus, the five numbers are 11,14,17,20,23.

Given :  $\frac{a^n + b ^n }{a ^{ n-1}+ b ^{n-1}}$  is the A.M. between a and b.

$\frac{a^n + b ^n }{a ^{ n-1}+ b ^{n-1}}=\frac{a+b}{2}$

$\Rightarrow \, 2(a^n + b ^n) =(a+b)(a ^{ n-1}+ b ^{n-1})$

$\Rightarrow \, 2a^n + 2b ^n =a ^{ n}+a. b ^{n-1}+b.a^{n-1}+b^n$

$\Rightarrow \, 2a^n + 2b ^n-a^n-b^n =a. b ^{n-1}+b.a^{n-1}$

$\Rightarrow \, a^n+b^n =a. b ^{n-1}+b.a^{n-1}$

$\Rightarrow \, a^n-b.a^{n-1} =a. b ^{n-1}-b^n$

$\Rightarrow \, a^{n-1}(a-b)= b ^{n-1}(a-b)$

$\Rightarrow \, a^{n-1}= b ^{n-1}$

$\Rightarrow \,\left [ \frac{a}{b} \right ]^{n-1}= 1$

$\Rightarrow \,n-1=0$

$\Rightarrow \,n=1$

Thus, value of n is 1.

Let A,B,C.........M be m numbers.

Then, $AP=1,A,B,C..........M,31$

Here we have,

$a=1,a_m_+_2=31,n=m+2$

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

$\Rightarrow \, \, 1+(m+2-1)d=31$

$\Rightarrow \, \, (m+1)d=30$

$\Rightarrow \, \, d=\frac{30}{m+1}$

Given : the ratio of $7 ^{th}$ and  $(m-1)^{th}$  numbers is 5 : 9.

$\Rightarrow \, \, \frac{a+(7)d}{a+(m-1)d}=\frac{5}{9}$

$\Rightarrow \, \, \frac{1+7d}{1+(m-1)d}=\frac{5}{9}$

$\Rightarrow \, \, 9(1+7d)=5(1+(m-1)d)$

$\Rightarrow \, \, 9+63d=5+5md-5d$

Putting value of d from above,

$\Rightarrow \, \, 9+63(\frac{30}{m+1})=5+5m\left ( \frac{30}{m+1} \right )-5\left ( \frac{30}{m+1} \right )$

$\Rightarrow \, \9(m+1)+1890=5(m+1)+150m-150$

$\Rightarrow \, \9m+9+1890=5m+5+150m-150$

$\Rightarrow \, 1890+9-5+150=155m-9m$

$\Rightarrow \, 2044=146m$

$\Rightarrow \, m=14$

Thus, value of m is 14.

The  first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

$AP= 100,105,110,115,.................$

We have ,$a=100\, \, and\, \, \, d=5$

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

$a_3_0=100+(30-1)5$

$a_3_0=100+(29)5$

$a_3_0=100+145$

$a_3_0=245$

Thus, he will pay  Rs. 245 in the 30th instalment.

The angles of polygon forms AP with common difference of $5 \degree$ and first term as $120 \degree$ .

We know that sum of angles of polygon with n sides is $180(n-2)$

$\therefore S_n=180(n-2)$

$\Rightarrow \frac{n}{2}[2a+(n-1)d]=180(n-2)$

$\Rightarrow \frac{n}{2}[2(120)+(n-1)5]=180(n-2)$

$\Rightarrow n[240+5n-5]=360n-720$

$\Rightarrow 235n+5n^2=360n-720$

$\Rightarrow 5n^2-125n+720=0$

$\Rightarrow n^2-25n+144=0$

$\Rightarrow n^2-16n-9n+144=0$

$\Rightarrow n(n-16)-9(n-16)=0$

$\Rightarrow (n-16)(n-9)=0$

$\Rightarrow n=9,16$

Sides of polygon are 9 or 16.

