Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series (Exercise 8.2) with comprehensive step-by-step explanations according to the latest CBSE and NCERT syllabus. This exercise focuses on Geometric Progression (G.P.), including important formulas for nth term, sum to n terms, infinite G.P., and relation between A.M., G.M., and H.M. The detailed solutions help students strengthen their conceptual understanding and excel in board as well as competitive exams like JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes in PDF format.
Question 1. Find the 20th and nth terms of the G.P.
$\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \dots$
Solution:
According to the question,
G.P.: $\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \dots$
So, first term $(a) = \dfrac{5}{2}$
The common ratio is
$$r = \dfrac{a_2}{a_1} = \dfrac{\frac{5}{4}}{\frac{5}{2}} = \dfrac{1}{2}$$
To find: 20th and nth terms of the given G.P.
The nth term of a G.P. is given by
$$a_n = ar^{n-1}$$
Now, the 20th term:
$$a_{20} = \left(\dfrac{5}{2}\right)\left(\dfrac{1}{2}\right)^{20-1} = \left(\dfrac{5}{2}\right)\left(\dfrac{1}{2}\right)^{19} = \dfrac{5}{2^{20}}$$
The nth term:
$$a_n = \dfrac{5}{2} \times \left(\dfrac{1}{2}\right)^{n-1} = \dfrac{5}{2^n}$$
Question 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
Given,
Common ratio $(r) = 2$
and 8th term $a_8 = 192$
Let $a$ be the first term.
Then,
$$a_8 = ar^{7}$$
$$a(2)^7 = 192$$
$$a = \dfrac{192}{128} = \dfrac{3}{2}$$
To find: 12th term $(a_{12})$
The nth term of a G.P. is
$$a_n = ar^{n-1}$$
So,
$$a_{12} = a r^{12-1} = a r^{11}$$
$$a_{12} = \left(\dfrac{3}{2}\right)(2)^{11}$$
$$a_{12} = \dfrac{3 \times 2048}{2} = 3072$$
Therefore, the 12th term is $3072$.
Question 3. The 5th, 8th and 11th terms of a G.P. are $p$, $q$ and $s$, respectively. Show that $q^2 = ps$.
Solution:
Let the first term be $a$ and common ratio be $r$.
Then,
$$a_5 = ar^4 = p \quad …(1)$$
$$a_8 = ar^7 = q \quad …(2)$$
$$a_{11} = ar^{10} = s \quad …(3)$$
Dividing equation (2) by (1):
$$\dfrac{ar^7}{ar^4} = \dfrac{q}{p}$$
$$r^3 = \dfrac{q}{p} \quad …(4)$$
Dividing equation (3) by (2):
$$\dfrac{ar^{10}}{ar^7} = \dfrac{s}{q}$$
$$r^3 = \dfrac{s}{q} \quad …(5)$$
From (4) and (5):
$$\dfrac{q}{p} = \dfrac{s}{q}$$
$$q^2 = ps$$
Hence proved.
Question 4. The 4th term of a G.P. is the square of its 2nd term, and the first term is –3. Determine its 7th term.
Solution:
Given,
First term $(a) = -3$
and 4th term is the square of its 2nd term.
Let the common ratio be $r$.
The nth term of a G.P. is
$$a_n = ar^{n-1}$$
So,
$$a_4 = ar^3 \quad \text{and} \quad a_2 = ar$$
It is given that
$$a_4 = (a_2)^2$$
Substituting,
$$ar^3 = (ar)^2$$
$$ar^3 = a^2r^2$$
$$r = a$$
Substituting $a = -3$, we get
$$r = -3$$
Now, find the 7th term:
$$a_7 = ar^{7-1} = ar^6$$
$$a_7 = (-3)(-3)^6 = (-3)(729) = -2187$$
Therefore, the 7th term is $-2187$.
Question 5. Which term of the following sequences is the given number?
(a) $2, 2\sqrt{2}, 4,\dots$. Which term is 128?
(b) $\sqrt{3}, 3, 3\sqrt{3},\dots. $ Which term is 729?
