Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 8 – 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:
$$
4, -8, 16, -32, \dots
$$
Question 17. 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.
Solution:
Let the first term of the G.P. be $a$ and the common ratio be $r$.
According to the question:
$$
a_4 = ar^3 = x \quad \text{…(1)}
$$
$$
a_{10} = ar^9 = y \quad \text{…(2)}
$$
$$
a_{16} = ar^{15} = z \quad \text{…(3)}
$$
Now, divide equation (2) by (1):
$$
\frac{a r^9}{a r^3} = \frac{y}{x}
$$
$$
r^6 = \frac{y}{x}
$$
Divide equation (3) by (2):
$$
\frac{a r^{15}}{a r^9} = \frac{z}{y}
$$
$$
r^6 = \frac{z}{y}
$$
Thus,
$$
\frac{y}{x} = \frac{z}{y}
$$
Hence, $x, y, z$ are in G.P.
Question 18. Find the sum to $n$ terms of the sequence $8, 88, 888, 8888, \dots$
Solution:
Given sequence: $8, 88, 888, 8888, \dots$
This sequence is neither an A.P. nor a G.P., but it can be transformed into a G.P.
Let
$$
S_n = 8 + 88 + 888 + 8888 + \dots \text{ (n terms)}
$$
We can rewrite as:
$$
S_n = 8(1 + 11 + 111 + 1111 + \dots)
$$
Now express each term as a series of 9’s:
$$
S_n = \frac{8}{9}(9 + 99 + 999 + 9999 + \dots)
$$
Each term inside the bracket can be written as:
$$
(10 – 1), (10^2 – 1), (10^3 – 1), (10^4 – 1), \dots
$$
So,
$$
S_n = \frac{8}{9}\Big[(10 + 10^2 + 10^3 + 10^4 + \dots + 10^n) – (1 + 1 + 1 + 1 + \dots + n \text{ terms})\Big]
$$
As we know, the sum of $n$ terms of a G.P. with first term $a$ and common ratio $r$ is
$$
S_n = \frac{a(r^n – 1)}{r – 1}
$$
Here, for the G.P. $10 + 10^2 + 10^3 + \dots$,
$a = 10$ and $r = 10$.
So,
$$
\text{Sum} = \frac{10(10^n – 1)}{10 – 1}
$$
Substitute this into the equation for $S_n$:
$$
S_n = \frac{8}{9} \left[\frac{10(10^n – 1)}{9} – n \right]
$$
Simplify:
$$
S_n = \frac{8}{9} \left[\frac{10^{n+1} – 10}{9} – n \right]
$$
Or equivalently,
$$
S_n = \frac{80(10^n – 1)}{9} – \frac{8n}{9}
$$
Question 19. Find the sum of the products of the corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \dfrac{1}{2}$.
Solution:
According to the question:
Sequence 1: $2, 4, 8, 16, 32$
Sequence 2: $128, 32, 8, 2, \dfrac{1}{2}$
The products of the corresponding terms are:
$$
2 \times 128, \quad 4 \times 32, \quad 8 \times 8, \quad 16 \times 2, \quad 32 \times \frac{1}{2}
$$
That is:
$$
256, 128, 64, 32, 16
$$
This is a G.P. with
- First term $a = 256$
- Common ratio $r = \dfrac{1}{2}$
The sum of $n$ terms of a G.P. is given by:
$$
S_n = \dfrac{a(1 – r^n)}{1 – r}
$$
For $n = 5$:
$$
S_5 = \dfrac{256(1 – (\dfrac{1}{2})^5)}{1 – \dfrac{1}{2}}
$$
Simplify:
$$
S_5 = 256 \times \dfrac{1 – \dfrac{1}{32}}{\dfrac{1}{2}} = 256 \times \dfrac{\dfrac{31}{32}}{\dfrac{1}{2}} = 256 \times \dfrac{31}{16} = 496
$$
Answer: $S_5 = 496$
Question 20. Show that the products of the corresponding terms of the sequences $a, ar, ar^2, \dots, ar^{n-1}$ and $A, AR, AR^2, \dots, AR^{n-1}$ form a G.P., and find the common ratio.
