Exercise Chapter 8 - Question 1

Question: Write the first five terms of the sequence whose $n^{\text{th}}$ term is given by $a_n = n(n + 2)$.

Solution:
To find the first five terms of the sequence, we substitute $n = 1, 2, 3, 4, 5$ into the given formula $a_n = n(n + 2)$:

• For $n = 1$: $a_1 = 1(1 + 2) = 1(3) = 3$
• For $n = 2$: $a_2 = 2(2 + 2) = 2(4) = 8$
• For $n = 3$: $a_3 = 3(3 + 2) = 3(5) = 15$
• For $n = 4$: $a_4 = 4(4 + 2) = 4(6) = 24$
• For $n = 5$: $a_5 = 5(5 + 2) = 5(7) = 35$

Thus, the first five terms of the sequence are $3, 8, 15, 24,$ and $35$.

Exercise Chapter 8 - Question 2

Question: Write the first five terms of the sequence whose $n^{\text{th}}$ term is given by $a_n = \frac{n}{n + 1}$.

Solution:
To find the first five terms of the sequence, we substitute $n = 1, 2, 3, 4, 5$ into the given formula $a_n = \frac{n}{n + 1}$:

• For $n = 1$: $a_1 = \frac{1}{1 + 1} = \frac{1}{2}$
• For $n = 2$: $a_2 = \frac{2}{2 + 1} = \frac{2}{3}$
• For $n = 3$: $a_3 = \frac{3}{3 + 1} = \frac{3}{4}$
• For $n = 4$: $a_4 = \frac{4}{4 + 1} = \frac{4}{5}$
• For $n = 5$: $a_5 = \frac{5}{5 + 1} = \frac{5}{6}$

Thus, the first five terms of the sequence are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5},$ and $\frac{5}{6}$.

Exercise Chapter 8 - Question 3

Question: Write the first five terms of the sequence whose $n^{\text{th}}$ term is given by $a_n = 2^n$.

Solution:
To find the first five terms of the sequence, we substitute $n = 1, 2, 3, 4, 5$ into the given formula $a_n = 2^n$:

• For $n = 1$: $a_1 = 2^1 = 2$
• For $n = 2$: $a_2 = 2^2 = 4$
• For $n = 3$: $a_3 = 2^3 = 8$
• For $n = 4$: $a_4 = 2^4 = 16$
• For $n = 5$: $a_5 = 2^5 = 32$

Thus, the first five terms of the sequence are $2, 4, 8, 16,$ and $32$.

Exercise Chapter 8 - Question 4

Question: Find the terms of the sequence whose $n^{\text{th}}$ term is given by $a_n = \frac{2n - 3}{6}$.

Solution:

The given sequence is defined by the $n^{\text{th}}$ term:

$$a_n = \frac{2n - 3}{6}$$

To find terms in the sequence, substitute natural numbers for $n$. For example, the first few terms are:

• For $n = 1$: $a_1 = \frac{2(1) - 3}{6} = \frac{2 - 3}{6} = -\frac{1}{6}$
• For $n = 2$: $a_2 = \frac{2(2) - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6}$
• For $n = 3$: $a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}$

Exercise - Question 5

Question: Find the first few terms of the sequence given by $a_n = (-1)^{n-1} 5^{n+1}$.

Solution:

The given sequence is:

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

Let us find the first few terms by substituting $n = 1, 2, 3, \dots$:

• For $n = 1$: $a_1 = (-1)^{1-1} 5^{1+1} = (-1)^0 5^2 = 1 \times 25 = 25$
• For $n = 2$: $a_2 = (-1)^{2-1} 5^{2+1} = (-1)^1 5^3 = -1 \times 125 = -125$
• For $n = 3$: $a_3 = (-1)^{3-1} 5^{3+1} = (-1)^2 5^4 = 1 \times 625 = 625$

Answer: The first three terms are $25, -125, 625$.

Exercise - Question 6

Question: Find the first few terms of the sequence given by $a_n = n\frac{n^2 + 5}{4}$.

Solution:

The given sequence is:

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

Let us compute the first few terms by substituting $n = 1, 2, 3, \dots$:

• For $n = 1$: $a_1 = 1 \cdot \frac{1^2 + 5}{4} = 1 \cdot \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}$
• For $n = 2$: $a_2 = 2 \cdot \frac{2^2 + 5}{4} = 2 \cdot \frac{4 + 5}{4} = 2 \cdot \frac{9}{4} = \frac{9}{2}$
• For $n = 3$: $a_3 = 3 \cdot \frac{3^2 + 5}{4} = 3 \cdot \frac{9 + 5}{4} = 3 \cdot \frac{14}{4} = \frac{21}{2}$

Answer: The first three terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}$.

Exercise - Question 7

Question: Find the indicated terms in the sequence whose $n^{\text{th}}$ term is $a_n = 4n - 3$; find $a_{17}$ and $a_{24}$.

Solution:

The given $n^{\text{th}}$ term is:

$$a_n = 4n - 3$$

Step 1: Find $a_{17}$

Substitute $n = 17$ into the formula:

$$a_{17} = 4(17) - 3 = 68 - 3 = 65$$

Step 2: Find $a_{24}$

Substitute $n = 24$ into the formula:

$$a_{24} = 4(24) - 3 = 96 - 3 = 93$$

Answer: $a_{17} = 65$ and $a_{24} = 93$.

Exercise - Question 8

Question: Given $a_n = \frac{n^2}{2^n}$; find $a_7$.

Solution:

Substitute $n = 7$ into the given expression for $a_n$:

$$a_7 = \frac{7^2}{2^7}$$ $$a_7 = \frac{49}{128}$$

Answer: $a_7 = \frac{49}{128}$

Exercise - Question 9

Question: Given $a_n = (-1)^{n-1} n^3$; find $a_9$.

Solution:

Substitute $n = 9$ into the given expression for $a_n$:

$$a_9 = (-1)^{9-1} (9)^3$$ $$a_9 = (-1)^8 \cdot 729$$ Since $(-1)^8 = 1$: $$a_9 = 1 \cdot 729 = 729$$

Answer: $a_9 = 729$

Exercise - Question 10

Question: Given $a_n = \frac{n(n - 2)}{n + 3}$; find $a_{20}$.

Solution:

Substitute $n = 20$ into the given expression for $a_n$:

$$a_{20} = \frac{20(20 - 2)}{20 + 3}$$ $$a_{20} = \frac{20(18)}{23}$$ $$a_{20} = \frac{360}{23}$$

Answer: $a_{20} = \frac{360}{23}$

Exercise - Question 11

Question: Write the first five terms of the sequence given by $a_1 = 3$, $a_n = 3a_{n-1} + 2$ for all $n > 1$, and obtain the corresponding series.

