NCERT Solutions Class 11 Maths Chapter-9 (Sequences and Series)

NCERT Solutions Class 11 Maths from class 11th Students will get the answers of Chapter-9 (Sequences and Series) . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answer

Class 11 Mathematics

Chapter-9 (Sequences and Series)

Questions and answers given in practice

Chapter-9 (Sequences and Series)

Exercise 9.1

Question 1:

Write the first five terms of the sequence whose nth term is an = n (n + 2).
Ans:
an = n(n +2)
Substituting n = 1, 2, 3, 4 and 5, we obtain
a1 = 1 (1 + 2) = 3,
a2 = 2 (2 + 2) = 8,
a3 = 3 (3 + 2) = 15,
a4 = 4 (4 + 2) = 24,
a5 = 5 (5 + 2) = 35
Therefore, the required terms are 3, 8, 15, 24 and 35.

Q2. Write the first five terms of the sequences whose $n^{th}$ term is $a_n=\frac{n}{n+1}$.
Ans:

 

$a_n=\frac{n}{n+1}$ Substituting n = 1, 2, 3, 4, 5, we obtain $a_1=\frac{1}{1+1}=\frac{1}{2},a_2=\frac{2}{2+1}=\frac{2}{3},a_3=\frac{3}{3+1}=\frac{3}{4},a_4=\frac{4}{4+1}=\frac{4}{5},a_5=\frac{5}{5+1}=\frac{5}{6}$ Therefore, the required terms are $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\text{and}\frac{5}{6}$

Question 3:
Write the first five terms of the sequence whose nth term is an = 2n.
Ans:
an = 2n
Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
a5 = 25 = 32
Therefore, the required terms are 2, 4, 8, 16 and 32.

Q4. Write the first five terms of the sequences whose $n^{th}$ term is $a_n=\frac{2n-3}{6}$
Answer. Substituting n = 1, 2, 3, 4, 5, we obtain $\begin{array}{cc}{a}_1& =\frac{2\times 1-3}{6}=\frac{-1}{6}\\ a_2& =\frac{2\times 2-3}{6}=\frac{1}{6}\\ a_3& =\frac{2\times 3-3}{6}=\frac{3}{6}=\frac{1}{2}\\ a_4& =\frac{2\times 4-3}{6}=\frac{5}{6}\\ a_5& =\frac{2\times 5-3}{6}=\frac{7}{6}\end{array}$ Therefore, the required terms are $\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\text{and}\frac{7}{6}$

Q5. Write the first five terms of the sequences whose $n^{th}$ term is $a_n=(-1{)}^{n-1}{5}^{n+1}$
Answer. Substituting n = 1, 2, 3, 4, 5, we obtain $\begin{array}{c}{a}_1=(-1{)}^{1-1}{5}^{1+1}=5^2=25\\ a_2=(-1{)}^{2-1}{5}^{2+1}=-5^3=-125\\ a_3=(-1{)}^{3-1}{5}^{3+1}=5^4=625\\ a_4=(-1{)}^{5-1}{5}^{4+1}=-5^5=-3125\\ a^5=(-1{)}^{5-1}{5}^{5+1}=5^6=15625\end{array}$ Therefore, the required terms are 25, –125, 625, –3125, and 15625.

Q6. Write the first five terms of the sequences whose $n^{th}$ term is $a_n=n\frac{n^2+5}{4}$
Answer. Substituting n = 1, 2, 3, 4, 5, we obtain $\begin{array}{cc}{a}_1& =1\cdot \frac{1^2+5}{4}=\frac{6}{4}=\frac{3}{2}\\ a_2& =2\cdot \frac{2^2+5}{4}=2\cdot \frac{9}{4}=\frac{9}{2}\\ a_3& =3\cdot \frac{3^2+5}{4}=3\cdot \frac{14}{4}=\frac{21}{2}\\ a_4& =4\cdot \frac{4^2+5}{4}=21\\ a_5=& 5\cdot \frac{5^2+5}{4}=5\cdot \frac{30}{4}=\frac{75}{2}\end{array}$ Therefore, the required terms are $\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\text{and}\frac{75}{2}$.

Q 7:Find the indicated terms in the following sequence whose nth term is an =4n – 3; a17, a24.
Answer. Substituting n = 17, we obtain $a_{17}=4\left(17\right)-3=68-3=65$ Substituting n = 24, we obtain $a_{24}=4\left(24\right)-3=96-3=93$

Q8. Find the $7^{th}$ term in the following sequence whose $n^{th}$ term is $a_n=\frac{n^2}{2^n};a_7$
Answer. Substituting n = 7, we obtain $a_7=\frac{7^2}{2\times 7}=\frac{7}{2}$ Here, an=n22n. Substituting n = 7, we obtain $a_7=\frac{7^2}{2^7}=\frac{49}{128}$

Q9. Find the $9^{th}$ term in the following sequence whose $n^{th}$ term is $a_n=(-1{)}^{n-1}{n}^3;a_9$
Answer. Substituting n = 9, we obtain $a_9=(-1{)}^{9-1}(9{)}^3=(9{)}^3=729$

Q10. $a_n=\frac{n(n-2)}{n+3};a_{20}$
Answer. Substituting n = 20, we obtain $a_{20}=\frac{20(20-2)}{20+3}=\frac{20\left(18\right)}{23}=\frac{360}{23}$

Q11:Write the first five terms of the following sequence and obtain the corresponding series:
a1 = 3, an = 3an – 1 + 2 for all n > 1.
Anwer:

a1 = 3, an = 3 an – 1 + 2 for all n > 1
=> a2 = 3a2 – 1 + 2
= 3a1 + 2
= 3(3) + 2 = 11

a3 = 3a3 – 1 + 2
= 3a2 + 2
= 3(11) + 2 = 35

a4 = 3a4 – 1 + 2
= 3a3 + 2
= 3(35) + 2 = 107

a5 = 3a5 – 1 + 2
= 3a4 + 2
= 3(107) + 2 = 323.
Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Q12. Write the first five terms of the following sequence and obtain the corresponding series: $a_1=-1,a_n=\frac{a_{n-1}}{n},n\ge 2$
Answer. $\begin{array}{c}{a}_1=-1,a_n=\frac{a_{n-1}}{n},n\ge 2\\ \Rightarrow a_2=\frac{a_1}{2}=\frac{-1}{2}\\ a_3=\frac{a_2}{3}=\frac{-1}{6}\\ a_4=\frac{a_3}{4}=\frac{-1}{24}\\ a_5=\frac{a_4}{4}=\frac{-1}{120}\end{array}$ Hence, the first five terms of the sequence are $-1,\frac{-1}{2},\frac{-1}{6},\frac{-1}{24},\text{and}\frac{-1}{120}$ The corresponding series is $(-1)+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right)+\dots$

Q13:
Write the first five terms of the following sequence and obtain the corresponding series:
a1 = a2 = 2, an = an – 1 – 1, n > 2.
Answer:

a1 = a2 = 2,
an = an – 1, n > 2
=> a3 = a2 – 1 = 2 – 1 = 1,
a4 = a3 – 1 = 1 – 1 = 0
a5 = a4 – 1 = 0 – 1 = – 1.
Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1.
The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….