(c) $\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27},\dots$. Which term is $\dfrac{1}{19683}$?
Solution:
(a) $2, 2\sqrt{2}, 4,\dots$. Which term is 128?
First term ($a=2$). Common ratio ($r=\sqrt{2}$). The nth term is ($a_n=ar^{,n-1}$).
We solve
$$2(\sqrt{2})^{,n-1}=128.$$
Divide by (2):
$$(\sqrt{2})^{,n-1}=64.$$
Write ($\sqrt{2}=2^{1/2}$) and ($64=2^6$):
$$2^{(n-1)/2}=2^6.$$
Equate exponents:
$$\frac{n-1}{2}=6\quad\Rightarrow\quad n-1=12\quad\Rightarrow\quad n=13.$$
Answer: (128) is the (13)th term.
(b) $\sqrt{3}, 3, 3\sqrt{3},\dots. $ Which term is 729?
First term ($a=\sqrt{3}=3^{1/2}$). Common ratio ($r=\sqrt{3}$). Solve
$$3^{1/2}(3^{1/2})^{,n-1}=729.$$
Left side is ($3^{1/2+(n-1)/2}=3^{n/2}$). Since ($729=3^6$),
$$3^{n/2}=3^6\quad\Rightarrow\quad \frac{n}{2}=6\quad\Rightarrow\quad n=12.$$
Answer: (729) is the (12)th term.
(c) $\dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27},\dots$. Which term is $\dfrac{1}{19683}$?
This is a G.P. with $a=\dfrac{1}{3}=(\dfrac{1}{3})^1$ and ($r=\tfrac{1}{3}$). Note ($19683=3^9$), so
$$\frac{1}{19683}=\frac{1}{3}.\frac{1}{3^{n-1}}=\frac{1}{3^n}=\Big(\frac{1}{3}\Big)^9.$$
Thus $\Big(\frac{1}{3}\Big)^9$ is the 9th term, so (n=9).
Answer: $\dfrac{1}{19683}$ is the (9)th term.
Question 6. For what values of $x$, the numbers $-\dfrac{2}{7}$, $x$, $-\dfrac{7}{2}$ are in G.P.?
Solution:
If the three numbers are in G.P., then their common ratios are equal.
$$\dfrac{x}{-\dfrac{2}{7}} = \dfrac{-\dfrac{7}{2}}{x}$$
Simplifying,
$$-\dfrac{7x}{2} = -\dfrac{7}{2x}$$
Multiply both sides by 2 and divide by 7,
$$x = \dfrac{1}{x}$$
Therefore,
$$x^2 = 1$$
Hence,
$$x = \pm 1$$
Question 7. Find the sum of 20 terms of the G.P. $0.15, 0.015, 0.0015, \dots$
Solution:
Here,
$$a = 0.15,\quad r = 0.1$$
Sum of $n$ terms of a G.P. is
$$S_n = \frac{a(1 – r^n)}{1 – r}$$
So,
$$S_{20} = \frac{0.15(1 – (0.1)^{20})}{1 – 0.1}$$
$$S_{20} = \frac{0.15}{0.9}(1 – (0.1)^{20})$$
Therefore,
$$S_{20} = \frac{1}{6}\left(1 – (0.1)^{20}\right)$$
Question 8. Find the sum of $n$ terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots$
Solution:
Here,
$$a = \sqrt{7},\quad r = \sqrt{3}$$
Sum of $n$ terms is
$$S_n = \frac{a(1 – r^n)}{1 – r}$$
Substitute the values:
$$S_n = \frac{\sqrt{7}(1 – (\sqrt{3})^n)}{1 – \sqrt{3}}$$
Rationalize the denominator by multiplying by $(1 + \sqrt{3})$:
$$S_n = \frac{\sqrt{7}(1 – (\sqrt{3})^n)(1 + \sqrt{3})}{(1 – \sqrt{3})(1 + \sqrt{3})}$$
$$S_n = \frac{\sqrt{7}(1 + \sqrt{3})(1 – (\sqrt{3})^n)}{1 – 3}$$
$$S_n = -\frac{\sqrt{7}(1 + \sqrt{3})(1 – (\sqrt{3})^n)}{2}$$
Therefore,
$$S_n = \frac{\sqrt{7}(1 + \sqrt{3})}{2}\left((\sqrt{3})^n – 1\right)$$
Question 9. Find the sum of $n$ terms of the G.P. $1, -a, a^2, -a^3, \dots$ (if $a \neq -1$)
Solution:
Here,
$$a = 1,\quad r = -a$$
Sum of $n$ terms:
$$S_n = \frac{a(1 – r^n)}{1 – r}$$
So,
$$S_n = \frac{1 – (-a)^n}{1 – (-a)}$$
Therefore,
$$S_n = \frac{1 – (-a)^n}{1 + a}$$
Question 10. Find the sum of $n$ terms of the G.P. $x^3, x^5, x^7, \dots$ (if $x \neq \pm 1$)
Solution:
Here,
$$a = x^3,\quad r = x^2$$
Sum of $n$ terms:
$$S_n = \frac{a(1 – r^n)}{1 – r}$$
So,
$$S_n = \frac{x^3(1 – (x^2)^n)}{1 – x^2}$$
Simplify:
$$S_n = \frac{x^3(1 – x^{2n})}{1 – x^2}$$
Therefore,
$$S_n = \frac{x^3(1 – x^{2n})}{1 – x^2}$$
Question 11. Evaluate $$\sum_{k=1}^{11} (2+3^k)$$
Solution:
We have
$$
\sum_{k=1}^{11} (2+3^k) = (2 + 3^1) + (2 + 3^2) + (2 + 3^3) + \dots + (2 + 3^{11})
$$
This can be separated as
$$
(2 + 2 + \dots + 2 \text{ [11 terms]}) + (3^1 + 3^2 + 3^3 + \dots + 3^{11})
$$
As we know, the sum of $n$ terms of a G.P. with first term $a$ and common ratio $r$ is given by
$$
S_n = \frac{a(1 – r^n)}{1-r}
$$
Here, the sum of the constant $2$ terms is
$$
2 \times 11 = 22
$$
The sum of the geometric series $3^1 + 3^2 + \dots + 3^{11}$ is
$$
S = \frac{3(3^{11} – 1)}{3 – 1} = \frac{3(3^{11} – 1)}{2}
$$
So, the total sum is
$$
\sum_{k=1}^{11} (2+3^k) = 22 + \frac{3}{2} \big(3^{11} – 1\big)
$$
Question 12. The sum of first three terms of a G.P. is $\dfrac{39}{10}$ and their product is $1$. Find the common ratio and the terms.
Solution:
Let the three terms be $\dfrac{a}{r}, a, ar$.
From the product of the terms:
$$
\dfrac{a}{r} \times a \times ar = a^3 = 1
$$
So,
$$
a = 1
$$
The sum of the terms is
$$
\dfrac{a}{r} + a + ar = \dfrac{1}{r} + 1 + r = \dfrac{39}{10}
$$
Multiply through by $r$ to simplify:
$$
r^2 + r + 1 = \dfrac{39}{10} r
$$
Multiply both sides by $10$:
$$
10r^2 + 10r + 10 = 39r
$$
Rearranging:
$$
10r^2 – 29r + 10 = 0
$$
Solving this quadratic equation, we get
$$
r = \dfrac{2}{5}, \dfrac{5}{2}
$$
Thus, the G.P. terms are
$$
\dfrac{5}{2}, 1, \dfrac{2}{5}
$$
Question 13. How many terms of the G.P. $3, 3^2, 3^3, \dots$ are needed to give the sum 120?