Solution:

Given the first term:

$$a_1 = 3$$

Now, compute the next terms using the recursive relation $a_n = 3a_{n-1} + 2$:

• For $n = 2$: $$a_2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11$$
• For $n = 3$: $$a_3 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35$$
• For $n = 4$: $$a_4 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107$$
• For $n = 5$: $$a_5 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323$$

Thus, the first five terms of the sequence are: $3, 11, 35, 107, 323$.

The corresponding series is obtained by adding the terms of the sequence:

$$3 + 11 + 35 + 107 + 323 + \dots$$

Answer: First five terms are $3, 11, 35, 107, 323$ and the corresponding series is $3 + 11 + 35 + 107 + 323 + \dots$

Exercise - Question 12

Question: Given $a_1 = -1$ and $a_n = \frac{a_{n-1}}{n}$ for $n \ge 2$, find the first five terms of the sequence.

Solution:
Given the recursive relation: $$a_1 = -1$$ $$a_n = \frac{a_{n-1}}{n} \quad \text{for } n \ge 2$$ Let us find the first few terms by substituting values of $n$:

• For $n = 2$: $$a_2 = \frac{a_1}{2} = \frac{-1}{2}$$
• For $n = 3$: $$a_3 = \frac{a_2}{3} = \frac{\frac{-1}{2}}{3} = -\frac{1}{6}$$
• For $n = 4$: $$a_4 = \frac{a_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24}$$
• For $n = 5$: $$a_5 = \frac{a_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120}$$

Thus, the first five terms of the sequence are $-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}$.

Answer: $-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}$

Sequence Problem - Question 13

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

Solution:
Given the recursive relation: $$a_1 = 2, \quad a_2 = 2$$ $$a_n = a_{n-1} - 1 \quad \text{for } n > 2$$ Let us find the subsequent terms:

• For $n = 3$: $$a_3 = a_2 - 1 = 2 - 1 = 1$$
• For $n = 4$: $$a_4 = a_3 - 1 = 1 - 1 = 0$$
• For $n = 5$: $$a_5 = a_4 - 1 = 0 - 1 = -1$$
• For $n = 6$: $$a_6 = a_5 - 1 = -1 - 1 = -2$$

Thus, the terms of the sequence are $2, 2, 1, 0, -1, -2, \dots$

Sequence Problem - Question 14

Question: The Fibonacci sequence is defined by $1 = a_1 = a_2$ and $a_n = a_{n-1} + a_{n-2}$, $n > 2$. Find $\frac{a_{n+1}}{a_n}$, for $n = 1, 2, 3, 4, 5$.

Solution:
First, let's write down the Fibonacci sequence terms using $a_1 = 1$, $a_2 = 1$, and $a_n = a_{n-1} + a_{n-2}$:

• $a_1 = 1$
• $a_2 = 1$
• $a_3 = a_2 + a_1 = 1 + 1 = 2$
• $a_4 = a_3 + a_2 = 2 + 1 = 3$
• $a_5 = a_4 + a_3 = 3 + 2 = 5$
• $a_6 = a_5 + a_4 = 5 + 3 = 8$

Now, we calculate the ratio $\frac{a_{n+1}}{a_n}$ for $n = 1, 2, 3, 4, 5$:

• For $n = 1$: $$\frac{a_2}{a_1} = \frac{1}{1} = 1$$
• For $n = 2$: $$\frac{a_3}{a_2} = \frac{2}{1} = 2$$
• For $n = 3$: $$\frac{a_4}{a_3} = \frac{3}{2} = 1.5$$
• For $n = 4$: $$\frac{a_5}{a_4} = \frac{5}{3}$$
• For $n = 5$: $$\frac{a_6}{a_5} = \frac{8}{5} = 1.6$$

Exercise 8.2 - Question 1

Question: Find the $20^{\text{th}}$ and $n^{\text{th}}$ terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

Solution:
Given Geometric Progression (G.P.): $$\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$$ Here, the first term $a = \frac{5}{2}$ and the common ratio $r$ is: $$r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2}$$ The $n^{\text{th}}$ term of a G.P. is given by the formula: $$a_n = a r^{n-1}$$ Substituting $a = \frac{5}{2}$ and $r = \frac{1}{2}$: $$a_n = \left(\frac{5}{2}\right) \left(\frac{1}{2}\right)^{n-1} = \frac{5}{2 \cdot 2^{n-1}} = \frac{5}{2^n}$$ To find the $20^{\text{th}}$ term ($a_{20}$): $$a_{20} = \frac{5}{2^{20}}$$ Thus, the $n^{\text{th}}$ term is $\frac{5}{2^n}$ and the $20^{\text{th}}$ term is $\frac{5}{2^{20}}$.

Exercise 8.2 - Question 2

Question: Find the $12^{\text{th}}$ term of a G.P. whose $8^{\text{th}}$ term is $192$ and the common ratio is $2$.

Solution:

Let the first term of the Geometric Progression (G.P.) be $a$ and the common ratio be $r$.

The general term ($n^{\text{th}}$ term) of a G.P. is given by the formula: $$a_n = a r^{n-1}$$

We are given:

• Common ratio, $r = 2$
• $8^{\text{th}}$ term, $a_8 = 192$

Using the formula for $a_8$: $$a_8 = a r^{8-1} = a r^7$$ $$192 = a (2)^7$$ $$192 = a \times 128$$ $$a = \frac{192}{128} = \frac{3}{2}$$

Now, we need to find the $12^{\text{th}}$ term ($a_{12}$): $$a_{12} = a r^{12-1} = a r^{11}$$ Substitute $a = \frac{3}{2}$ and $r = 2$: $$a_{12} = \left(\frac{3}{2}\right) (2)^{11}$$ $$a_{12} = 3 \times 2^{10} = 3 \times 1024 = 3072$$

Answer: The $12^{\text{th}}$ term of the G.P. is $3072$.

Exercise - Question 3

Question: The $5^{\text{th}}$, $8^{\text{th}}$ and $11^{\text{th}}$ terms of a G.P. are $p$, $q$ and $s$, respectively. Show that $q^2 = ps$.

Solution:

Let $a$ be the first term and $r$ be the common ratio of the given G.P.

Using the $n^{\text{th}}$ term formula $a_n = a r^{n-1}$, we have:

• $5^{\text{th}}$ term: $p = a_5 = a r^{5-1} = a r^4$    --- (Equation 1)
• $8^{\text{th}}$ term: $q = a_8 = a r^{8-1} = a r^7$    --- (Equation 2)
• $11^{\text{th}}$ term: $s = a_{11} = a r^{11-1} = a r^{10}$    --- (Equation 3)

Now, consider the right-hand side (R.H.S.) of the relation to be proved, which is $ps$:

$$ps = (a r^4) (a r^{10}) = a^2 r^{4+10} = a^2 r^{14}$$

Next, consider the left-hand side (L.H.S.) of the relation, which is $q^2$:

$$q^2 = (a r^7)^2 = a^2 (r^7)^2 = a^2 r^{14}$$

Since L.H.S. = R.H.S. ($q^2 = a^2 r^{14}$ and $ps = a^2 r^{14}$), we conclude that: $$q^2 = ps$$ Hence proved.