Q14. The Fibonacci sequence is defined by $\begin{array}{c}1=a_1=a_2\text{and}{a}_n=a_{n-1}+a_{n-2},n>2\\ \text{Find}\frac{a_{n+1}}{a_n},\text{for}n=1,2,3,4,5\end{array}$

Answer. $\begin{array}{c}1=a_1=a_2\\ a_n=a_{n-1}+a_{n-2},n>2\\ \therefore a_3=a_2+a_1=1+1=2\\ a_4=a_3+a_2=2+1=3\\ a_5=a_4+a_3=3+2=5\\ a_6=a_5+a_4=5+3=8\end{array}$ $\begin{array}{c}\therefore \text{For}n=1,\frac{a_n+1}{a_n}=\frac{a_2}{a_1}=\frac{1}{1}=1\\ \text{For}n=2,\frac{a_n+1}{a_n}=\frac{a_3}{a_1}=\frac{2}{1}=2\\ \text{For}n=3,\frac{a_n+1}{a_n}=\frac{a_4}{a_3}=\frac{3}{2}\\ \text{For}n=4,\frac{a_n+1}{a_n}=\frac{a_5}{a_4}=\frac{5}{3}\\ \text{For}n=5,\frac{a_n+1}{a_n}=\frac{a_6}{a_5}=\frac{8}{5}\end{array}$

an+1an=a6a5=85

Exercise 9.2

Q1:Find the sum of odd integers from 1 to 2001.
Answer. The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001. This sequence forms an A.P. Here, first term, a = 1 Common difference, d = 2 $\begin{array}{c}\text{Here,}a+(n-1)d=2001\\ \Rightarrow 1+(n-1)\left(2\right)=2001\\ \Rightarrow 2n-2=2000\\ \Rightarrow n=1001\\ S_n=\frac{n}{2}[2a+(n-1\left)d\right]\end{array}$ $\begin{array}{cc}\therefore S_n& =\frac{1001}{2}[2\times 1+(1001-1)\times 2]\\ & =\frac{1001}{2}[2+1000\times 2]\\ & =\frac{1001}{2}\times 2002\\ & =1001\times 1001\\ & =1002001\end{array}$ Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Q2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer. The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995. $\begin{array}{c}\text{Here,}a=105\text{and}d=5\\ a+(n-1)d=995\\ \Rightarrow 105+(n-1)5=995\\ \Rightarrow (n-1)5=995-105=890\\ \Rightarrow n-1=178\\ \Rightarrow n=179\end{array}$ $\begin{array}{cc}\therefore S_n& =\frac{179}{2}\left[2\right(105)+(179-1\left)\right(5\left)\right]\\ & =\frac{179}{2}\left[2\right(105)+(178\left)\right(5\left)\right]\\ & =179[105+(89\left)5\right]\\ & =\left(179\right)(105+445)\\ & =\left(179\right)\left(550\right)\\ & =98450\end{array}$ Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Q3:In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is – 112.
Answer. First term = 2 Let d be the common difference of the A.P. Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, … Sum of first five terms = 10 + 10d Sum of next five terms = 10 + 35d According to the given condition, $\begin{array}{c}10+10d=\frac{1}{4}(10+35d)\\ \Rightarrow 40+40d=10+35d\\ \Rightarrow 30=-5d\\ \Rightarrow d=-6\\ \therefore a_{20}=a+(20-1)d=2+\left(19\right)(-6)=2-114=-112\end{array}$ Thus, the 20th term of the A.P. is

Q4. How many terms of the A.P. $-6,-\frac{11}{2},-5,\dots$ are needed to give the sum –25?
Answer. Let the sum of n terms of the given A.P. be –25. It is known that, 

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$, where n = number of terms, a = first term, and d = common difference Here, a = –6 $d=-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}$ Therefore, we obtain $\begin{array}{c}-25=\frac{n}{2}[2\times (-6)+(n-1\left)\left(\frac{1}{2}\right)\right]\\ \Rightarrow -50=n[-12+\frac{n}{2}-\frac{1}{2}]\\ \Rightarrow -50=n[-\frac{25}{2}+\frac{n}{2}]\\ \Rightarrow -100=n(-25+n)\\ \Rightarrow n^2-55n+100=0\\ \Rightarrow n^2-5n-20(n-5)=0\\ \Rightarrow n=20\text{or}5\end{array}$

Q5. In an A.P., if pth term is $\frac{1}{q}$ and qth term is $\frac{1}{p}$, prove that the sum of first pq terms is $\frac{1}{2}(pq+1),\text{where}p\ne q$
Answer. It is known that the general term of an A.P. is an = a + (n – 1)d ∴ According to the given information, $\begin{array}{c}{p}^{\text{th}}\mathrm{term}=a_p=a+(p-1)d=\frac{1}{q}...\left(1\right)\\ q^{\text{th}}\mathrm{term}=a_q=a+(q-1)d=\frac{1}{p}...\left(2\right)\end{array}$ Subtracting (2) from (1), we obtain $\begin{array}{c}(p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}\\ \Rightarrow (p-1-q+1)d=\frac{p-q}{pq}\\ \Rightarrow (p-q)d=\frac{p-q}{pq}\\ \Rightarrow d=\frac{1}{pq}\end{array}$ Putting the value of d in (1), we obtain $\begin{array}{c}a+(p-1)\frac{1}{pq}=\frac{1}{q}\\ \Rightarrow a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq}\\ \therefore S_{pq}=\frac{pq}{2}[2a+(pq-1\left)d\right]\end{array}$ $\begin{array}{cc}& =\frac{pq}{2}[\frac{2}{pq}+(pq-1\left)\frac{1}{pq}\right]\\ & =1+\frac{1}{2}(pq-1)\\ & =\frac{1}{2}pq+1-\frac{1}{2}=\frac{1}{2}pq+\frac{1}{2}\\ & =\frac{1}{2}(pq+1)\end{array}$ Thus, the sum of first pq terms of the A.P. is $\frac{1}{2}(pq+1)$.

Q6:If the sum of a certain number, of terms of the A.P. 25, 22, 19, …is 116. Find the last term.
Answer. Let the sum of n terms of the given A.P. be 116. $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ Here, a = 25 and d = 22 – 25 = – 3 $\begin{array}{c}\therefore S_n=\frac{n}{2}[2\times 25+(n-1\left)\right(-3\left)\right]\\ \Rightarrow 116=\frac{n}{2}[50-3n+3]\\ \Rightarrow 232=n(53-3n)=53n-3n^2\\ \Rightarrow 3n^2-53n+232=0\\ \Rightarrow 3n^2-24n-29n+232=0\\ \Rightarrow 3n(n-8)-29(n-8)=0\\ \Rightarrow (n-8)(3n-29)=0\\ \Rightarrow n=8\text{or}n=\frac{29}{3}\end{array}$ However, n cannot be equal to $\frac{29}{9}$. Therefore, n = 8 ∴ a8 = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3) = 25 + (7) (– 3) = 25 – 21 = 4 Thus, the last term of the A.P. is 4.

Q7. Find the sum to n terms of the A.P., whose $k^{th}$ term is 5k + 1.
Answer. It is given that the $k^{th}$ term of the A.P. is 5k + 1. $k^{th}$ term = ak = a + (k – 1)d ∴ a + (k – 1)d = 5k + 1 a + kd – d = 5k + 1 Comparing the coefficient of k, we obtain d = 5 a – d = 1 ⇒ a – 5 = 1 ⇒ a = 6 $\begin{array}{cc}{S}_n& =\frac{n}{2}[2a+(n-1\left)d\right]\\ & =\frac{n}{2}\left[2\right(6)+(n-1\left)\right(5\left)\right]\\ & =\frac{n}{2}[12+5n-5]\\ & =\frac{n}{2}(5n+7)\end{array}$

Q8. If the sum of n terms of an A.P. is $(pn+q{n}^2)$, where p and q are constants,find the common difference.
Answer. It is known that, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ According to the given condition, $\begin{array}{c}\frac{n}{2}[2a+(n-1\left)d\right]=pn+{qn}^2\\ \Rightarrow \frac{n}{2}[2a+nd-d]=pn+{qn}^2\\ \Rightarrow na+n^2\frac{d}{2}-n\cdot \frac{d}{2}=pn+{qn}^2\end{array}$ Comparing the coefficients of n2 on both sides, we obtain $\frac{d}{2}=q$ ∴ d = 2 q Thus, the common difference of the A.P. is 2q.