CBSE NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.3

G.P :

$\frac{5}{2},\frac{5}{4},\frac{5}{8},....$

first term = a

$a=\frac{5}{2}$

common ratio =r

$r=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}$

$a_n=a.r^{n-1}$

$a_2_0=\frac{5}{2}.(\frac{1}{2})^{20-1}$

$a_2_0=\frac{5}{2}.(\frac{1}{2^{19}})$

$a_2_0=\frac{5}{2^{20}}$

$a_n=a.r^{n-1}$

$a_n=\frac{5}{2}.\left ( \frac{1}{2} \right )^{n-1}$

$a_n=\frac{5}{2}. \frac{1}{2^{n-1}}$

$a_n=\frac{5}{2^n}$   the nth term of G.P

First term = a

common ratio =r=2

$8 ^{th}$ term is 192

$a_n=a.r^{n-1}$

$a_8=a.(2)^{8-1}$

$192=a.(2)^{7}$

$a=\frac{2^6.3}{2^7}$

$a=\frac{3}{2}$

$a_n=a.r^{n-1}$

$a_1_2=\frac{3}{2}. ( 2 )^{12-1}$

$a_1_2=\frac{3}{2}. ( 2 )^{11}$

$a_1_2= 3. ( 2 )^{10}$

$a_1_2= 3072$ is the $12 ^{th}$ term of a G.P.

To prove : $q ^2 = ps$

Let first term=a and common ratio = r

$a_5=a.r^4=p..................(1)$

$a_8=a.r^7=q..................(2)$

$a_1_1=a.r^1^0=s..................(3)$

Dividing equation 2 by 1, we have

$\frac{a.r^7}{a.r^4}=\frac{q}{p}$

$\Rightarrow r^3=\frac{q}{p}$

Dividing equation 3 by 2, we have

$\frac{a.r^1^0}{a.r^7}=\frac{s}{q}$

$\Rightarrow r^3=\frac{s}{q}$

Equating values of $r^3$ ,  we have

$\frac{q}{p}=\frac{s}{q}$

$\Rightarrow q^2=ps$

Hence proved

First term =a= -3

$4^{th}$ term of a G.P. is square of its second term

$\Rightarrow a_4=(a_2)^2$

$\Rightarrow a.r^{4-1}=(a.r^{2-1})^2$

$\Rightarrow a.r^{3}=a^2.r^{2}$

$\Rightarrow r=a=-3$

$a_7=a.r^{7-1}$

$\Rightarrow a_7=(-3).(-3)^{6}$

$\Rightarrow a_7=(-3)^{7}=-2187$

Thus, seventh term is -2187.

Given :  $GP = 2,2\sqrt 2 , 4 .,............$

$a=2\, \, \, \, \, and \, \, \, \, \, r=\frac{2\sqrt{2}}{2}=\sqrt{2}$

n th term is given as 128.

$a_n=a.r^{n-1}$

$\Rightarrow 128=2.(\sqrt{2})^{n-1}$

$\Rightarrow 64=(\sqrt{2})^{n-1}$

$\Rightarrow 2^6=(\sqrt{2})^{n-1}$

$\Rightarrow \sqrt{2}^1^2=(\sqrt{2})^{n-1}$

$\Rightarrow n-1=12$

$\Rightarrow n=12+1=13$

The, 13 th term is 128.

Given :  $GP=\sqrt 3 ,3 , 3 \sqrt 3 ,........$

$a=\sqrt{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{3}{\sqrt{3}}=\sqrt{3}$

n th term is given as 729.

$a_n=a.r^{n-1}$

$\Rightarrow 729=\sqrt{3}.(\sqrt{3})^{n-1}$

$\Rightarrow 729=(\sqrt{3})^{n}$

$\Rightarrow( \sqrt{3})^1^2=(\sqrt{3})^{n}$

$\Rightarrow n=12$

The, 12 th term is 729.

Given :  $GP=\frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,............$

$a=\frac{1}{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$

n th term is given as  $\frac{1}{19683}$

$a_n=a.r^{n-1}$

$\Rightarrow \frac{1}{19683}=\frac{1}{3}.(\frac{1}{3})^{n-1}$

$\Rightarrow \frac{1}{19683}=\frac{1}{3^n}$

$\Rightarrow \frac{1}{3^9}=\frac{1}{3^n}$

$\Rightarrow n=9$

Thus, n=9.

$GP=-\frac{2}{7} ,x, -\frac{7}{2}$

Common ratio=r.