Solution:
Given sequences:
- $a, ar, ar^2, \dots, ar^{n-1}$
- $A, AR, AR^2, \dots, AR^{n-1}$
The corresponding terms’ products are:
$$
aA, ar \cdot AR, ar^2 \cdot AR^2, \dots, ar^{n-1} \cdot AR^{n-1}
$$
That is:
$$
aA, aA rR, aA (rR)^2, \dots, aA (rR)^{n-1}
$$
This is clearly a G.P. with:
- First term $aA$
- Common ratio $rR$
To verify,
$$
\dfrac{arAR}{aA} = rR \quad \text{and} \quad \dfrac{ar^2AR^2}{arAR} = rR
$$
Hence, the products form a G.P. and the common ratio is $rR$.
Question 21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the fourth by 18.
Solution:
Let the first term be $a$ and the common ratio be $r$.
The four numbers in G.P. are:
$$a, ar, ar^2, ar^3$$
According to the question:
- The third term is greater than the first term by 9:
$$
ar^2 = a + 9
$$
$$
ar^2 – a = 9 \quad \text{…(1)}
$$ - The second term is greater than the fourth by 18:
$$
ar = ar^3 + 18
$$
$$
ar – ar^3 = 18 \quad \text{…(2)}
$$
Now divide equation (2) by equation (1):
$$
\frac{ar – ar^3}{ar^2 – a} = \frac{18}{9}
$$
Simplify:
$$
\frac{ar(1 – r^2)}{a(r^2 – 1)} = 2
$$
Since $(1 – r^2) = -(r^2 – 1)$,
$$
\frac{ar(-1)}{a} = 2
$$
$$
-r = 2
$$
$$
r = -2
$$
Now, put $r = -2$ in equation (1):
$$
a(-2)^2 – a = 9
$$
$$
4a – a = 9
$$
$$
3a = 9
$$
$$
a = 3
$$
Hence, the four numbers in G.P. are:
$$
3, 3(-2), 3(-2)^2, 3(-2)^3
$$
$$
= 3, -6, 12, -24
$$
Answer: The required numbers are $3, -6, 12, -24$.
Question 22. If the $p^{\text{th}}$, $q^{\text{th}}$, and $r^{\text{th}}$ terms of a G.P. are $a, b,$ and $c$ respectively, prove that $$a^{q-r} b^{r-p} c^{p-q} = 1.$$
Solution:
Let the first term of the G.P. be $k$ and the common ratio be $x$.
Then, according to the definition of the $n^{\text{th}}$ term:
$$
k x^{p-1} = a \quad \text{…(1)}
$$
$$
k x^{q-1} = b \quad \text{…(2)}
$$
$$
k x^{r-1} = c \quad \text{…(3)}
$$
We need to prove that:
$$
a^{q-r} b^{r-p} c^{p-q} = 1
$$
Proof:
Substitute the values of $a, b, c$ from equations (1), (2), and (3):
$$
a^{q-r} b^{r-p} c^{p-q} = (k x^{p-1})^{q-r} (k x^{q-1})^{r-p} (k x^{r-1})^{p-q}
$$
Simplify:
$$
= k^{(q-r)+(r-p)+(p-q)} x^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}
$$
Now, the powers of $k$ and $x$ simplify as:
$$
(q – r) + (r – p) + (p – q) = 0
$$
and
$$
(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) = 0
$$
Therefore:
$$
a^{q-r} b^{r-p} c^{p-q} = k^0 x^0 = 1
$$
Hence proved. ✅
Question 23. If the first and the $n^{\text{th}}$ term of a G.P. are $a$ and $b$, respectively, and if $P$ is the product of $n$ terms, prove that $$P^2 = (ab)^n.$$
Solution:
According to the question,
The first term of the G.P. is $a$ and the last term is $b$.
Let the G.P. be:
$$a, ar, ar^2, ar^3, \ldots, ar^{n-1}$$
where $r$ is the common ratio.