Q9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their ${18}^{th}$ terms.
Answer. Let $a_1,a_2,d_1{d}_2$ be the first terms and the common difference of the first and second arithmetic progression respectively. According to the given condition, $\begin{array}{c}\frac{\text{Sum of}n\text{terms of first A.P.}}{\text{Sum of}n\text{terms of second A.P.}}=\frac{5n+4}{9n+6}\\ \Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{5n+4}{9n+6}\\ \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}...\left(1\right)\end{array}$ Substituting n = 35 in (1), we obtain $\begin{array}{c}\frac{2a_1+34d_1}{2a_2+34d_2}=\frac{5\left(35\right)+4}{9\left(35\right)+6}\\ \Rightarrow \frac{a_1+17d_1}{a_2+17d_2}=\frac{179}{321}...\left(2\right)\\ \frac{{18}^{\text{th}}\text{term of first A.P.}}{{18}^{\text{th}}\text{term of second A.P.}}=\frac{a_1+17d_1}{a_2+17d_2}\dots \text{(3)}\end{array}$ From (2) and (3), we obtain $\frac{{18}^{\text{th}}\text{term of first A.P.}}{{18}^{\text{th}}\text{term of second A.P.}}=\frac{179}{321}$ Thus, the ratio of 18th term of both the A.P.s is 179: 321.

Q10:If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms.
Answer. Let a and d be the first term and the common difference of the A.P. respectively. Here, $\begin{array}{cc}{S}_p& =\frac{p}{2}[2a+(p-1\left)d\right]\\ S_q& =\frac{q}{2}[2a+(q-1\left)d\right]\end{array}$ According to the given condition, $\begin{array}{c}\frac{p}{2}[2a+(p-1\left)d\right]=\frac{q}{2}[2a+(q-1\left)d\right]\\ \Rightarrow p[2a+(p-1\left)d\right]=q[2a+(q-1\left)d\right]\\ \Rightarrow 2ap+pd(p-1)=2aq+qd(q-1)\\ \Rightarrow 2a(p-q)+d\left[p\right(p-1)-q(q-1\left)\right]=0\\ \Rightarrow 2a(p-q)+d[p^2-p-q^2+q]=0\end{array}$ $\begin{array}{cc}& \Rightarrow 2a(p-q)+d\left[\right(p-q\left)\right(p+q)-(p-q\left)\right]=0\\ & \Rightarrow 2a(p-q)+d\left[\right(p-q\left)\right(p+q-1\left)\right]=0\\ & \Rightarrow 2a+d(p+q-1)=0\\ & \Rightarrow d=\frac{-2a}{p+q-1}...\left(1\right)\\ & \therefore S_{p+q}=\frac{p+q}{2}[2a+(p+q-1)\cdot d]\end{array}$ $\begin{array}{cc}\Rightarrow S_{p+q}& =\frac{p+q}{2}[2a+(p+q-1\left)\left(\frac{-2a}{p+q-1}\right)\right]\left[\text{From}\right(1\left)\right]\\ & =\frac{p+q}{2}[2a-2a]\\ & =0\end{array}$ Thus, the sum of the first (p + q) terms of the A.P. is

Q11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
Answer. Let a1 and d be the first term and the common difference of the A.P. respectively. According to the given information, $\begin{array}{c}{S}_p=\frac{p}{2}[2a_1+(p-1\left)d\right]=a\\ \Rightarrow 2a_1+(p-1)d=\frac{2a}{p}...\left(1\right)\\ S_q=\frac{q}{2}[2a_1+(q-1\left)d\right]=b\\ \Rightarrow 2a_1+(q-1)d=\frac{2b}{q}...\left(2\right)\\ S_r=\frac{r}{2}[2a_1+(r-1\left)d\right]=c\\ \Rightarrow 2a_1+(r-1)d=\frac{2c}{r}\dots \left(3\right)\end{array}$ Subtracting (2) from (1), we obtain p – 1d – q – 1d = 2ap – 2bq ⇒dp – 1 – q + 1 = 2aq – 2bppq ⇒dp – q = 2aq – 2bppq ⇒d = 2aq – bppqp – q …….(4) Subtracting (3) from (2), we obtain $\begin{array}{c}(q-1)d-(r-1)d=\frac{2b}{q}-\frac{2c}{r}\\ \Rightarrow d(q-1-r+1)=\frac{2b}{q}-\frac{2c}{r}\\ \Rightarrow d(q-r)=\frac{2br-2qc}{qr}\\ \Rightarrow d=\frac{2(br-qc)}{qr(q-r)}...\left(5\right)\end{array}$ Equating both the values of d obtained in (4) and (5), we obtain aq – bppqp – q = br – qcqrq – r ⇒aq – bppp – q = br – qcrq – r ⇒rq – raq – bp = pp – qbr – qc ⇒raq – bpq – r = pbr – qcp – q ⇒aqr – bprq – r = bpr – cpqp – q Dividing both sides by pqr, we obtain $\begin{array}{c}(\frac{a}{p}-\frac{b}{q})(q-r)=(\frac{b}{q}-\frac{c}{r})(p-q)\\ \Rightarrow \frac{a}{p}(q-r)-\frac{b}{q}(q-r+p-q)+\frac{c}{r}(p-q)=0\\ \Rightarrow \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\end{array}$ Thus, the given result is proved.

Q12. The ratio of the sums of m and n terms of an A.P. is $m^2:n^2$. Show that the ratio of $m^{th}$ and $n^{th}$ term is (2m – 1) : (2n – 1).
Answer. Let a and b be the first term and the common difference of the A.P. respectively. According to the given condition, $\begin{array}{c}\frac{\text{Sum of}m\text{terms}}{\text{Sum of}n\text{terms}}=\frac{m^2}{n^2}\\ \Rightarrow \frac{\frac{m}{2}[2a+(m-1\left)d\right]}{\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]}=\frac{m^2}{n^2}\\ \Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}...\left(1\right)\end{array}$ Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain $\begin{array}{c}\frac{2a+(2m-2)d}{2a+(2n-2)d}=\frac{2m-1}{2n-1}\\ \Rightarrow \frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}...\left(2\right)\\ \frac{m^{\text{th}}\text{term of A.P.}}{n^{\text{th}}\text{term of A.P.}}=\frac{a+(m-1)d}{a+(n-1)d}...\left(3\right)\end{array}$ From (2) and (3), we obtain $\frac{m^{\text{th}}\text{term of}A.P}{n^{\text{th}}\text{term of}A.P}=\frac{2m-1}{2n-1}$ Thus, the given result is proved.

Q13. If the sum of n terms of an A.P. is $3n^2+5n$ and its mth term is 164, find the value of m.
Answer. Let a and b be the first term and the common difference of the A.P. respectively. am = a + (m – 1)d = 164 … (1) Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ Here, $\begin{array}{c}\frac{n}{2}[2a+nd-d]=3n^2+5n\\ \Rightarrow na+n^2\cdot \frac{d}{2}=3n^2+5n\end{array}$ Comparing the coefficient of n2 on both sides, we obtain $\begin{array}{c}\frac{d}{2}=3\\ \Rightarrow d=6\end{array}$ Comparing the coefficient of n on both sides, we obtain $\begin{array}{c}a-\frac{d}{2}=5\\ \Rightarrow a-3=5\\ \Rightarrow a=8\end{array}$ Therefore, from (1), we obtain 8 + (m – 1) 6 = 164 ⇒ (m – 1) 6 = 164 – 8 = 156 ⇒ m – 1 = 26 ⇒ m = 27 Thus, the value of m is 27.