Exercise - Question 4

Question: The $4^{\text{th}}$ term of a G.P. is square of its second term, and the first term is $-3$. Determine its $7^{\text{th}}$ term.

Solution:

Let $a$ be the first term and $r$ be the common ratio of the G.P.

We are given:

• First term, $a = -3$
• $4^{\text{th}}$ term is the square of its second term: $a_4 = (a_2)^2$

Using the formula $a_n = a r^{n-1}$, we express $a_4$ and $a_2$ in terms of $a$ and $r$:

• $a_2 = a r^{2-1} = ar$
• $a_4 = a r^{4-1} = ar^3$

According to the given condition: $$ar^3 = (ar)^2$$ $$ar^3 = a^2 r^2$$

Since $a = -3 \neq 0$ and assuming $r \neq 0$, we can divide both sides by $ar^2$: $$r = a$$ Therefore, the common ratio $r = -3$.

Now, we need to find the $7^{\text{th}}$ term ($a_7$): $$a_7 = a r^{7-1} = a r^6$$ Substitute $a = -3$ and $r = -3$: $$a_7 = (-3)(-3)^6 = (-3)^1 (-3)^6 = (-3)^{1+6} = (-3)^7$$ $$(-3)^7 = -2187$$

Answer: The $7^{\text{th}}$ term of the G.P. is $-2187$.

Exercise - Question 5

Question: Which term of the following sequences:
(a) $2, 2\sqrt{2}, 4, \dots$ is $128$?
(b) $\sqrt{3}, 3, 3\sqrt{3}, \dots$ is $729$?
(c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ is $\frac{1}{19683}$?

Solution:

(a) Sequence: $2, 2\sqrt{2}, 4, \dots$

First term, $a = 2$
Common ratio, $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Let $128$ be the $n^{\text{th}}$ term of this sequence. Using $a_n = a r^{n-1}$: $$2 (\sqrt{2})^{n-1} = 128$$ $$(\sqrt{2})^{n-1} = \frac{128}{2} = 64$$ Since $\sqrt{2} = 2^{1/2}$, we can write: $$(2^{1/2})^{n-1} = 2^6$$ $$2^{\frac{n-1}{2}} = 2^6$$

Equating the exponents on both sides: $$\frac{n-1}{2} = 6$$ $$n - 1 = 12 \implies n = 13$$

Answer (a): $128$ is the $13^{\text{th}}$ term.

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(b) Sequence: $\sqrt{3}, 3, 3\sqrt{3}, \dots$

First term, $a = \sqrt{3}$
Common ratio, $r = \frac{3}{\sqrt{3}} = \sqrt{3}$

Let $729$ be the $n^{\text{th}}$ term. Using $a_n = a r^{n-1}$: $$\sqrt{3} (\sqrt{3})^{n-1} = 729$$ $$(\sqrt{3})^n = 729$$ Converting $729$ in terms of base $\sqrt{3}$:

• $3 = (\sqrt{3})^2$
• $729 = 3^6 = ((\sqrt{3})^2)^6 = (\sqrt{3})^{12}$

Therefore: $$(\sqrt{3})^n = (\sqrt{3})^{12} \implies n = 12$$

Answer (b): $729$ is the $12^{\text{th}}$ term.

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(c) Sequence: $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$

First term, $a = \frac{1}{3}$
Common ratio, $r = \frac{1/9}{1/3} = \frac{1}{3}$

Let $\frac{1}{19683}$ be the $n^{\text{th}}$ term. Using $a_n = a r^{n-1}$: $$\left(\frac{1}{3}\right) \left(\frac{1}{3}\right)^{n-1} = \frac{1}{19683}$$ $$\left(\frac{1}{3}\right)^n = \frac{1}{19683}$$

We know that $3^9 = 19683$, so: $$\left(\frac{1}{3}\right)^n = \left(\frac{1}{3}\right)^9 \implies n = 9$$

Answer (c): $\frac{1}{19683}$ is the $9^{\text{th}}$ term.

Exercise - Question 6

Question: For what values of $x$, the numbers $-\frac{2}{7}, x, -\frac{7}{2}$ are in G.P.?

Solution:

If three numbers $a, b, c$ are in Geometric Progression (G.P.), then the square of the middle term is equal to the product of the first and the third terms (i.e., $b^2 = ac$).

Here, the numbers are $-\frac{2}{7}$, $x$, and $-\frac{7}{2}$. Therefore,

$$x^2 = \left(-\frac{2}{7}\right) \cdot \left(-\frac{7}{2}\right)$$ $$x^2 = 1$$

Taking the square root on both sides, we get:

$$x = \pm 1$$

Answer: The values of $x$ are $1$ and $-1$.

Exercise - Question 7

Question: Find the sum to $20$ terms of the geometric progression: $0.15, 0.015, 0.0015, \dots$

Solution:

Given G.P. is $0.15, 0.015, 0.0015, \dots$

• First term, $a = 0.15 = \frac{15}{100} = \frac{3}{20}$
• Common ratio, $r = \frac{0.015}{0.15} = 0.1 = \frac{1}{10} < 1$
• Number of terms, $n = 20$

The sum of $n$ terms of a G.P. when $r < 1$ is given by the formula:

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

Substituting the values for $S_{20}$:

$$S_{20} = \frac{0.15 \left(1 - (0.1)^{20}\right)}{1 - 0.1}$$ $$S_{20} = \frac{0.15 \left(1 - (0.1)^{20}\right)}{0.9}$$

Simplify the fraction $\frac{0.15}{0.9}$:

$$\frac{0.15}{0.9} = \frac{15}{90} = \frac{1}{6}$$

Thus, the sum is:

$$S_{20} = \frac{1}{6} \left(1 - (0.1)^{20}\right)$$

Exercise - Question 8

Question: Find the sum to $n$ terms of the geometric progression: $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots$

Solution:

Given G.P. is $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots$

• First term, $a = \sqrt{7}$
• Common ratio, $r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}} = \sqrt{3} > 1$

Since $r > 1$, the sum of $n$ terms is given by:

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

Substituting the values of $a$ and $r$:

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

Rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{3} + 1)$:

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

Exercise - Question 9

Question: Find the sum to $n$ terms of the geometric progression: $1, -a, a^2, -a^3, \dots$ (if $a \neq -1$).

Solution:

Given G.P. is $1, -a, a^2, -a^3, \dots$

• First term, $a_{\text{first}} = 1$
• Common ratio, $r = \frac{-a}{1} = -a$
• Number of terms = $n$

The sum of $n$ terms of a G.P. is given by:

$$S_n = \frac{a_{\text{first}}(1 - r^n)}{1 - r}$$

Substituting $a_{\text{first}} = 1$ and $r = -a$:

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

Exercise - Question 10

Question: Find the sum to $n$ terms of the geometric progression: $x^3, x^5, x^7, \dots$ (if $x \neq \pm 1$).