$r=\frac{x}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{x}$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

Thus, for $x=\pm 1$ ,given numbers will be in GP.

geometric progressions is  0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

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

$S_2_0=\frac{0.15(1-(0.1)^{20})}{1-0.1}$

$S_2_0=\frac{0.15(1-(0.1)^{20})}{0.9}$

$S_2_0=\frac{0.15}{0.9}(1-(0.1)^{20})$

$S_2_0=\frac{15}{90}(1-(0.1)^{20})$

$S_2_0=\frac{1}{6}(1-0.1^{20})$

$GP=\sqrt 7 , \sqrt {21} , 3 \sqrt 7 ,...............$

$a=\sqrt{7}\, \, \, \, and\, \, \, \, \, r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$

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

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-\sqrt{3}}$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{1-3} (1+\sqrt{3})$

$S_n=\frac{\sqrt{7}(1-\sqrt{3}^n)}{-2} (1+\sqrt{3})$

$S_n=\frac{\sqrt{7}(1+\sqrt{3})}{2} (\sqrt{3}^n-1)$

The sum to the indicated number of terms in each of the geometric progressions is:

$GP=1,-a , a^2 , - a ^3 , .............$

$a=1\, \, \, and\, \, \, \, r=-a$

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

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

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

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

$GP=x ^3 , x^5 , x^7 .....................$

$a=x^3\, \, \, and\, \, r=\frac{x^5}{x^3}=x^2$

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

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

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

Given :

$\sum_{k = 1}^{11} ( 2+ 3 ^k )$

$\sum_{k = 1}^{11} ( 2+ 3 ^k )=\sum _{k=1}^{11}2 +\sum _{k=1}^{11} 3^k$

$=22 +\sum _{k=1}^{11} 3^k...............(1)$

$\sum _{k=1}^{11} 3^k=3^1+3^2+3^3+....................3^1^1$

These terms form GP with a=3 and r=3.

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

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

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

$S_n=\frac{3(3^1^1-1)}{2}$$=\sum _{k=1}^{11} 3^k$

$\sum_{k = 1}^{11} ( 2+ 3 ^k )$$=22+\frac{3(3^1^1-1)}{2}$

Given : The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1.

Let three terms be   $\frac{a}{r},a,ar$.

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

$S_3=\frac{a(1-r^3)}{1-r}=\frac{39}{10}$

$\frac{a}{r}+a+ar=\frac{39}{10}.........1$

Product of 3 terms is 1.

$\frac{a}{r}\times a\times ar=1$

$\Rightarrow a^3=1$

$\Rightarrow a=1$

Put value of a in equation 1,

$\frac{1}{r}+1+r=\frac{39}{10}$

$10(1+r+r^2)=39(r)$

$\Rightarrow 10r^2-29r+10=0$

$\Rightarrow 10r^2-25r-4r+10=0$

$\Rightarrow 5r(2r-5)-2(2r-5)=0$

$\Rightarrow (2r-5)(5r-2)=0$

$\Rightarrow r=\frac{5}{2},r=\frac{2}{5}$

The three terms of AP are $\frac{5}{2},1,\frac{2}{5}$.

G.P.=    $3 , 3 ^ 2 , 3 ^ 3$, …............

Sum =120

These terms are GP with a=3 and r=3.

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

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

$120\times \frac{-2}{3}=(1-3^n)$

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

$\Rightarrow 3^n=1+80=81$

$\Rightarrow 3^n=81$

$\Rightarrow 3^n=3^4$

$\Rightarrow n=4$

Hence, we have value of n as 4 to get sum of 120.

Let GP be $a,ar,ar^2,ar^3,ar^4,ar^5,ar^6................................$

Given :  The sum of first three terms of a G.P. is 16

$a+ar+ar^2=16$

$\Rightarrow a(1+r+r^2)=16...............................(1)$

Given :  the sum of the next three terms is128.

$ar^3+ar^4+ar^5=128$

$\Rightarrow ar^3(1+r+r^2)=128...............................(2)$

Dividing equation (2) by (1), we have

$\Rightarrow \frac{ar^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}$

$\Rightarrow r^3=8$

$\Rightarrow r^3=2^3$

$\Rightarrow r=2$

Putting value of r =2 in equation 1,we have

$\Rightarrow a(1+2+2^2)=16$

$\Rightarrow a(7)=16$

$\Rightarrow a=\frac{16}{7}$

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

$S_n=\frac{\frac{16}{7}(1-2^n)}{1-2}$

$S_n=\frac{16}{7}(2^n-1)$

Given a G.P. with a = 729 and $7 ^{th}$ term 64.