Q14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer:
Let A1, A2, A3, A4 and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P.
Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1)d
6d = 26 – 8 = 18
d = 3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23.
Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

Q15. If $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b, then find the value of n.
Answer. A.M. of a and b $=\frac{a+b}{2}$ According to the given condition, $\begin{array}{c}\frac{a+b}{2}=\frac{a^n+b^n}{a^{n-1}+b^{n-1}}\\ \Rightarrow (a+b)(a^{n-1}+b^{n-1})=2(a^n+b^n)\\ \Rightarrow a^n+a{b}^{n-1}+b{a}^{n-1}+b^n=2a^n+2b^n\\ \Rightarrow a{b}^{n-1}+a^{n-1}b=a^n+b^n\\ \Rightarrow a{b}^{n-1}-b^n=a^n-a^{n-1}b\end{array}$ $\begin{array}{c}\Rightarrow b^{n-1}(a-b)=a^{n-1}(a-b)\\ \Rightarrow b^{n-1}=a^{n-1}\\ \Rightarrow {\left(\frac{a}{b}\right)}^{n-1}=1={\left(\frac{a}{b}\right)}^0\\ \Rightarrow n-1=0\\ \Rightarrow n=1\end{array}$

Q16:
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
Answer. Let $A_1,A_2,\dots A_m$ be m numbers such that 1, $A_1,A_2,\dots A_m$, 31 is an A.P. Here, a = 1, b = 31, n = m + 2 ∴ 31 = 1 + (m + 2 – 1) (d) ⇒ 30 = (m + 1) d $\Rightarrow d=\frac{30}{m+1}...\left(1\right)$ A1 = a + d A2 = a + 2d A3 = a + 3d … ∴ A7 = a + 7d Am–1 = a + (m – 1) d According to the given condition, $\begin{array}{c}\frac{a+7d}{a+(m-1)d}=\frac{5}{9}\\ \Rightarrow \frac{1+7\left(\frac{30}{(m+1)}\right)}{1+(m-1)\left(\frac{30}{m+1}\right)}=\frac{5}{9}\left[\text{From}\right(1\left)\right]\end{array}$ $\begin{array}{c}\Rightarrow \frac{m+1+7\left(30\right)}{m+1+30(m-1)}=\frac{5}{9}\\ \Rightarrow \frac{m+1+210}{m+1+30m-30}=\frac{5}{9}\\ \Rightarrow \frac{m+211}{31m-29}=\frac{5}{9}\\ \Rightarrow 9m+1899=155m-145\end{array}$ $\begin{array}{c}\Rightarrow 155m-9m=1899+145\\ \Rightarrow 146m=2044\\ \Rightarrow m=14\end{array}$ Thus, the value of m is 14.

Q17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?
Answer. The first installment of the loan is Rs 100. The second installment of the loan is Rs 105 and so on. The amount that the man repays every month forms an A.P. The A.P. is 100, 105, 110, … First term, a = 100 Common difference, d = 5 A30 = a + (30 – 1)d = 100 + (29) (5) = 100 + 145 = 245 Thus, the amount to be paid in the 30th installment is Rs 245.

Q18:
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Answer. The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°. It is known that the sum of all angles of a polygon with n sides is 180° (n – 2). $\begin{array}{c}\therefore S_n={180}^{\circ }(n-2)\\ \Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]={180}^{\circ }(n-2)\\ \Rightarrow \frac{n}{2}[{240}^{\circ }+(n-1\left)5^{\circ }\right]=180(n-2)\\ \Rightarrow n[240+(n-1\left)5\right]=360(n-2)\\ \Rightarrow 240n+5n^2-5n=360n-720\end{array}$ $\begin{array}{c}\Rightarrow 5n^2+235n-360n+720=0\\ \Rightarrow 5n^2-125n+720=0\\ \Rightarrow n^2-25n+144=0\\ \Rightarrow n^2-16n-9n+144=0\\ \Rightarrow n(n-16)-9(n-16)=0\\ \Rightarrow (n-9)(n-16)=0\\ \Rightarrow n=9\text{or}16\end{array}$

Exercise 9.3

Q1. Find the ${20}^{th}$ and $n^{th}$ terms of the G.P. $\frac{5}{2},\frac{5}{4},\frac{5}{8},\dots$
Answer. The given G.P. is $\frac{5}{2},\frac{5}{4},\frac{5}{8},\dots$ Here, a = First term = $\frac{5}{2}$ r = Common ratio $=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}$ $\begin{array}{c}{a}_{20}=a{r}^{20-1}=\frac{5}{2}{\left(\frac{1}{2}\right)}^{19}=\frac{5}{\left(2\right)(2{)}^{19}}=\frac{5}{(2{)}^{20}}\\ a_n=a{r}^{n-1}=\frac{5}{2}{\left(\frac{1}{2}\right)}^{n-1}=\frac{5}{\left(2\right)(2{)}^{n-1}}=\frac{5}{(2{)}^n}\end{array}$

Q2. Find the ${12}^{th}$ term of a G.P. whose $8^{th}$ term is 192 and the common ratio is 2.
Answer. Common ratio, r = 2 Let a be the first term of the G.P. ∴ a8 = ar 8–1 = ar7 ⇒ ar7 = 192 a(2)7 = 192 a(2)7 = (2)6 (3) $\begin{array}{c}\Rightarrow a=\frac{(2{)}^6\times 3}{(2{)}^7}=\frac{3}{2}\\ \therefore a_{12}=a{r}^{12-1}=\left(\frac{3}{2}\right)(2{)}^{11}=(3\left)\right(2{)}^{10}=3072\end{array}$

Q3. The $5^{th},8^{th}$ and ${11}^{th}$ terms of a G.P. are p, q and s, respectively. Show that $q^2=ps$.
Answer. Let a be the first term and r be the common ratio of the G.P. According to the given condition, a5 = a r5–1 = a r4 = p … (1) a8 = a r8–1 = a r7 = q … (2) a11 = a r11–1 = a r10 = s … (3) Dividing equation (2) by (1), we obtain $\begin{array}{c}\frac{a{r}^7}{a{r}^4}=\frac{q}{p}\\ r^3=\frac{q}{p}...\left(4\right)\end{array}$ Dividing equation (3) by (2), we obtain $\begin{array}{cc}\frac{a{r}^{10}}{a{r}^7}& =\frac{s}{q}\\ \Rightarrow r^3& =\frac{s}{q}...\left(5\right)\end{array}$ Equating the values of r3 obtained in (4) and (5), we obtain $\begin{array}{c}\frac{q}{p}=\frac{s}{q}\\ \Rightarrow q^2=ps\end{array}$ Thus, the given result is proved.

Q4. The $4^{th}$ term of a G.P. is square of its second term, and the first term is – 3.Determine its $7^{th}$ term.
Answer. Let a be the first term and r be the common ratio of the G.P. ∴ a = –3 $\begin{array}{c}\text{It is known that,}{a}_n=a{r}^{n-1}\\ \therefore a_4=a{r}^3=(-3)r^3\\ a_2=a{r}^2=(-3)r\\ \text{According to the given condition,}\\ (-3)r^3=\left[\right(-3)r{]}^2\\ \Rightarrow r=9r^2\\ \Rightarrow r=-3\\ a_7=a{r}^{7-1}=a{r}^6=(-3)(-3{)}^6=-(3{)}^7=-2187\end{array}$ Thus, the seventh term of the G.P. is –2187.