Solution:

Given the series: $x^3, x^5, x^7, \dots$ up to $n$ terms.

This is a geometric progression (G.P.) where:

• First term, $a = x^3$
• Common ratio, $r = \frac{x^5}{x^3} = x^2$

Since $x \neq \pm 1$, $r = x^2 \neq 1$. The sum of $n$ terms of a G.P. is given by the formula:

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

Substituting the values of $a$ and $r$:

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

Exercise - Question 11

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

Solution:

We need to evaluate the summation:

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

Evaluating the first sum (constant term repeated 11 times):

$$\sum_{k=1}^{11} 2 = 2 \times 11 = 22$$

Evaluating the second sum (a geometric series with first term $3^1 = 3$, common ratio $3$, and $11$ terms):

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

Combining both parts:

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

Exercise - Question 12

Question: 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.

Solution:

Let the first three terms of the G.P. be $\frac{a}{r}$, $a$, and $ar$.

Step 1: Use the product condition

$$\left(\frac{a}{r}\right) \cdot (a) \cdot (ar) = 1$$ $$a^3 = 1 \implies a = 1$$

Step 2: Use the sum condition

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

Substitute $a = 1$:

$$\frac{1}{r} + 1 + r = \frac{39}{10}$$ $$\frac{1 + r + r^2}{r} = \frac{39}{10}$$ $$10(1 + r + r^2) = 39r$$ $$10 + 10r + 10r^2 = 39r$$ $$10r^2 - 29r + 10 = 0$$

Step 3: Solve the quadratic equation for $r$

$$10r^2 - 25r - 4r + 10 = 0$$ $$5r(2r - 5) - 2(2r - 5) = 0$$ $$(5r - 2)(2r - 5) = 0$$

Thus, $r = \frac{2}{5}$ or $r = \frac{5}{2}$.

Step 4: Find the terms

• If $a = 1$ and $r = \frac{2}{5}$, the terms are: $\frac{1}{2/5}, 1, 1\left(\frac{2}{5}\right) \implies \frac{5}{2}, 1, \frac{2}{5}$.
• If $a = 1$ and $r = \frac{5}{2}$, the terms are: $\frac{2}{5}, 1, \frac{5}{2}$.

In both cases, the terms are $\frac{2}{5}, 1, \frac{5}{2}$ and the common ratios are $\frac{2}{5}$ or $\frac{5}{2}$.

Exercise - Question 13

Question: How many terms of G.P. $3, 3^2, 3^3, \dots$ are needed to give the sum $120$?

Solution:

Given the G.P.: $3, 3^2, 3^3, \dots$

• First term, $a = 3$
• Common ratio, $r = 3$

Let $n$ be the number of terms required to give a sum of $120$. Using the sum formula:

$$S_n = \frac{a(r^n - 1)}{r - 1} = 120$$ $$\frac{3(3^n - 1)}{3 - 1} = 120$$ $$\frac{3(3^n - 1)}{2} = 120$$ $$3(3^n - 1) = 240$$ $$3^n - 1 = 80$$ $$3^n = 81$$

Since $81 = 3^4$:

$$3^n = 3^4 \implies n = 4$$

Answer: $4$ terms are needed to give the sum $120$.

Exercise - Question 14

Question: 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 term of the G.P. be $a$ and the common ratio be $r$.

The sum of the first three terms ($S_3$) is given by: $$S_3 = a + ar + ar^2 = 16 \implies a(1 + r + r^2) = 16 \quad \text{--- (1)}$$

The next three terms are the $4^{\text{th}}$, $5^{\text{th}}$, and $6^{\text{th}}$ terms: $$ar^3 + ar^4 + ar^5 = 128 \implies ar^3(1 + r + r^2) = 128 \quad \text{--- (2)}$$

Dividing equation (2) by equation (1), we get: $$\frac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \frac{128}{16}$$ $$r^3 = 8 \implies r = 2$$

Substitute $r = 2$ into equation (1): $$a(1 + 2 + 2^2) = 16 \implies a(1 + 2 + 4) = 16 \implies 7a = 16 \implies a = \frac{16}{7}$$

Now, the sum to $n$ terms ($S_n$) of a G.P. when $r > 1$ is given by: $$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{\frac{16}{7}(2^n - 1)}{2 - 1} = \frac{16}{7}(2^n - 1)$$

Final Answer: First term $a = \frac{16}{7}$, common ratio $r = 2$, and sum to $n$ terms $S_n = \frac{16}{7}(2^n - 1)$.

Exercise - Question 15

Question: Given a G.P. with $a = 729$ and $7^{\text{th}}$ term $64$, determine $S_7$.

Solution:

We are given the first term $a = 729$.

The $n^{\text{th}}$ term of a G.P. is given by $a_n = ar^{n-1}$.

Therefore, the $7^{\text{th}}$ term is: $$a_7 = ar^{6} = 64$$

Substitute $a = 729$: $$729 r^6 = 64 \implies r^6 = \frac{64}{729} = \left(\frac{2}{3}\right)^6 \implies r = \frac{2}{3}$$

Since $r < 1$, the sum of the first 7 terms ($S_7$) is given by: $$S_7 = \frac{a(1 - r^7)}{1 - r} = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}}$$

$$S_7 = \frac{729 \left(1 - \frac{128}{2187}\right)}{\frac{1}{3}} = 729 \times 3 \times \left(\frac{2187 - 128}{2187}\right)$$

Since $729 \times 3 = 2187$, we can cancel it out: $$S_7 = 2187 \times \left(\frac{2059}{2187}\right) = 2059$$

Final Answer: $S_7 = 2059$.

Exercise - Question 16

Question: Find a G.P. for which 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 the common ratio be $r$.

Given that the sum of the first two terms is $-4$: $$a + ar = -4 \implies a(1 + r) = -4 \quad \text{--- (1)}$$

Also, the fifth term is $4$ times the third term: $$a_5 = 4 a_3 \implies ar^4 = 4 ar^2$$

Since $a \neq 0$, dividing by $ar^2$: $$r^2 = 4 \implies r = 2 \quad \text{or} \quad r = -2$$

Case 1: When $r = 2$

Substitute $r = 2$ into equation (1): $$a(1 + 2) = -4 \implies 3a = -4 \implies a = -\frac{4}{3}$$

The G.P. is: $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$

Case 2: When $r = -2$

Substitute $r = -2$ into equation (1): $$a(1 - 2) = -4 \implies -a = -4 \implies a = 4$$

The G.P. is: $4, -8, 16, -32, \dots$

Final Answer: The possible G.P.s are $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$ or $4, -8, 16, -32, \dots$.