Then,
$$b = ar^{n-1}$$
Now, the product of $n$ terms is:
$$P = a \times ar \times ar^2 \times \ldots \times ar^{n-1}$$
Combine the powers of $a$ and $r$:
$$P = a^n r^{1 + 2 + 3 + \ldots + (n-1)}$$
We know that the sum of the first $(n-1)$ natural numbers is:
$$1 + 2 + 3 + \ldots + (n-1) = \frac{n(n-1)}{2}$$
Hence,
$$P = a^n , r^{\frac{n(n-1)}{2}}$$
Now,
$$P^2 = a^{2n} r^{n(n-1)}$$
But from $b = ar^{n-1}$
$$r^{n-1} = \frac{b}{a}$$
Substitute this value:
$$P^2 = a^{2n} \left(\frac{b}{a}\right)^n = (ab)^n$$
Hence Proved. ✅
Question 24. Show that the ratio of the sum of first $n$ terms of a G.P. to the sum of terms from $(n + 1)^{\text{th}}$ to $(2n)^{\text{th}}$ term is $\dfrac{1}{r^n}$.
Solution:
Let the first term of the G.P. be $a$ and the common ratio be $r$
Step 1: Write the formula for the sum of n terms of a G.P.
The sum of the first $n$ terms of a G.P. is given by
$$
S_{1n} = \dfrac{a(r^n – 1)}{r – 1}
$$
Step 2: Write the sum of the first 2n terms
Similarly, the sum of the first $2n$ terms is
$$
S_{2n} = \dfrac{a(r^{2n} – 1)}{r – 1}
$$
Step 3: Find the sum of terms from $(n + 1)^{\text{th}}$ to $(2n)^{\text{th}}$
The sum of terms from $(n + 1)^{\text{th}}$ to $(2n)^{\text{th}}$ is obtained by subtracting the sum of the first $n$ terms from the sum of the first $2n$ terms
$$
S_{(n+1) \text{ to } 2n} = S_{2n} – S_{1n}
$$
Substituting the values of $S_{2n}$ and $S_{1n}$, we get
$$
S_{(n+1) \text{ to } 2n} = \dfrac{a(r^{2n} – 1)}{r – 1} – \dfrac{a(r^n – 1)}{r – 1}
$$
Simplify:
$$
S_{(n+1) \text{ to } 2n} = \dfrac{a(r^{2n} – 1 – r^n + 1)}{r – 1}
$$
$$
S_{(n+1) \text{ to } 2n} = \dfrac{a(r^{2n} – r^n)}{r – 1}
$$
$$
S_{(n+1) \text{ to } 2n} = \dfrac{ar^n(r^n – 1)}{r – 1}
$$
Step 4: Find the required ratio
The required ratio is
$$
\dfrac{S_{1n}}{S_{(n+1) \text{ to } 2n}} = \dfrac{\dfrac{a(r^n – 1)}{r – 1}}{\dfrac{ar^n(r^n – 1)}{r – 1}}
$$
Simplify by canceling common terms:
$$
\dfrac{S_{1n}}{S_{(n+1) \text{ to } 2n}} = \dfrac{a(r^n – 1)}{r – 1} \times \dfrac{r – 1}{ar^n(r^n – 1)}
$$
$$
\dfrac{S_{1n}}{S_{(n+1) \text{ to } 2n}} = \dfrac{1}{r^n}
$$
✅ Hence proved that
$$
\dfrac{S_{1n}}{S_{(n+1) \text{ to } 2n}} = \dfrac{1}{r^n}
$$
Question 25 : If $a, b, c,$ and $d$ are in G.P., show that $$
(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$$
Solution
Let first term be $a$ and the common ratio of the G.P. be $r$. Then:
$$
b = ar, \quad c = ar^2, \quad d = ar^3
$$
Simplifying LHS:
$$
(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (a^2 + a^2 r^2 + a^2 r^4)(a^2 r^2 + a^2 r^4 + a^2 r^6)
$$
$$
= a^4 r^2 (1 + r^2 + r^4)^2
$$
Simplifying RHS:
$$
(ab + bc + cd)^2 = (a^2 r + a^2 r^3 + a^2 r^5)^2
$$
$$
= a^4 r^2 (1 + r^2 + r^4)^2
$$
Hence,
$$
(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$$
$$
\text{LHS} = \text{RHS}
$$
The relation is proved.
Question 26 : Insert two numbers between $3$ and $81$ so that the resulting sequence is G.P.