$a_n=a.r^{n-1}$

$\Rightarrow 64=729.r^{7-1}$

$\Rightarrow r^6=\frac{64}{729}$

$\Rightarrow r^6=\left ( \frac{2}{3} \right )^6$

$\Rightarrow r=\frac{2}{3}$

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

$S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{1-\frac{2}{3}}$

$S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{\frac{1}{3}}$

$S_7=3\times 729 \left ( \frac{3^7-2^7}{3^7} \right )$

$S_7= \left ( 3^7-2^7 \right )$

$S_7= 2187-128$

$S_7= 2059$ (Answer)

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

$a_5=4.a_3$

$\Rightarrow a.r^{5-1}=4.a.r^{3-1}$

$\Rightarrow a.r^{4}=4.a.r^{2}$

$\Rightarrow r^{2}=4$

$\Rightarrow r=\pm 2$

If r=2, then

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

$\Rightarrow \frac{a(1-2^2)}{1-2}=-4$

$\Rightarrow \frac{a(1-4)}{-1}=-4$

$\Rightarrow a(-3)=4$

$\Rightarrow a=\frac{-4}{3}$

If r= - 2, then

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

$\Rightarrow \frac{a(1-(-2)^2)}{1-(-2)}=-4$

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

$\Rightarrow a(-3)=-12$

$\Rightarrow a=\frac{-12}{-3}=4$

Thus, required GP is $\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},.........$      or    $4,-8,-16,-32,..........$

Let x,y, z are in G.P.

Let first term=a and common ratio = r

$a_4=a.r^3=x..................(1)$

$a_1_0=a.r^9=y..................(2)$

$a_1_6=a.r^1^5=z..................(3)$

Dividing equation 2 by 1, we have

$\frac{a.r^9}{a.r^3}=\frac{y}{x}$

$\Rightarrow r^4=\frac{y}{x}$

Dividing equation 3 by 2, we have

$\frac{a.r^1^5}{a.r^9}=\frac{z}{y}$

$\Rightarrow r^4=\frac{z}{y}$

Equating values of $r^4$ ,  we have

$\frac{y}{x}=\frac{z}{y}$

Thus, x,y,z are in GP

8, 88, 888, 8888… is not a GP.

It can be changed in GP by writing terms as

$S_n=8+88+888+8888+.............$ to n terms

$S_n=\frac{8}{9}[9+99+999+9999+................]$

$S_n=\frac{8}{9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+................]$

$S_n=\frac{8}{9}[(10+10^2+10^3+........)-(1+1+1.....................)]$

$S_n=\frac{8}{9}[\frac{10(10^n-1)}{10-1}-(n)]$

$S_n=\frac{8}{9}[\frac{10(10^n-1)}{9}-(n)]$

$S_n=\frac{80}{81}(10^n-1)-\frac{8n}{9}$

$Required \, \, sum=2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}$

$Required \, \, sum=64\left [ 4+2+1+\frac{1}{2}+\frac{1}{2^2} \right ]$

Here, $4,2,1,\frac{1}{2},\frac{1}{2^2}$    is a GP.

first term =a=4

common ratio =r

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

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

$S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{1-\frac{1}{2}}$

$S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{\frac{1}{2}}$

$S_5=8(1-\left ( \frac{1}{32} \right ))$

$S_5=8(\frac{31}{32})$

$S_5=\frac{31}{4}$

$Required \, \, sum=64\left [ \frac{31}{4} \right ]$

$Required \, \, sum=16\times 31=496$

To prove : $aA,arAR,ar^2AR^2,...................$     is a GP.

$\frac{second \, \, term}{first\, \, term}=\frac{arAR}{aA}=rR$

$\frac{third \, \, term}{second\, \, term}=\frac{ar^2AR^2}{arAR}=rR$

Thus, the above sequence is a GP with common ratio of rR.

Let first term be a and common ratio be r.

$a_1=a,a_2=ar,a_3=ar^2,a_4=ar^3$

Given : the third term is greater than the first term by 9, and  the second term is greater than the $4 ^{th}$ by 18.

$a_3=a_1+9$

$\Rightarrow ar^2=a+9$

$\Rightarrow a(r^2-1)=9.................1$

$a_2=a_4+18$

$\Rightarrow ar=ar^3+18$

$\Rightarrow ar(1-r^2)=18......................2$

Dividing equation 2 by 1 , we get

$\frac{ ar(1-r^2)}{ -a(1-r^2)}=\frac{18}{9}$

$\Rightarrow r=-2$

Putting value of r , we get

$4a=a+9$

$\Rightarrow 4a-a=9$

$\Rightarrow 3a=9$

$\Rightarrow a=3$

Thus, four terms of GP are $3,-6,12,-24.$

To prove : $a ^{ q-r } b ^{r- p } C ^{p-q} = 1$

Let A  be the first term and R be common ratio.