Exercise - Question 17

Question: If the $4^{\text{th}}$, $10^{\text{th}}$ and $16^{\text{th}}$ 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 given information: $$x = \text{4}^{\text{th}}\text{ term} = AR^3$$ $$y = \text{10}^{\text{th}}\text{ term} = AR^9$$ $$z = \text{16}^{\text{th}}\text{ term} = AR^{15}$$

To prove that $x, y, z$ are in G.P., we need to show that $y^2 = xz$.

Consider the LHS ($y^2$): $$y^2 = (AR^9)^2 = A^2 R^{18}$$

Consider the RHS ($xz$): $$xz = (AR^3)(AR^{15}) = A^2 R^{3 + 15} = A^2 R^{18}$$

Since $\text{LHS} = \text{RHS}$ ($y^2 = xz$), the terms $x, y, z$ satisfy the condition of a Geometric Progression.

Final Answer: Hence proved that $x, y, z$ are in G.P.

Exercise - Question 18

Question: Find the sum to $n$ terms of the sequence, $8, 88, 888, 8888\dots$.

Solution:

Let $S_n$ denote the sum of the first $n$ terms of the given sequence.

$$S_n = 8 + 88 + 888 + 8888 + \dots + n \text{ terms}$$

Factor out $8$ from each term:

$$S_n = 8 (1 + 11 + 111 + 1111 + \dots + n \text{ terms})$$

Multiply and divide the expression inside the bracket by $9$:

$$S_n = \frac{8}{9} (9 + 99 + 999 + 9999 + \dots + n \text{ terms})$$

Rewrite each term as a power of $10$ minus $1$:

$$S_n = \frac{8}{9} \left[ (10 - 1) + (10^2 - 1) + (10^3 - 1) + (10^4 - 1) + \dots + n \text{ terms} \right]$$

Group the powers of $10$ together and the $-1$ terms together:

$$S_n = \frac{8}{9} \left[ (10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times}) \right]$$

The first part forms a geometric progression with first term $a = 10$ and common ratio $r = 10$. The sum of $n$ terms of this G.P. is $\frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$. The second part is simply $n$. Thus:

$$S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$$ $$S_n = \frac{8}{81} [10(10^n - 1) - 9n]$$

Final Answer: $S_n = \frac{8}{81} [10(10^n - 1) - 9n]$.

Exercise - Question 19

Question: Find the sum of the products of the corresponding terms of the sequences $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \frac{1}{2}$.

Solution:

The given sequences are:

Sequence 1: $2, 4, 8, 16, 32$

Sequence 2: $128, 32, 8, 2, \frac{1}{2}$

The products of the corresponding terms are:

$$P_1 = 2 \times 128 = 256$$ $$P_2 = 4 \times 32 = 128$$ $$P_3 = 8 \times 8 = 64$$ $$P_4 = 16 \times 2 = 32$$ $$P_5 = 32 \times \frac{1}{2} = 16$$

We need to find the sum of these products:

$$S = 256 + 128 + 64 + 32 + 16$$

Observe that these terms form a geometric progression with first term $a = 256$, common ratio $r = \frac{128}{256} = \frac{1}{2}$, and number of terms $n = 5$.

Using the sum formula for a G.P., $S_n = \frac{a(1 - r^n)}{1 - r}$:

$$S = \frac{256 \left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}} = \frac{256 \left(1 - \frac{1}{32}\right)}{\frac{1}{2}}$$ $$S = 2 \times 256 \times \left(\frac{31}{32}\right) = 512 \times \frac{31}{32} = 16 \times 31 = 496$$

Answer: $496$

Exercise - Question 20

Question: 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:

The given sequences are:

Sequence 1: $a_k = ar^{k-1}$ for $k = 1, 2, \dots, n$

Sequence 2: $A_k = AR^{k-1}$ for $k = 1, 2, \dots, n$

The product of the corresponding terms, let's call it $T_k$, is:

$$T_k = a_k \cdot A_k = (ar^{k-1})(AR^{k-1}) = (aA)(rR)^{k-1}$$

Writing out the first few terms of the new sequence:

• $T_1 = aA$
• $T_2 = (aA)(rR)$
• $T_3 = (aA)(rR)^2$
• $\vdots$
• $T_n = (aA)(rR)^{n-1}$

To check if this sequence forms a G.P., we find the ratio of any term to its preceding term:

$$\frac{T_k}{T_{k-1}} = \frac{(aA)(rR)^{k-1}}{(aA)(rR)^{k-2}} = rR$$

Since the ratio $\frac{T_k}{T_{k-1}} = rR$ is constant for all $k \ge 2$, the products of the corresponding terms form a geometric progression.

Common Ratio: $rR$

Exercise - Question 21

Question: 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 $4^{\text{th}}$ term by $18$.

Solution:

Let the four numbers of the G.P. be $a, ar, ar^2, ar^3$.

According to the given conditions:

1. The third term is greater than the first term by $9$:

$$ar^2 = a + 9 \implies a(r^2 - 1) = 9 \quad \text{--- (1)}$$

2. The second term is greater than the fourth term by $18$:

$$ar = ar^3 + 18 \implies ar(1 - r^2) = 18 \quad \text{--- (2)}$$

Divide equation (2) by equation (1):

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

Substitute $r = -2$ into equation (1):

$$a((-2)^2 - 1) = 9$$ $$a(4 - 1) = 9 \implies 3a = 9 \implies a = 3$$

Now, we find the four numbers:

• First term: $a = 3$
• Second term: $ar = 3(-2) = -6$
• Third term: $ar^2 = 3(-2)^2 = 12$
• Fourth term: $ar^3 = 3(-2)^3 = -24$

Answer: The four numbers are $3, -6, 12, -24$.

Exercise - Question 22

Question: 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 $A$ and the common ratio be $R$. The $n^{\text{th}}$ term of a G.P. is given by $T_n = A R^{n-1}$.

According to the given condition: $$p^{\text{th}} \text{ term: } A R^{p-1} = a \quad \text{--- (1)}$$ $$q^{\text{th}} \text{ term: } A R^{q-1} = b \quad \text{--- (2)}$$ $$r^{\text{th}} \text{ term: } A R^{r-1} = c \quad \text{--- (3)}$$
We need to evaluate the expression $a^{q-r} b^{r-p} c^{p-q}$. Substituting the values of $a$, $b$, and $c$ from (1), (2), and (3): $$a^{q-r} b^{r-p} c^{p-q} = (A R^{p-1})^{q-r} (A R^{q-1})^{r-p} (A R^{r-1})^{p-q}$$
Separating the terms of $A$ and $R$: $$= A^{(q-r) + (r-p) + (p-q)} \times R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)}$$
Simplifying the exponent of $A$: $$(q - r + r - p + p - q) = 0$$ So, $A^0 = 1$.