Solution:
Given G.P.: $3, 3^2, 3^3, \dots$
- First term: $a = 3$
- Common ratio: $r = 3$
- Required sum: $S_n = 120$
The sum of $n$ terms of a G.P. is given by:
$$
S_n = \frac{a(1 – r^n)}{1 – r}
$$
Substitute the values:
$$
120 = \frac{3(1 – 3^n)}{1 – 3} = \frac{3(1 – 3^n)}{-2}
$$
Multiply both sides by $-2$:
$$
-240 = 3(1 – 3^n)
$$
Divide by $3$:
$$
-80 = 1 – 3^n
$$
Subtract $1$ from both sides:
$$
-81 = -3^n
$$
Multiply both sides by $-1$:
$$
3^n = 81
$$
Since $81 = 3^4$, we get
$$
n = 4
$$
Answer: 4 terms of the G.P. are needed to give the sum 120.
Question 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.
Solution:
Let the first three terms of the G.P. be $a, ar, ar^2$, where $a$ is the first term and $r$ is the common ratio.
From the question:
$$
a + ar + ar^2 = 16
$$
Factor out $a$:
$$
a(1 + r + r^2) = 16 \quad \text{…(1)}
$$
The next three terms are $ar^3, ar^4, ar^5$ with sum 128:
$$
ar^3 + ar^4 + ar^5 = 128
$$
Factor out $ar^3$:
$$
ar^3(1 + r + r^2) = 128 \quad \text{…(2)}
$$
Divide equation (2) by equation (1):
$$
\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{128}{16}
$$
Simplify:
$$
r^3 = 8 \implies r = 2
$$
Substitute $r = 2$ into equation (1) to find $a$:
$$
a(1 + 2 + 2^2) = 16
$$
$$
a(1 + 2 + 4) = 16
$$
$$
a \cdot 7 = 16
$$
$$
a = \frac{16}{7}
$$
The sum of $n$ terms of a G.P. is:
$$
S_n = \frac{a(r^n – 1)}{r – 1}
$$
Substitute $a = \frac{16}{7}$ and $r = 2$:
$$
S_n = \frac{\frac{16}{7}(2^n – 1)}{2 – 1} = \frac{16(2^n – 1)}{7}
$$
Answer:
- First term: $a = \frac{16}{7}$
- Common ratio: $r = 2$
- Sum of $n$ terms: $S_n = \frac{16(2^n – 1)}{7}$
Question 15. Given a G.P. with $a = 729$ and 7th term $64$, determine $S_7$.
Solution:
Given:
- First term: $a = 729$
- 7th term: $a_7 = 64$
Let the common ratio be $r$. The $n$-th term of a G.P. is
$$
a_n = ar^{n-1}
$$
So, for the 7th term:
$$
a_7 = ar^6 = 64
$$
$$
729 r^6 = 64
$$
$$
r^6 = \dfrac{64}{729} = \left(\dfrac{2}{3}\right)^6
$$
$$
r = \pm \dfrac{2}{3}
$$
The sum of $n$ terms of a G.P. is
$$
S_n = \dfrac{a(r^n – 1)}{r – 1}
$$
Case 1: $r = \frac{2}{3}$
$$
S_7 = \dfrac{729\left(\left(\dfrac{2}{3}\right)^7 – 1\right)}{\dfrac{2}{3} – 1} = 2059
$$
Case 2: $r = -\dfrac{2}{3}$
$$
S_7 = \dfrac{729\left(\left(-\dfrac{2}{3}\right)^7 – 1\right)}{-\dfrac{2}{3} – 1} = 463
$$
Question 16. Find a G.P. for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
Solution:
Let the first term be $a$ and common ratio be $r$.
From the question:
$$
a + ar = -4
$$
The fifth term is 4 times the third term:
$$
a r^4 = 4(a r^2)
$$
$$
r^4 = 4 r^2
$$
$$
r^2 = 4 \implies r = \pm 2
$$
Case 1: $r = 2$
$$
a + ar = a + 2a = 3a = -4
$$
$$
a = -\frac{4}{3}
$$
So, the G.P. is:
$$
-\frac{4}{3}, -\frac{8}{3}, \frac{16}{3}, \dots
$$
Case 2: $r = -2$
$$
a + ar = a – 2a = -a = -4
$$
$$
a = 4
$$
So, the G.P. is:
$$
4, -8, 16, -32, \dots
$$