According to the given information, we have

$a_p=A.R^{p-1}=a$

$a_q=A.R^{q-1}=b$

$a_r=A.R^{r-1}=c$

L.H.S : $a ^{ q-r } b ^{r- p } C ^{p-q}$

$=A^{q-r}.R^{(q-r)(p-1)}.A^{r-p}.R^{(r-p)(q-1)}.A^{p-q}.R^{(p-q)(r-1)}$

$=A^{q-r+r-p+p-q}.R^{(qp-rp-q+r)+(rq-pq+p-r)+(pr-p-qr+q)}$

$=A^0.R^0=1$=RHS

Thus, LHS = RHS.

Hence proved.

Given : First term =a  and n th term = b.

Common ratio = r.

To prove : $P^2 = ( ab)^n$

Then , $GP = a,ar,ar^2,ar^3,ar^4,..........................$

$a_n=a.r^{n-1}=b..................................1$

P = product of n terms

$P=(a).(ar).(ar^2).(ar^3)..............(ar^{n-1})$

$P=(a.a.a...............a)((1).(r).(r^2).(r^3)..............(r^{n-1}))$

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

Here, $1+2+.........(n-1)$   is a AP.

$\therefore\, \, \, sum= \frac{n}{2}\left [2a+(n-1)d \right ]$

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

$= \frac{n-1}{2}\left [2+n-2 \right ]$

$= \frac{n-1}{2}\left [n \right ]$

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

Put in equation (2),

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

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

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

$P^2=(a.b)^n$

Hence proved .

Let first term =a  and common ratio = r.

$sum \, \, of\, \, n\, \, terms=\frac{a(1-r^n)}{1-r}$

Since there are n terms from (n+1) to 2n  term.

Sum of terms from (n+1) to 2n.

$S_n=\frac{a_(_n+_1_)(1-r^n)}{1-r}$

$a_(_n+_1)=a.r^{n+1-1}=ar^n$

Thus, the required ratio  = $\frac{a(1-r^n)}{1-r}\times \frac{1-r}{ar^n(1-r^n)}$

$=\frac{1}{r^n}$

Thus,  the common ratio of the sum of first n terms of a G.P. to the sum of terms from  $( n+1)^{th} \: \: to\: \: (2n)^{th}$ term is  $\frac{1}{r^n}$.

If a, b, c and d are in G.P.

$bc=ad....................(1)$

$b^2=ac....................(2)$

$c^2=bd....................(3)$

To prove : $(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .$

RHS : $(ab + bc + cd)^2 .$

$=(ab + ad + cd)^2 .$

$=(ab + d (a+ c))^2 .$

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

$=a^2b^2 + d^2 (a^2+ c^2+2ac) + 2a^2bd+2bcd$

Using equation (1) and (2),

$=a^2b^2 + 2a^2c^2+ 2b^2c^2+d^2a^2+2d^2b^2+d^2c^2$

$=a^2b^2 + a^2c^2+ a^2c^2+b^2c^2+b^2c^2+d^2a^2+d^2b^2+d^2b^2+d^2c^2$

$=a^2b^2 + a^2c^2+ a^2d^2+b^2.b^2+b^2c^2+b^2d^2+c^2b^2+c^2.c^2+d^2c^2$

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

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

Hence proved

Let A, B be two numbers between 3 and 81 such that series    3, A, B,81  forms a GP.

Let a=first term and common ratio =r.

$\therefore a_4=a.r^{4-1}$

$81=3.r^{3}$

$27=r^{3}$

$r=3$

For $r=3$,

$A=ar=(3)(3)=9$

$B=ar^2=(3)(3)^2=27$

The, required numbers are 9,27.