Now, simplifying the exponent of $R$: $$(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)$$ $$= (pq - pr - q + r) + (qr - pq - r + p) + (rp - rq - p + q)$$ $$= pq - pr - q + r + qr - pq - r + p + rp - rq - p + q = 0$$ So, $R^0 = 1$.

Therefore, $$a^{q-r} b^{r-p} c^{p-q} = 1 \times 1 = 1.$$ Hence proved.

Exercise - Question 23

Question: 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: Let the common ratio of the G.P. be $r$. The terms of the G.P. are: $$a, ar, ar^2, ar^3, \dots, ar^{n-1}$$
Given that the first term is $a$ and the $n^{\text{th}}$ term is $b$: $$b = ar^{n-1} \quad \text{--- (1)}$$
$P$ is the product of $n$ terms: $$P = (a) \cdot (ar) \cdot (ar^2) \cdot (ar^3) \cdots (ar^{n-1})$$ $$P = a^n \cdot (r^{1 + 2 + 3 + \dots + (n-1)})$$
The sum of the arithmetic progression in the exponent of $r$ is: $$1 + 2 + 3 + \dots + (n-1) = \frac{n(n-1)}{2}$$ Therefore, $$P = a^n r^{\frac{n(n-1)}{2}}$$
Squaring both sides to find $P^2$: $$P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 = a^{2n} r^{n(n-1)}$$ $$P^2 = (a^2)(r^{n-1})^n = (a \cdot ar^{n-1})^n$$
From equation (1), the $n^{\text{th}}$ term is $b = ar^{n-1}$. Substituting this: $$P^2 = (a \cdot b)^n = (ab)^n$$ Hence proved.

Exercise - Question 24

Question: 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 $\frac{1}{r^n}$.

Solution: Let the first term of the G.P. be $a$ and the common ratio be $r$.

1. The sum of the first $n$ terms ($S_n$) is given by: $$S_n = \frac{a(r^n - 1)}{r - 1}$$
2. The sum of terms from $(n+1)^{\text{th}}$ to $(2n)^{\text{th}}$ term is: $$S' = T_{n+1} + T_{n+2} + T_{n+3} + \dots + T_{2n}$$ This is also a G.P. with $n$ terms, starting with $T_{n+1} = ar^n$ and common ratio $r$: $$S' = \frac{ar^n(r^n - 1)}{r - 1}$$
3. We need to find the ratio of $S_n$ to $S'$: $$\text{Ratio} = \frac{S_n}{S'} = \frac{\frac{a(r^n - 1)}{r - 1}}{\frac{ar^n(r^n - 1)}{r - 1}}$$
Canceling out the common terms $\frac{a(r^n - 1)}{r - 1}$: $$\text{Ratio} = \frac{1}{r^n}$$ Hence proved.

Exercise - Question 25

Question: 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: Since $a, b, c, d$ are in G.P., let the common ratio be $r$. Then: $$b = ar, \quad c = ar^2, \quad d = ar^3$$
L.H.S.: $(a^2 + b^2 + c^2)(b^2 + c^2 + d^2)$
Substitute $b, c,$ and $d$ in terms of $a$ and $r$: $$L.H.S. = (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6)$$ Take $a^2$ common from the first bracket and $a^2r^2$ from the second bracket: $$L.H.S. = a^2(1 + r^2 + r^4) \cdot a^2r^2(1 + r^2 + r^4)$$ $$L.H.S. = a^4 r^2 (1 + r^2 + r^4)^2 \quad \text{--- (1)}$$
R.H.S.: $(ab + bc + cd)^2$
Substitute $b, c,$ and $d$: $$R.H.S. = (a(ar) + (ar)(ar^2) + (ar^2)(ar^3))^2$$ $$R.H.S. = (a^2r + a^2r^3 + a^2r^5)^2$$ Take $a^2r$ common from inside the parenthesis: $$R.H.S. = [a^2r(1 + r^2 + r^4)]^2$$ $$R.H.S. = a^4 r^2 (1 + r^2 + r^4)^2 \quad \text{--- (2)}$$
From (1) and (2), we have: $$L.H.S. = R.H.S.$$ Therefore, $(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$. Hence proved.

Exercise - Question 26

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

Solution:

Let the two numbers be $G_1$ and $G_2$ such that the sequence $3, G_1, G_2, 81$ forms a geometric progression (G.P.).

Let the first term of the G.P. be $a = 3$ and the common ratio be $r$.

Since there are 4 terms in total, the 4th term ($a_4$) is given by:

$$a_4 = ar^3 = 81$$

Substitute $a = 3$ into the equation:

$$3r^3 = 81$$ $$r^3 = \frac{81}{3} = 27$$ $$r = 3$$

Now, we can find the inserted numbers $G_1$ and $G_2$:

$$G_1 = ar = 3 \times 3 = 9$$ $$G_2 = ar^2 = 3 \times (3)^2 = 27$$

Answer: The two numbers are $9$ and $27$.

Exercise Chapter 9 - Question 27

Question: 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.) between two numbers $a$ and $b$ is given by $\sqrt{ab}$ or $(ab)^{1/2}$.

According to the given condition, we have:

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

Cross-multiplying gives:

$$a^{n+1} + b^{n+1} = (a^n + b^n)a^{1/2}b^{1/2}$$ $$a^{n+1} + b^{n+1} = a^{n + 1/2}b^{1/2} + a^{1/2}b^{n + 1/2}$$

Rearranging the terms to group terms containing $a$ on one side and terms containing $b$ on the other side:

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

Factor out $a^{n + 1/2}$ on the left-hand side and $b^{n + 1/2}$ on the right-hand side:

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

Assuming $a \neq b$, we can divide both sides by $(a^{1/2} - b^{1/2})$:

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

Equating the exponents:

$$n + \frac{1}{2} = 0 \implies n = -\frac{1}{2}$$

Answer: $n = -\frac{1}{2}$

Exercise Chapter 9 - Question 28

Question: The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$.

Solution:

Let the two positive numbers be $a$ and $b$.

Their sum is $a + b$ and their geometric mean is $\sqrt{ab}$. Given that:

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

Dividing the equation by $b$:

$$\frac{a}{b} + 1 = 6\sqrt{\frac{a}{b}}$$

Let $t = \sqrt{\frac{a}{b}}$. The equation becomes:

$$t^2 - 6t + 1 = 0$$

Using the quadratic formula to solve for $t$:

$$t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$$

Thus, $\sqrt{\frac{a}{b}} = 3 \pm 2\sqrt{2}$.