Q5. Which term of the following sequences: (a) $2,2\sqrt{2},4,\dots \text{is}128?$ (b) $\sqrt{3},3,3\sqrt{3},\dots .is729?$ (c) $\frac{1}{3},\frac{1}{9},\frac{1}{27},\dots \text{is}\frac{1}{19683}?$
Answer. (a) The given sequence is $2,2\sqrt{2},4,\dots$ Here, a = 2 and r = $\frac{2\sqrt{2}}{2}=\sqrt{2}$ Let the nth term of the given sequence be 128. $\begin{array}{c}{a}_n=a{r}^{n-1}\\ \Rightarrow \left(2\right)(\sqrt{2}{)}^{n-1}=128\\ \Rightarrow \left(2\right)(2{)}^{\frac{n-1}{2}}=(2{)}^7\\ \Rightarrow (2{)}^{\frac{n-1}{2}+1}=(2{)}^7\\ \therefore \frac{n-1}{2}=6\\ \Rightarrow n-1=12\\ \Rightarrow n=13\end{array}$ Thus, the 13th term of the given sequence is 128. (b) The given sequence is $\sqrt{3},3,3\sqrt{3}$ Here, $a=\sqrt{3}\text{and}r=\frac{3}{\sqrt{3}}=\sqrt{3}$ Let the nth term of the given sequence be 729. $\begin{array}{c}{a}_n=a{r}^{n-1}\\ \therefore a{r}^{n-1}=729\\ \Rightarrow \left(\sqrt{3}\right)(\sqrt{3}{)}^{n-1}=729\\ \Rightarrow \left(3{)}^{\frac{1}{2}}\right(3{)}^{\frac{n-1}{2}}=(3{)}^6\\ \Rightarrow (3{)}^{\frac{1}{2}+\frac{n-1}{2}}=(3{)}^6\end{array}$ $\begin{array}{c}\therefore \frac{1}{2}+\frac{n-1}{2}=6\\ \Rightarrow \frac{1+n-1}{2}=6\\ \Rightarrow n=12\end{array}$ Thus, the 12th term of the given sequence is 729. (c) The given sequence is $\frac{1}{3},\frac{1}{9},\frac{1}{27},\dots$ Here, $a=\frac{1}{3}\text{and}r=\frac{1}{9}÷\frac{1}{3}=\frac{1}{3}$ Let the nth term of the given sequence be $\frac{1}{19683}$. $\begin{array}{c}{a}_n=a{r}^{n-1}\\ \therefore a{r}^{n-1}=\frac{1}{19683}\\ \Rightarrow \left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^{n-1}=\frac{1}{19683}\\ \Rightarrow {\left(\frac{1}{3}\right)}^n={\left(\frac{1}{3}\right)}^9\\ \Rightarrow n=9\end{array}$ Thus, the 9th term of the given sequence is $\frac{1}{19683}$.

Q6. For what values of x, the numbers $-\frac{2}{7},x,-\frac{7}{2}a$ are in G.P.?
Answer. The given numbers are $-\frac{2}{7},x,-\frac{7}{2}a$. Common ratio $=\frac{\frac{x}{-2}}{\frac{-2}{7}}=\frac{-7x}{2}$ Also, common ratio = $\frac{\frac{-7}{2}}{x}=\frac{-7}{2x}$ $\begin{array}{c}\therefore \frac{-7x}{2}=\frac{-7}{2x}\\ \Rightarrow x^2=\frac{-2\times 7}{-2\times 7}=1\\ \Rightarrow x=\sqrt{1}\\ \Rightarrow x=\pm 1\end{array}$ Thus, for x = ± 1, the given numbers will be in G.P.

Q7. Find the sum to n terms in the geometric progression 0.15, 0.015, 0.0015, ... 20 terms.
Answer. The given G.P. is 0.15, 0.015, 0.00015, … Here, a = 0.15 and $r=\frac{0.015}{0.15}=0.1$ $\begin{array}{cc}{S}_n=& \frac{a(1-r^n)}{1-r}\\ \therefore S_{20}& =\frac{0.15[1-(0.1{)}^{20}]}{1-0.1}\\ & =\frac{0.15}{0.9}[1-(0.1{)}^{20}]\\ & =\frac{15}{90}[1-(0.1{)}^{20}]\\ & =\frac{1}{6}[1-(0.1{)}^{20}]\end{array}$

Q8. Find the sum to n terms in the geometric progression $\sqrt{7},\sqrt{21},3\sqrt{7},\dots ,n$
Answer. $\begin{array}{c}\text{The given G.P. is}\sqrt{7},\sqrt{21},3\sqrt{7},\dots \\ \text{Here,}a=\sqrt{7}\\ r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\\ S_n=\frac{a(1-r^n)}{1-r}\end{array}$ $\begin{array}{cc}\therefore S_n& =\frac{\sqrt{7}[1-(\sqrt{3}{)}^n]}{1-\sqrt{3}}\\ & =\frac{\sqrt{7}[1-(\sqrt{3}{)}^n]}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}\text{[by rationalizing]}\end{array}$ $\begin{array}{cc}& =\frac{\sqrt{7}(1+\sqrt{3})[1-(\sqrt{3}{)}^n]}{1-3}\\ & =\frac{-\sqrt{7}(1+\sqrt{3})}{2}[1-(3{)}^{\frac{n}{2}}]\\ & =\frac{\sqrt{7}(1+\sqrt{3})}{2}\left[\right(3{)}^{\frac{n}{2}}-1]\end{array}$

Q9. $\begin{array}{c}\text{Find the sum to}n\text{terms in the geometric progression}\\ 1,-a,a^2,-a^3\dots (\text{if}a\ne -1)\end{array}$

Answer. $\begin{array}{c}\text{The given G.P. is}1,-a,a^2,-a^3,\dots \dots \dots \dots \\ \text{Here, first term}=a_1=1\\ \text{Common ratio}=r=-a\end{array}$ $\begin{array}{c}{S}_n=\frac{a_1(1-r^n)}{1-r}\\ \therefore S_n=\frac{1[1-(-a{)}^n]}{1-(-a)}=\frac{[1-(-a{)}^n]}{1+a}\end{array}$

Q10:
Find the sum to n terms in the geometric progression x3, x5, x7 … (if x ≠ ± 1).
Answer. $\begin{array}{c}\text{The given G.P. is}{x}^3,x^5,x^7,\dots \\ \text{Here,}a=x^3\text{and}r=x^2\\ S_n=\frac{a(1-r^n)}{1-r}=\frac{x^3[1-{\left(x^2\right)}^n]}{1-x^2}=\frac{x^3(1-x^{2n})}{1-x^2}\end{array}$

Q11. Evaluate ${\sum }_{k=1}^{11}(2+3^k)$
Answer. ${\sum }_{k=1}^{11}(2+3^k)={\sum }_{k=1}^{11}\left(2\right)+{\sum }_{k=1}^{11}{3}^k$$=2\left(11\right)+{\sum }_{k=1}^{11}{3}^k=22+{\sum }_{k=1}^{11}{3}^k\dots \left(1\right)$ ${\sum }_{k=1}^{11}{3}^k=3^1+3^2+3^3+\dots +3^{11}$ $\begin{array}{c}\text{The terms of this sequence}3,3^2,3^3,\dots \text{forms a G.P.}\\ S_n=\frac{a(r^n-1)}{r-1}\\ \Rightarrow S_n=\frac{3\left[\right(3{)}^{11}-1]}{3-1}\\ \Rightarrow S_n=\frac{3}{2}(3^{11}-1)\end{array}$ $\begin{array}{c}\therefore {\sum }_{k=1}^{11}{3}^k=\frac{3}{2}(3^{11}-1)\\ \text{Substituting this value in equation}\left(1\right),\text{we obtain}\\ {\sum }_{k=1}^{11}(2+3^k)=22+\frac{3}{2}(3^{11}-1)\end{array}$