M of a and b is $\sqrt{ab}.$

Given :

$\frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}=\sqrt{ab}$

Squaring both sides ,

$\left ( \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n} \right )^2=ab$

$\left (a^{n+1}+ b ^{n+1})^2=({a^n+b^n} \right )^2ab$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^2^n+b^2^n+2.a^n.b^n} \right )ab$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^{2n+1}.b+a.b^{2n+1}+2.a^{n+1}.b^{n+1}} \right )$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2})=({a^{2n+1}.b+a.b^{2n+1}} \right )$

$\Rightarrow a^{2n+2}-{a^{2n+1}.b=a.b^{2n+1}} \right )- b ^{2n+2}$

$\Rightarrow a^{2n+1}(a-b)=b^{2n+1}( a-b)$

$\Rightarrow a^{2n+1}=b^{2n+1}$

$\Rightarrow \left ( \frac{a}{b} \right )^{2n+1}=1$

$\Rightarrow \left ( \frac{a}{b} \right )^{2n+1}=1=\left ( \frac{a}{b} \right )^0$

$\Rightarrow 2n+1=0$

$\Rightarrow 2n=-1$

$\Rightarrow n=\frac{-1}{2}$

Let there be two numbers a and b

geometric mean $=\sqrt{ab}$

According to the given condition,

$a+b=6\sqrt{ab}$

$(a+b)^2=36(ab)$.............................................................(1)

Also,$(a-b)^2=(a+b)^2-4ab=36ab-4ab=32ab$

$(a-b)=\sqrt{32}\sqrt{ab}$

$(a-b)=4\sqrt{2}\sqrt{ab}$.......................................................(2)

From (1) and (2), we get

$2a=(6+4\sqrt{2})\sqrt{ab}$

$a=(3+2\sqrt{2})\sqrt{ab}$

Putting the value of 'a' in (1),

$b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}$

$b=(3-2\sqrt{2})\sqrt{ab}$

$\frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}$

$\frac{a}{b}=\frac{(3+2\sqrt{2})}{(3-2\sqrt{2})}$

Thus, the ratio is $( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )$

If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.

$AM=A=\frac{a+b}{2}$

$\Rightarrow a+b=2A$...................................................................1

$GM=G=\sqrt{ab}$

$\Rightarrow ab=G^2$...........................................................................2

We know $(a-b)^2=(a+b)^2-4ab$

Put values from equation 1 and 2,

$(a-b)^2=4A^2-4G^2$

$(a-b)^2=4(A^2-G^2)$

$(a-b)^2=4(A+G)(A-G)$

$(a-b)=4\sqrt{(A+G)(A-G)}$..................................................................3

From 1 and 3 , we have

$2a=2A+2\sqrt{(A+G)(A-G)}$

$\Rightarrow a=A+\sqrt{(A+G)(A-G)}$

Put value of a in equation 1, we get

$b=2A-A-\sqrt{(A+G)(A-G)}$

$\Rightarrow b=A-\sqrt{(A+G)(A-G)}$

Thus, numbers are $A \pm \sqrt{( A+G)(A-G)}$

The number of bacteria in a certain culture doubles every hour.It forms GP.

Given :      a=30   and  r=2.

$a_3=a.r^{3-1}=30(2)^2=120$

$a_5=a.r^{5-1}=30(2)^4=480$

$a_n+_1=a.r^{n+1-1}=30(2)^n$

Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and $30(2)^n$ respectively.

Given: Bank pays an annual interest rate of 10%  compounded annually.

Rs 500 amounts are deposited in the bank.

At the end of the first year, the amount

$=500\left ( 1+\frac{1}{10} \right )=500(1.1)$

At the end of the second year, the amount $=500(1.1)(1.1)$

At the end of the third year, the amount $=500(1.1)(1.1)(1.1)$

At the end of 10 years, the amount $=500(1.1)(1.1)(1.1)........(10times)$

$=500(1.1)^{10}$

Thus, at the end of 10 years, amount $=Rs. 500(1.1)^{10}$

Let roots of the quadratic equation be a and b.

According to given condition,

$AM=\frac{a+b}{2}=8$

$\Rightarrow (a+b)=16$

$GM=\sqrt{ab}=5$

$\Rightarrow ab=25$

We know that   $x^2-x(sum\, of\, roots)+(product\, of\, roots)=0$

$x^2-x(16)+(25)=0$

$x^2-16x+25=0$

Thus, the quadratic equation = $x^2-16x+25=0$

NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.4

the series = $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ...$

n th term  = $n(n+1)=a_n$

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

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

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

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

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

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

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

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

the series =  $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...$

n th term  = $n(n+1)(n+2)=a_n$

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

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

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

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

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

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

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

$=\left [ \frac{n(n+1)}{4} \right ] \left ( n^2+5n+6 \right )$

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

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

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

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

Thus, sum is

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

the series  $3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2$

nth term  = $(2n+1)(n^2)=2n^3+n^2=a_n$

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

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

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

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

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

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

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

$= \frac{n(n+1)(3n^2+5n+1)}{6}$

Thus, the sum is

$= \frac{n(n+1)(3n^2+5n+1)}{6}$

Series =

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+ ...$

$n^{th}\, term=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$a_1=\frac{1}{1}-\frac{1}{2}$

$a_2=\frac{1}{2}-\frac{1}{3}$

$a_3=\frac{1}{3}-\frac{1}{4}$.................................