Squaring both sides to find the ratio $\frac{a}{b}$ (taking the positive root $3 + 2\sqrt{2}$):

$$\frac{a}{b} = (3 + 2\sqrt{2})^2 = 3^2 + (2\sqrt{2})^2 + 2(3)(2\sqrt{2}) = 9 + 8 + 12\sqrt{2} = 17 + 12\sqrt{2}$$

Alternatively, expressing the ratio directly in factored form:

$$\frac{a}{b} = \frac{(3 + 2\sqrt{2})^2}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} = \frac{(3 + 2\sqrt{2})^2}{9 - 8} = (3 + 2\sqrt{2})^2$$

Thus, the numbers are in the ratio $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$. Hence proved.

Exercise Chapter 9 - Question 29

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

Solution:

Let the two positive numbers be $a$ and $b$.

The Arithmetic Mean ($A$) and Geometric Mean ($G$) are given by:

$$A = \frac{a + b}{2} \implies a + b = 2A$$ $$G = \sqrt{ab} \implies ab = G^2$$

We know the algebraic identity relating $(a - b)$, $(a + b)$, and $ab$:

$$(a - b)^2 = (a + b)^2 - 4ab$$

Substitute $a + b = 2A$ and $ab = G^2$ into the identity:

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

Taking the square root on both sides:

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

Now we have a system of linear equations in $a$ and $b$:

1. $a + b = 2A$
2. $a - b = \pm 2\sqrt{(A - G)(A + G)}$

Adding (1) and (2) to find $2a$:

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

Similarly, subtracting (2) from (1) gives $b$:

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

Thus, the two numbers are given by $A \pm \sqrt{(A + G)(A - G)}$. Hence proved.

Exercise Chapter 8 - Question 30

Question: 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 $2^{\text{nd}}$ hour, $4^{\text{th}}$ hour and $n^{\text{th}}$ hour?

Solution:

Let the initial number of bacteria present initially ($t = 0$) be $a = 30$.

Since the number of bacteria doubles every hour, the number of bacteria at the end of each successive hour forms a Geometric Progression (GP) where the common ratio is $r = 2$.

• Number of bacteria at the end of $1^{\text{st}}$ hour: $a_1 = 30 \times 2 = 60$
• Number of bacteria at the end of $2^{\text{nd}}$ hour: $a_2 = 30 \times 2^2 = 30 \times 4 = 120$
• Number of bacteria at the end of $4^{\text{th}}$ hour: $a_4 = 30 \times 2^4 = 30 \times 16 = 480$
• Number of bacteria at the end of $n^{\text{th}}$ hour: $a_n = a \cdot r^n = 30 \cdot 2^n$

Thus, the number of bacteria present at the end of the $2^{\text{nd}}$ hour, $4^{\text{th}}$ hour, and $n^{\text{th}}$ hour are $120$, $480$, and $30 \cdot 2^n$ respectively.

Exercise Chapter 8 - Question 31

Question: What will Rs $500$ amount to in $10$ years after its deposit in a bank which pays annual interest rate of $10\%$ compounded annually?

Solution:

The initial principal amount is $P = \text{Rs } 500$, and the rate of compound interest is $r = 10\%$ per annum.

The amount $A_n$ at the end of $n$ years under compound interest is given by the formula:

$$A_n = P\left(1 + \frac{R}{100}\right)^n$$

Here, $P = 500$, $R = 10$, and $n = 10$ years. Substituting these values into the formula:

$$A_{10} = 500 \left(1 + \frac{10}{100}\right)^{10}$$ $$A_{10} = 500 \left(1 + \frac{1}{10}\right)^{10} = 500 (1.1)^{10}$$

Thus, the amount after $10$ years will be $\text{Rs } 500(1.1)^{10}$.

Exercise Chapter 8 - Question 32

Question: If A.M. and G.M. of roots of a quadratic equation are $8$ and $5$, respectively, then obtain the quadratic equation.

Solution:

Let the roots of the quadratic equation be $\alpha$ and $\beta$.

Given that the Arithmetic Mean (A.M.) of the roots is $8$:

$$\frac{\alpha + \beta}{2} = 8 \implies \alpha + \beta = 16$$

Also, the Geometric Mean (G.M.) of the roots is $5$:

$$\sqrt{\alpha \beta} = 5 \implies \alpha \beta = 5^2 = 25$$

A quadratic equation with roots $\alpha$ and $\beta$ is given by:

$$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$$

Substituting the values of the sum and product of the roots:

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

Hence, the required quadratic equation is $x^2 - 16x + 25 = 0$.

Miscellaneous Exercise - Question 1

Question: If $f$ is a function satisfying $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{N}$ such that $f(1) = 3$ and $\sum_{x=1}^{n} f(x) = 120$, find the value of $n$.

Solution:

Given the functional equation $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{N}$ and $f(1) = 3$.

Let us evaluate $f(x)$ for different natural numbers:

• $f(2) = f(1 + 1) = f(1) f(1) = 3 \times 3 = 3^2$
• $f(3) = f(2 + 1) = f(2) f(1) = 3^2 \times 3 = 3^3$

In general, by mathematical induction, $f(x) = 3^x$ for all $x \in \mathbb{N}$.

We are given the sum:

$$\sum_{x=1}^{n} f(x) = 120$$ $$\sum_{x=1}^{n} 3^x = 120$$

The terms $3, 3^2, 3^3, \dots, 3^n$ form a geometric progression with first term $a = 3$ and common ratio $r = 3$. The sum of the first $n$ terms is given by:

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

Equating this to $120$:

$$\frac{3(3^n - 1)}{2} = 120$$ $$3(3^n - 1) = 240$$ $$3^n - 1 = 80$$ $$3^n = 81$$ $$3^n = 3^4 \implies n = 4$$

Thus, the value of $n$ is $4$.

Exercise - Question 2

Question: The sum of some terms of G.P. is $315$ whose first term and the common ratio are $5$ and $2$, respectively. Find the last term and the number of terms.

Solution:

Given data for the Geometric Progression (G.P.):

• First term, $a = 5$
• Common ratio, $r = 2$
• Sum of $n$ terms, $S_n = 315$

The formula for the sum of $n$ terms of a G.P. when $r > 1$ is:

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

Substitute the given values into the formula:

$$315 = \frac{5(2^n - 1)}{2 - 1}$$ $$315 = \frac{5(2^n - 1)}{1}$$ $$315 = 5(2^n - 1)$$

Divide both sides by $5$:

$$\frac{315}{5} = 2^n - 1$$ $$63 = 2^n - 1$$ $$2^n = 63 + 1$$ $$2^n = 64$$

Since $64 = 2^6$, we get:

$$n = 6$$

Thus, the number of terms is $6$.

Now, we find the last term ($l$), which is the $6^{\text{th}}$ term ($a_6$):

$$l = a_n = a r^{n-1}$$ $$a_6 = 5 \cdot (2)^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160$$

Answer: The last term is $160$ and the number of terms is $6$.

Exercise - Question 3

Question: The first term of a G.P. is $1$. The sum of the third term and fifth term is $90$. Find the common ratio of G.P.

Solution:

Let the first term of the G.P. be $a$ and the common ratio be $r$.