Q12. The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.
Answer. $\begin{array}{c}\text{Let}{r}^{\frac{a}{r},a,ar}\text{be the first three terms of the G.P.}\\ \frac{a}{r}+a+ar=\frac{39}{10}...\left(1\right)\\ \left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=1...\left(2\right)\end{array}$ From (2), we obtain $a^3=1$ ⇒ a = 1 (Considering real roots only) Substituting a = 1 in equation (1), we obtain $\begin{array}{c}\frac{1}{r}+1+r=\frac{39}{10}\\ \Rightarrow 1+r+r^2=\frac{39}{10}r\\ \Rightarrow 10+10r+10r^2-39r=0\\ \Rightarrow 10r^2-29r+10=0\\ \Rightarrow 10r^2-25r-4r+10=0\\ \Rightarrow 5r(2r-5)-2(2r-5)=0\\ \Rightarrow (5r-2)(2r-5)=0\\ \Rightarrow r=\frac{2}{5}\text{or}\frac{5}{2}\end{array}$ Thus, the three terms of G.P. are $\frac{5}{2},1,\text{and}\frac{2}{5}$

Q13:
How many terms of G.P. 3, 32, 33, …………… are needed to give the sum 120?
Answer. The given G.P. is $3,3^2,3^3,\dots$ Let n terms of this G.P. be required to obtain the sum as 120. $S_n=\frac{a(r^n-1)}{r-1}$ Here, a = 3 and r = 3 $\begin{array}{c}\therefore S_n=120=\frac{3(3^n-1)}{3-1}\\ \Rightarrow 120=\frac{3(3^n-1)}{2}\\ \Rightarrow \frac{120\times 2}{3}=3^n-1\\ \Rightarrow 3^n-1=80\\ \Rightarrow 3^n=81\\ \therefore n=4\end{array}$ Thus, four terms of the given G.P. are required to obtain the sum as 120.

Q 14:
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Answer. $\begin{array}{c}\text{Let the G.P. be a, ar,}a{r}^2,a{r}^3,\dots \\ \text{According to the given condition,}\\ a+ar+a{r}^2=16\text{and}a{r}^3+a{r}^4+a{r}^5=128\end{array}$ $\begin{array}{c}\Rightarrow a(1+r+r^2)=16\dots \left(1\right)\\ a{r}^3(1+r+r^2)=128\dots \left(2\right)\end{array}$ Dividing equation (2) by (1), we obtain $\begin{array}{c}\frac{a{r}^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}\\ \Rightarrow r^3=8\\ \therefore r=2\end{array}$ Substituting r = 2 in (1), we obtain a (1 + 2 + 4) = 16< ⇒ a (7) = 16 $\begin{array}{c}\Rightarrow a=\frac{16}{7}\\ S_n=\frac{a(r^n-1)}{r-1}\\ \Rightarrow S_n=\frac{16}{7}\frac{(2^n-1)}{2-1}=\frac{16}{7}(2^n-1)\end{array}$

Q15:
Given a G.P. with a = 729 and 7th term 64, determine S7.
Answer. a = 729 a7 = 64 Let r be the common ratio of the G.P. It is known that,$a_n=a{r}^{n-1}$ $\begin{array}{c}{a}_7=a{r}^{7-1}=\left(729\right)r^6\\ \Rightarrow 64=729r^6\\ \Rightarrow r^6=\frac{64}{729}\\ \Rightarrow r^6={\left(\frac{2}{3}\right)}^6\\ \Rightarrow r=\frac{2}{3}\end{array}$ Also, it is known that, $S_n=\frac{a(1-J^n)}{1-r}$ $\begin{array}{cc}\therefore S_7& =\frac{729[1-{\left(\frac{2}{3}\right)}^7]}{1-\frac{2}{3}}\\ & =3\times 729[1-{\left(\frac{2}{3}\right)}^7]\\ & =(3{)}^7-\left(\frac{(3{)}^7-(2{)}^7}{(3{)}^7}\right]\\ & =2187-128\\ & =2059\end{array}$

Q16:
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Answer. Let a be the first term and r be the common ratio of the G.P. According to the given conditions, $\begin{array}{c}{S}_2=-4=\frac{a(1-r^2)}{1-r}...\left(1\right)\\ a_5=4\times a_3\\ a_5^4=4{\mathrm{ar}}^2\\ \Rightarrow r^2=4\\ \therefore r=\pm 2\end{array}$ From (1), we obtain $\begin{array}{c}-4=\frac{a[1-(2{)}^2]}{1-2}\text{for}r=2\\ \Rightarrow -4=\frac{a(1-4)}{-1}\\ \Rightarrow -4=a\left(3\right)\\ \Rightarrow a=\frac{-4}{3}\end{array}$ $\begin{array}{c}\text{Also,}-4=\frac{a[1-(-2{)}^2]}{1-(-2)}\text{for}r=-2\\ \Rightarrow -4=\frac{a(1-4)}{1+2}\\ \Rightarrow -4=\frac{a(-3)}{3}\\ \Rightarrow a=4\end{array}$ Thus, the required G.P. is $\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},\dots$ 4,-8,16,-32,...

Q17:
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Answer. Let a be the first term and r be the common ratio of the G.P. According to the given condition, $\begin{array}{c}{a}_4=a{r}^3=x\dots \left(1\right)\\ a_{10}=a{r}^9=y\dots \left(2\right)\\ a_{16}=a{r}^{15}=z\dots \left(3\right)\end{array}$ Dividing (2) by (1), we obtain $\frac{y}{x}=\frac{a{r}^9}{a{r}^3}\Rightarrow \frac{y}{x}=r^6$ Dividing (3) by (2), we obtain $\begin{array}{c}\frac{z}{y}=\frac{a{r}^{15}}{a{r}^9}\Rightarrow \frac{z}{y}=r^6\\ \therefore \frac{y}{x}=\frac{z}{y}\end{array}$ Thus, x, y, z are in G. P.

Q18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Answer. The given sequence is 8, 88, 888, 8888… This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as Sn = 8 + 88 + 888 + 8888 + …………….. to n terms $\begin{array}{cc}& =\frac{8}{9}[9+99+999+9999+\dots \dots \dots \dots \text{to}n\text{terms}]\\ & =\frac{8}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+({10}^4-1)+\dots \dots \text{. to}n\text{terms}]\end{array}$ $\begin{array}{cc}& =\frac{8}{9}[(10+{10}^2+\dots ..n\text{terms})-(1+1+1+\dots n\text{terms}\left)\right]\\ & =\frac{8}{9}[\frac{10({10}^n-1)}{10-1}-n]\\ & =\frac{8}{9}[\frac{10({10}^n-1)}{9}-n]\\ & =\frac{80}{81}({10}^n-1)-\frac{8}{9}n\end{array}$

Q19. Find the sum of the products of the corresponding terms of the sequences 2,4,8,$16,32\text{and}128,32,8,2,\frac{1}{2}$.
Answer. $\begin{array}{c}\text{Required sum}=2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}\\ =64[4+2+1+\frac{1}{2}+\frac{1}{2^2}]\\ \text{Here,}4,2,1,\frac{1}{2},\frac{1}{2^2}\text{is a G.P.}\\ \text{First term,}a=4\end{array}$ $\begin{array}{c}\text{Common ratio,}r=\frac{1}{2}\\ S_n=\frac{a(1-r^n)}{1-r}\\ \therefore S_5=\frac{4[1-{\left(\frac{1}{2}\right)}^5]}{1-\frac{1}{2}}=\frac{4[1-\frac{1}{32}]}{\frac{1}{2}}=8\left(\frac{32-1}{32}\right)=\frac{31}{4}\end{array}$ $\therefore \text{Required sum}=64\left(\frac{31}{4}\right)=\left(16\right)\left(31\right)=496$

Q20:
Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, …….. ARn – 1 form a G.P. and find the common ratio.
Answer. It has to be proved that the sequence,a, ar, $a{r}^2$ $\dots a{r}^{n-1}\text{and}A,AR,A{R}^2,\dots A{R}^{n-1}$ form a G.P, $\frac{\text{Second term}}{\text{First term}}=\frac{arAR}{aA}=rR$ $\frac{\text{Third term}}{\text{Second term}}=\frac{a{r}^{2}A{R}^2}{arAR}=rR$ Thus, the above sequence forms a G.P. and the common ratio is rR.