$a_n=\frac{1}{n}-\frac{1}{n+1}$

$a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} +\frac{1}{2}+\frac{1}{3}+............\frac{1}{n}\right ]-\left [ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.............\frac{1}{n+1} \right ]$

$a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} \right ]-\left [ \frac{1}{n+1} \right ]$

$S_n=\frac{n+1-1}{n+1}$

$S_n=\frac{n}{n+1}$

Hence, the sum is

$S_n=\frac{n}{n+1}$

series =    $5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2$

n th term  = $(n+4)^2=n^2+8n+16=a_n$

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

$=\sum _{k=1}^{n} k^2+8\sum _{k=1}^{n} k+\sum _{k=1}^{n}16$

$=\frac{n(n+1)(2n+1)}{6}+\frac{8.n(n+1)}{2}+16n$

16th term is $(16+4)^2=20^2$

$S_1_6=\frac{16(16+1)(2(16)+1)}{6}+\frac{8.(16)(16+1)}{2}+16(16)$

$S_1_6=\frac{16(17)(33)}{6}+\frac{8.(16)(17)}{2}+16(16)$

$S_1_6=1496+1088+256$

$S_1_6=2840$

Hence, the sum of the series $5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2$  is 2840.

series = $3 \times 8 + 6 \times 11 + 9\times 14+...$

=(n th term of 3,6,9,...........)$\times$(nth terms of 8,11,14,..........)

n th term  = $3n(3n+5)=a_n=9n^2+15n$

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} 3k(3k+5)$

$=9\sum _{k=1}^{n} k^2+15\sum _{k=1}^{n} k$

$=\frac{9.n(n+1)(2n+1)}{6}+\frac{15.n(n+1)}{2}$

$=\frac{3.n(n+1)(2n+1)}{2}+\frac{15.n(n+1)}{2}$

$=\frac{n(n+1)}{2}\left (3(2n+1)+15 \right )$

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

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

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

$=3.n(n+1)\left (n+3\right )$

Hence, sum is  $=3.n(n+1)\left (n+3\right )$

series =    $1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...$

n th term  = $a_n=1^2+2^2+3^2+...................n^2=\frac{n(n+1)(2n+1)}{6}$

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

$S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{2k^3+3k^2+k}{6}$

$=\frac{1}{3}\sum _{k=1}^{n} k^3+\frac{1}{2}\sum _{k=1}^{n} k^2+\frac{1}{6}\sum _{k=1}^{n} k$

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

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

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

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

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

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

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

nth terms is given by $n (n+1) ( n + 4 )$

$a_n=n (n+1) ( n + 4 )=n(n^2+5n+4)=n^3+5n^2+4n$

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

$=\sum _{k=1}^{n} k^3+5\sum _{k=1}^{n} k^2+4\sum _{k=1}^{n} k$

$=\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+\frac{4.n(n+1)}{2}$

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

$=\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+20n+10+24}{6} \right )$

$=\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+23n+34}{6} \right )$

$=\left [ \frac{n(n+1)}{24} \right ] \left ( 3n^2+23n+34 \right )$

nth terms are given by $n^2 + 2 ^ n$

$a_n=n^2 + 2 ^ n$

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

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

This term is a GP with first term =a =2 and common ratio =r =2.

$\sum _{k=1}^{n} 2^k$$=\frac{2(2^n-1)}{2-1}=2(2^n-1)$

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

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

Thus, the sum is

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

nth terms is given by  $( 2n-1) ^2$.

$a_n=( 2n-1) ^2=4n^2+1-4n$

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

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

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

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

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

$=n(\frac{4n^2+6n+2-6n-6+3}{3})$

$=n(\frac{4n^2-1}{3})$

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

## NCERT solutions for class 11 maths chapter 9 sequences and series-Miscellaneous Exercise

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.