Solution
Let the two numbers be $x_1$ and $x_2$. Then the G.P. is:
$$
3, x_1, x_2, 81
$$
Let the common ratio be $r$. First term $a = 3$, fourth term $a_4 = 81$:
$$
a r^3 = 81
$$
Substitute $a = 3$:
$$
3 r^3 = 81
$$
$$
r^3 = 27
$$
$$
r = 3
$$
Now:
$$
x_1 = a r = 3 \cdot 3 = 9
$$
$$
x_2 = a r^2 = 3 \cdot 3^2 = 27
$$
Hence, the G.P. is:
$$
3, 9, 27, 81
$$
Question 27 : 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 $b$.
Solution
The geometric mean (G.M.) is:
$$
\sqrt{ab} = \frac{a^{n+1} + b^{n+1}}{a^n + b^n}
$$
Squaring both sides:
$$
ab = \frac{(a^{n+1} + b^{n+1})^2}{(a^n + b^n)^2}
$$
Multiply both sides by $(a^n + b^n)^2$:
$$
ab(a^n + b^n)^2 = (a^{n+1} + b^{n+1})^2
$$
Expand both sides:
$$
ab(a^{2n} + b^{2n} + 2 a^n b^n) = a^{2n+2} + b^{2n+2} + 2 a^{n+1} b^{n+1}
$$
$$
a^{2n+1} b +a b^{2n+1} + 2 a^{n+1} b^{n+1} = a^{2n+2} + b^{2n+2} + 2 a^{n+1} b^{n+1}
$$
Cancel $2 a^{n+1} b^{n+1}$ from both sides:
$$
a^{2n+1} b + a b^{2n+1} = a^{2n+2} + b^{2n+2}
$$
Rewriting:
$$
b^{2n+1}(a-b) = a^{2n+1}(a-b)
$$
If $a \ne b$, cancel $(a-b)$:
$$
b^{2n+1} = a^{2n+1}
$$
$$
\left(\frac{a}{b}\right)^{2n+1} = 1
$$
$$
\left(\frac{a}{b}\right)^{2n+1} = \left(\frac{a}{b}\right)^{0}
$$
$$
2n + 1 = 0
$$
$$
\boxed{n = -\frac{1}{2}}
$$
Question 28 : The sum of two numbers is 6 times their geometric mean. Show that the numbers are in the ratio $$(3 + 2\sqrt{2}) : (3 – 2\sqrt{2})$$
Solution
Let the two numbers be $a$ and $b$.
Geometric mean (G.M.) is:
$$
\text{G.M.} = \sqrt{ab}
$$
According to the question:
$$
a + b = 6 \sqrt{ab} \quad \text{(1)}
$$
Squaring both sides:
$$
(a + b)^2 = 36 ab
$$
We know:
$$
(a – b)^2 = (a + b)^2 – 4ab
$$
Substitute $(a + b)^2 = 36ab$:
$$
(a – b)^2 = 36 ab – 4 ab
$$
$$
(a – b)^2 = 32 ab
$$
$$
a – b = 4 \sqrt{2} \sqrt{ab} \quad \text{(2)}
$$
Adding equations (1) and (2):
$$
(a + b) + (a – b) = 6 \sqrt{ab} + 4 \sqrt{2} \sqrt{ab}
$$
$$
2a = \sqrt{ab} (6 + 4 \sqrt{2})
$$
$$
a = \sqrt{ab} (3 + 2 \sqrt{2})
$$
From (1), solve for $b$:
$$
\sqrt{ab} (3 + 2 \sqrt{2}) + b = 6 \sqrt{ab}
$$
$$
b = \sqrt{ab} (3 – 2 \sqrt{2})
$$
Hence, the ratio:
$$
\frac{a}{b} = \frac{\sqrt{ab} (3 + 2 \sqrt{2})}{\sqrt{ab} (3 – 2 \sqrt{2})} = \frac{3 + 2\sqrt{2}}{3 – 2\sqrt{2}}
$$
Question 29 : If $A$ and $G$ be the A.M. and G.M., respectively, between two positive numbers, prove that the numbers are $$ A \pm \sqrt{(A + G)(A – G)}$$
Solution
Let the numbers be $a$ and $b$.