Q21:
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Answer. Let a be the first term and r be the common ratio of the G.P. $\begin{array}{c}{a}_1=a,a_2=ar,a_3=a{r}^2,a_4=a{r}^3\\ \text{By the given condition,}\\ a_3=a_1+9\\ \Rightarrow a{r}^2=a+9\dots \left(1\right)\\ a_2=a_4+18\\ \Rightarrow ar=a{r}^3+18\dots \left(2\right)\end{array}$ $\begin{array}{c}\text{From}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ a(r^2-1)=9\dots \left(3\right)\\ \text{ar}(1-r^2)=18\dots \left(4\right)\\ \text{Dividing}\left(4\right)\text{by}\left(3\right),\text{we obtain}\end{array}$ $\begin{array}{c}\frac{ar(1-r^2)}{a(r^2-1)}=\frac{18}{9}\\ \Rightarrow -r=2\\ \Rightarrow r=-2\end{array}$ Substituting the value of r in (1), we obtain 4a = a + 9 ⇒ 3a = 9 ∴ a = 3 Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3­(–2)3 i.e., 3¸–6, 12, and –24.

Q22. If the $p^{th},q^{th}and{r}^{th}$ terms of a G.P. are a, b and c, respectively. Prove that $a^{q-r}{b}^{r-p}{c}^{P-q}=1$

Answer. Let A be the first term and R be the common ratio of the G.P. According to the given information, $\begin{array}{c}{AR}^{p-1}=a\\ {AR}^{q-1}=b\\ {AR}^{r-1}=c\\ a^{q-r}{b}^{r-p}{c}^{p-q}\end{array}$ $=A^{q-r}\times R^{(p-1)(q-r)}\times A^{r-p}\times R^{(q-1)(r-p)}\times A^{p-q}\times R^{(r-1)(p-q)}$ $\begin{array}{cc}& ={Aq}^{-r+r-p+p-q}\times R(pr-pr-q+r)+(rq-r+p-pq)+(pr-p-qr+q)\end{array}$ $\begin{array}{c}=A^0\times R^0\\ =1\end{array}$ Thus, the given result is proved.

Q23:
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
Answer. The first term of the G.P is a and the last term is b. Therefore, the G.P. is a,$a{r}^2,a{r}^3,\dots a{r}^{n-1}$ where r is the common ratio. $\begin{array}{c}b={ar}^{n-1}\dots \left(1\right)\\ P=\text{Product of}n\text{terms}\end{array}$ $\begin{array}{cc}& =\left(a\right)\left(ar\right)\left({ar}^2\right)\dots \left({ar}^{n-1}\right)\\ & =(a\times a\times \dots a)(r\times r^2\times \dots r^{n-1})\\ & =a^n{r}^{1+2+\dots (n-1)}\dots \text{(2)}\end{array}$ Here, 1, 2, …(n – 1) is an A.P. ∴1 + 2 + ……….+ (n – 1) $\begin{array}{c}=\frac{n-1}{2}[2+(n-1-1)\times 1]=\frac{n-1}{2}[2+n-2]=\frac{n(n-1)}{2}\\ P=a^n{r}^{\frac{n(n-1)}{2}}\\ \therefore P^2=a^{2n}{r}^{n(n-1)}\end{array}$ $\begin{array}{cc}& ={\left[a^2{r}^{(n-1)}\right]}^n\\ & ={[a\times {\mathrm{ar}}^{n-1}]}^n\\ & =(ab{)}^n\text{[using(1)]}\end{array}$ Thus, the given result is proved.

Q24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n+1{)}^{\text{th}}\text{to}(2n{)}^{\text{th}}\text{term is}\frac{1}{r^n}$
Answer. Let a be the first term and r be the common ratio of the G.P. Sum of 1st $n\text{terms}=\frac{a(1-r^n)}{(1-r)}$ Since there are n terms from (n +1)th to (2n)th term,$=\frac{a_{n+1}(1-r^n)}{(1-r)}$ Thus, required ratio = $\frac{a(1-r^n)}{(1-r)}\times \frac{(1-r)}{a{r}^n(1-r^n)}=\frac{1}{r^n}$ Thus, the 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 $\frac{1}{r^n}$ .

Q25. If a, b, c and d are in G.P. show that $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd{)}^2$Answer. a, b, c, d are in G.P. Therefore, $\begin{array}{c}bc=\text{ad}\dots .\left(1\right)\\ b^2=ac\dots \left(2\right)\\ c^2=bd\dots \left(3\right)\\ \text{It has to be proved that,}\\ (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc-cd{)}^2\end{array}$ R.H.S. $\begin{array}{cc}& =(ab+bc+cd{)}^2\\ & =(ab+ad+cd{)}^2[U\mathrm{sing}\left(1\right)]\\ & =[ab+d(a+c){]}^2\\ & =a^2{b}^2+2abd(a+c)+d^2(a+c{)}^2\\ & =a^2{b}^2+2a^2{c}^2+2acbd+d^2(a^2+2ac+c^2)\\ & =a^2{b}^2+2a^2{c}^2+2b^2{c}^2+d^2{a}^2+2a^2{b}^2+d^2{c}^2\left[U\mathrm{sing}\right(1)\text{and}\left(2\right)]\end{array}$ $\begin{array}{cc}& =a^2{b}^2+a^2{c}^2+a^2{c}^2+b^2{c}^2+b^2{c}^2+d^2{a}^2+d^2{b}^2+d^2{b}^2\\ & +d^2{c}^2\\ & =a^2{b}^2+a^2{c}^2+a^2{d}^2+b^2\times b^2+b^2{c}^2+b^2{d}^2+c^2{b}^2+c^2\times c^2+c^2{d}^2\end{array}$ [Using (2) and (3) and rearranging terms] $\begin{array}{c}=a^2(b^2+c^2+d^2)+b^2(b^2+c^2+d^2)+c^2(b^2+c^2+d^2)\\ =(a^2+b^2+c^2)(b^2+c^2+d^2)\end{array}$ = L.H.S. ∴ L.H.S. = R.H.S. $\therefore (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd{)}^2$

Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer. $\begin{array}{c}\text{Let}{G}_1\text{and}{G}_2\text{be two numbers between}3\text{and}81\text{such}\\ \text{that the series,}3,G_1,G_2,81,\text{forms a G.P.}\end{array}$ Let a be the first term and r be the common ratio of the G.P. $\begin{array}{c}\therefore 81=\left(3\right)(r{)}^3\\ \Rightarrow r^3=27\\ \therefore r=3\text{(Taking real roots only)}\\ \text{For}r=3\end{array}$ $\begin{array}{c}{G}_1=ar=\left(3\right)\left(3\right)=9\\ G_2={ar}^2=\left(3\right)(3{)}^2=27\end{array}$ Thus, the required two numbers are 9 and 27.

Q27. Find the value of n so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between a and b.

Answer. $\begin{array}{c}\text{G. M. of a and b is}\sqrt{ab}\\ \text{By the given condition,}\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}\end{array}$ Squaring both sides, we obtain $\begin{array}{c}\frac{{(a^{n+1}+b^{n+1})}^2}{{(a^n+b^n)}^2}=ab\\ \Rightarrow a^{2n+2}+2a^{n+1}{b}^{n+1}+b^{2n+2}=\left(ab\right)(a^{2n}+2a^n{b}^n+b^{2n})\\ \Rightarrow a^{2n+2}+2a^{n+1}{b}^{n+1}+b^{2n+1}b+2a^{n+1}{b}^{n+1}+a{b}^{2n+1}\\ \Rightarrow a^{2n+2}+b^{2n+1}b+a{b}^{2n+1}\end{array}$ $\begin{array}{cc}& \Rightarrow a^{2n+2}-a^{2n+1}b=a{b}^{2n+1}-b^{2n+2}\\ & \Rightarrow a^{2n+1}(a-b)=b^{2n+1}(a-b)\\ & \Rightarrow {\left(\frac{a}{b}\right)}^{2n+1}=1={\left(\frac{a}{b}\right)}^0\\ & \Rightarrow 2n+1=0\\ & \Rightarrow n=\frac{-1}{2}\end{array}$

Q28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio ($(3+2\sqrt{2}):(3-2\sqrt{2})$
Answer. Let the two numbers be a and b. $G.M.=\sqrt{ab}$ According to the given condition, $\begin{array}{c}a+b=6\sqrt{ab}\\ \Rightarrow (a+b{)}^2=36(ab)...\left(1\right)\end{array}$ Also, $\begin{array}{c}(a-b{)}^2=(a+b{)}^2-4ab=36ab-4ab=32ab\\ \Rightarrow a-b=\sqrt{32}\sqrt{ab}\\ =4\sqrt{2}\sqrt{ab}...\left(2\right)\end{array}$ Adding (1) and (2), we obtain $\begin{array}{c}2a=(6+4\sqrt{2})\sqrt{ab}\\ \Rightarrow a=(3+2\sqrt{2})\sqrt{ab}\end{array}$ Substituting the value of a in (1), we obtain $\begin{array}{c}b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}\\ \Rightarrow b=(3-2\sqrt{2})\sqrt{ab}\\ \frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\end{array}$ Thus, the required ratio is $(3+2\sqrt{2}):(3-2\sqrt{2})$

Q29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A\pm \sqrt{(A+G)(A-G)}$.

Answer. It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b. $\begin{array}{cc}\therefore AM=A=\frac{a+b}{2}& \dots \left(1\right)\\ GM=G=\sqrt{ab}& \dots \left(2\right)\end{array}$ $\begin{array}{c}\text{From}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ a+b=2A\dots \text{(3)}\\ ab=G^2\dots \left(4\right)\\ \text{Substituting the value of a and b from (3) and (4) in the}\\ \text{identity}(a-b{)}^2=(a+b{)}^2-4ab,\text{we obtain}\end{array}$ $\begin{array}{c}(a-b{)}^2=4A^2-4G^2=4(A^2-G^2)\\ (a-b{)}^2=4(A+G\left)\right(A-G)\\ (a-b)=2\sqrt{(A+G)(A-G)}...\left(5\right)\end{array}$ $\begin{array}{c}\text{From}\left(3\right)\text{and}\left(5\right),\text{we obtain}\\ 2a=2A+2\sqrt{(A+G)(A-G)}\\ \Rightarrow a=A+\sqrt{(A+G)(A-G)}\\ \text{Substituting the value of a in (3), we obtain}\end{array}$ $b=2A-A-\sqrt{(A+G)(A-G)}=A-\sqrt{(A+G)(A-G)}$ Thus, the two numbers are $A\pm \sqrt{(A+G)(A-G)}$

Q30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
Answer. It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P. Here, a = 30 and r = 2 $\begin{array}{c}\text{Here,}a=30\text{and}r=2\\ \therefore a_3=a{r}^2=\left(30\right)(2{)}^2=120\end{array}$ Therefore, the number of bacteria at the end of 2nd hour will be 120. $\begin{array}{c}\text{The number of bacteria at the end of}{4}^{\text{th}}\text{hour will be}480.\\ a_{n+1}={ar}^n=\left(30\right)2^n\\ \text{Thus, number of bacteria at the end of}{n}^{\text{th}}\text{hour will be}\\ 30(2{)}^n.\end{array}$

Q31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Answer. The amount deposited in the bank is Rs 500. At the end of first year, amount =$\mathrm{Rs}500(1+\frac{1}{10})$= Rs 500 (1.1) At the end of 2nd year, amount = Rs 500 (1.1) (1.1) At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on ∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times) = Rs $500(1.1{)}^{10}$

Q32:
If AM. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Answer. Let the root of the quadratic equation be a and b. According to the given condition, $\begin{array}{cc}A\cdot M.& =\frac{a+b}{2}=8\Rightarrow a+b=16...\left(1\right)\\ G.M& =\sqrt{ab}=5\Rightarrow ab=25...\left(2\right)\end{array}$ The quadratic equation is given by, $\begin{array}{c}{x}^2-x(\text{sum of roots)}+(\text{Product of roots})=0\\ x^2-x(a+b)+\left(\text{Product of roots}\right)=0\\ x^2-16x+25=0\text{[using}\left(1\right)\text{and}\left(2\right)]\end{array}$ Thus, the required quadratic equation is $x^2-16x+25=$ 0 .

Exercise 9.4

Question 1:
Find the sum to n terms of the series
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …………

Answer.Given:1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +...to nth terms

The nth term of the series is an = n(n+1)

Question2. Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

 Answer.

an =n(n+1)(n+2)

=(n2 + n)(n + 2)

=n3 + 3n2 + 2n
The sum of n terms of a series is given by the equation 

=

Question3. Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7×32+…

Answer.
Given: 3 × 12 + 5 × 22 + 7×32+…to   nth terms

Question4.

Answer.
Given:to nth terms

Question5. Find the sum to n terms of the series 52 + 62 + 72+…+202

Answer. 

Given: 52 + 62 + 72+…+202

= The nth term of the series is

= an=(n+4)2

=n2+8n+16

= The sum of n terms of a series is given by the equation

= 

Question6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 + …

Answer.
Given: 3 × 8 + 6 × 11 + 9 × 14 + …to nth terms

an= (nth
term of 3,6,9...
)×
(nth
term of 8,11,14...

)

=(3n)(3n+5)

=9n2+15n
The sum of n terms of a series is given by the equation 

 The sum of  n terms of the given series is
 .

Question7. Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32)+…

 Answer.
Given: 12 + (12 + 22) + (12 + 22 + 32)+…to nth terms

Question8. Find the sum to n terms of the series n(n+1)(n+4)

Answer.
Given:n(n+1)(n+4)

 

Question9. n2+2n

Answer.
Given:n2+2n

The nth term of the series is an = n2+2n
The sum of n terms of a series is given by the equation 

The sum of first n terms of the given series is

 

Question10. Find the sum to n terms of the series  (2n−1)2

Answer.
Given:(2n−1)2

the nth term of the series is an=(2n−1)2

=4n2−4n+1The sum of n terms of a series is given by the equation 

 .The sum of first  nth  terms of the given series is