$$
A = \frac{a + b}{2} \quad \Rightarrow \quad a + b = 2A \quad \text{(1)}
$$
Geometric mean:
$$
G = \sqrt{ab} \quad \Rightarrow \quad ab = G^2 \quad \text{(2)}
$$
We know:
$$
(a – b)^2 = (a + b)^2 – 4ab
$$
Substitute (1) and (2):
$$
(a – b)^2 = (2A)^2 – 4G^2
$$
$$
(a – b)^2 = 4(A^2 – G^2)
$$
$$
(a – b)^2 = 4 (A + G)(A – G)
$$
$$
a – b = 2 \sqrt{(A + G)(A – G)} \quad \text{(3)}
$$
Adding (1) and (3):
$$
(a + b) + (a – b) = 2A + 2 \sqrt{(A + G)(A – G)}
$$
$$
2a = 2A + 2 \sqrt{(A + G)(A – G)}
$$
$$
a = A + \sqrt{(A + G)(A – G)}
$$
From (1), solve for $b$:
$$
A + \sqrt{(A + G)(A – G)} + b = 2A
$$
$$
b = A – \sqrt{(A + G)(A – G)}
$$
Hence, the two numbers are:
$$
\boxed{A \pm \sqrt{(A + G)(A – G)}}
$$
Question 30 : The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and $n$th hour?
Solution
The bacteria count doubles every hour, so this forms a G.P. with:
$$
a = 30, \quad r = 2
$$
End of 2nd hour:
$$
a_3 = a r^2 = 30 \cdot 2^2 = 120
$$
So, at the end of 2nd hour, there are 120 bacteria.
End of 4th hour:
$$
a_5 = a r^4 = 30 \cdot 2^4 = 480
$$
So, at the end of 4th hour, there are 480 bacteria.
End of $n$th hour:
$$
a_{n+1} = a r^n = 30 \cdot 2^n
$$
So, at the end of $n$th hour, there are $30 \cdot 2^n$ bacteria.
Question 31 : What will Rs. 500 amount to in 10 years after deposit in a bank which pays an annual interest rate of 10% compounded annually?
Solution
Let the principal amount be $P = 500$ and
the annual interest rate r = 10% = 0.1.
The amount after each year is obtained by multiplying the previous year’s amount by $(1 + r)$. This forms a geometric progression (G.P.) with:
$$
a = 500, \quad r_{\text{G.P.}} = 1.1
$$
- End of 1st year:
$$
A_1 = 500 \cdot 1.1 = 550
$$
- End of 2nd year:
$$
A_2 = 500 \cdot 1.1 \cdot 1.1 = 500 \cdot 1.1^2 = 605
$$
- End of 3rd year:
$$
A_3 = 500 \cdot 1.1 \cdot 1.1 \cdot 1.1 = 500 \cdot 1.1^3 \approx 665.5
$$
- Continuing this pattern, the amount at the end of $n$ years is:
$$
A_n = 500 \cdot (1.1)^n
$$
- For $n = 10$ years:
$$
A_{10} = 500 \cdot (1.1)^{10}
$$
Now, calculating $(1.1)^{10}$ approximately:
$$
(1.1)^{10} \approx 2.5937
$$
$$
A_{10} \approx 500 \cdot 2.5937 = 1296.85
$$
Answer:
$$
\boxed{A_{10} \approx 1296.85 \text{ Rs.}}
$$
Question 32 : If the A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, obtain the quadratic equation.
Solution
Let the roots be $a$ and $b$.
Given:
$$
\text{A.M.} = \frac{a + b}{2} = 8 \quad \Rightarrow \quad a + b = 16
$$
$$
\text{G.M.} = \sqrt{ab} = 5 \quad \Rightarrow \quad ab = 25
$$
The quadratic equation with roots $a$ and $b$ is:
$$
x^2 – (a + b)x + ab = 0
$$
Substitute the values of $a + b$ and $ab$:
$$
x^2 – 16x + 25 = 0
$$
Hence, the required quadratic equation is:
$$
\boxed{x^2 – 16x + 25 = 0}
$$