Given:

• $a = 1$
• $a_3 + a_5 = 90$

The general term of a G.P. is given by $a_n = a r^{n-1}$. Therefore:

• Third term, $a_3 = a r^{3-1} = a r^2 = 1 \cdot r^2 = r^2$
• Fifth term, $a_5 = a r^{5-1} = a r^4 = 1 \cdot r^4 = r^4$

According to the given condition:

$$r^2 + r^4 = 90$$

Let $r^2 = x$. The equation becomes a quadratic in terms of $x$:

$$x + x^2 = 90 \implies x^2 + x - 90 = 0$$

Factorizing the quadratic equation:

$$(x + 10)(x - 9) = 0$$

This gives two possible values for $x$:

$$x = -10 \quad \text{or} \quad x = 9$$

Since $x = r^2$ and the square of a real number cannot be negative ($r^2 \geq 0$):

$$r^2 = 9$$ $$r = \pm 3$$

Answer: The common ratio of the G.P. is $\pm 3$.

Exercise - Question 4

Question: The sum of three numbers in G.P. is $56$. If we subtract $1, 7, 21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Solution:

Let the three numbers in G.P. be $\frac{a}{r}$, $a$, and $ar$.

Given that their sum is $56$:

$$\frac{a}{r} + a + ar = 56 \implies a\left(\frac{1}{r} + 1 + r\right) = 56 \quad \text{--- (1)}$$

If we subtract $1, 7, 21$ from these numbers in that order, the new numbers are:

$$\left(\frac{a}{r} - 1\right), (a - 7), (ar - 21)$$

These numbers form an Arithmetic Progression (A.P.). Therefore, twice the middle term equals the sum of the outer terms:

$$2(a - 7) = \left(\frac{a}{r} - 1\right) + (ar - 21)$$ $$2a - 14 = \frac{a}{r} + ar - 22$$ $$2a + 8 = \frac{a}{r} + ar \quad \text{--- (2)}$$

From equation (1), we have:

$$\frac{a}{r} + ar = 56 - a$$

Substitute this into equation (2):

$$2a + 8 = 56 - a$$ $$2a + a = 56 - 8$$ $$3a = 48 \implies a = 16$$

Now, substitute $a = 16$ back into equation (1):

$$16\left(\frac{1}{r} + 1 + r\right) = 56$$ $$\frac{1}{r} + 1 + r = \frac{56}{16} = \frac{7}{2}$$ $$\frac{1}{r} + r = \frac{7}{2} - 1 = \frac{5}{2}$$

Multiply through by $2r$:

$$2(1 + r^2) = 5r$$ $$2r^2 - 5r + 2 = 0$$

Factorize the quadratic equation:

$$2r^2 - 4r - r + 2 = 0$$ $$2r(r - 2) - 1(r - 2) = 0$$ $$(2r - 1)(r - 2) = 0$$

Thus, $r = 2$ or $r = \frac{1}{2}$.

• If $a = 16$ and $r = 2$, the numbers are $\frac{16}{2}, 16, 16(2) \implies 8, 16, 32$.
• If $a = 16$ and $r = \frac{1}{2}$, the numbers are $\frac{16}{1/2}, 16, 16\left(\frac{1}{2}\right) \implies 32, 16, 8$.

Answer: The three numbers are $8, 16, 32$ (or $32, 16, 8$).

Exercise - Question 5

Question: A G.P. consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.

Solution:

Let the G.P. consist of $2n$ terms (an even number of terms), with first term $a$ and common ratio $r$:

$$\text{G.P.: } a, ar, ar^2, ar^3, \dots, ar^{2n-1}$$

The sum of all $2n$ terms is given by:

$$S_{2n} = \frac{a(r^{2n} - 1)}{r - 1} \quad \text{--- (1)}$$

The terms occupying odd places are: $a_1, a_3, a_5, \dots, a_{2n-1}$, which are:

$$a, ar^2, ar^4, \dots, ar^{2n-2}$$

This is also a G.P. with $n$ terms, first term $a$, and common ratio $r^2$. The sum of these terms ($S_{\text{odd}}$) is:

$$S_{\text{odd}} = \frac{a((r^2)^n - 1)}{r^2 - 1} = \frac{a(r^{2n} - 1)}{r^2 - 1} \quad \text{--- (2)}$$

According to the given condition, the sum of all terms is $5$ times the sum of terms at odd places:

$$S_{2n} = 5 \cdot S_{\text{odd}}$$

Substitute equations (1) and (2):

$$\frac{a(r^{2n} - 1)}{r - 1} = 5 \cdot \frac{a(r^{2n} - 1)}{r^2 - 1}$$

Since $a \neq 0$ and $r^{2n} - 1 \neq 0$, we can cancel $\frac{a(r^{2n} - 1)}{r - 1}$ from both sides (noting that $r^2 - 1 = (r-1)(r+1)$):

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

Answer: The common ratio of the G.P. is $4$.

Exercise - Question 6

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

Solution:

Given:

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

Applying componendo and dividendo to each term (i.e., if $\frac{A}{B} = \frac{C}{D}$, then $\frac{A+B}{A-B} = \frac{C+D}{C-D}$):

For the first equality:

$$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)}$$ $$\frac{2a}{2bx} = \frac{2b}{2cx} \implies \frac{a}{bx} = \frac{b}{cx}$$

Cancelling $x$ from denominators:

$$\frac{a}{b} = \frac{b}{c} \implies b^2 = ac$$

Similarly, taking the second equality:

$$\frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$$ $$\frac{2b}{2cx} = \frac{2c}{2dx} \implies \frac{b}{cx} = \frac{c}{dx}$$

Cancelling $x$:

$$\frac{b}{c} = \frac{c}{d} \implies c^2 = bd$$

Since $\frac{a}{b} = \frac{b}{c}$ and $\frac{b}{c} = \frac{c}{d}$, we have:

$$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$$

Thus, $a, b, c,$ and $d$ are in G.P. Hence proved.

Exercise - Question 7

Question: Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $P^2 R^n = S^n$.

Solution: Let the G.P. consist of $n$ terms with first term $a$ and common ratio $r$: $$a, ar, ar^2, \dots, ar^{n-1}$$ 1. **Sum of $n$ terms ($S$):** $$S = \frac{a(1 - r^n)}{1 - r} \quad (\text{if } r \neq 1)$$ 2. **Product of $n$ terms ($P$):** $$P = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{1 + 2 + \dots + (n-1)} = a^n r^{\frac{n(n-1)}{2}}$$ 3. **Sum of reciprocals ($R$):** $$R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{n-1}}$$ This is also a G.P. with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$: $$R = \frac{\frac{1}{a}\left(1 - \frac{1}{r^n}\right)}{1 - \frac{1}{r}} = \frac{1}{a} \cdot \frac{r^n - 1}{r^n} \cdot \frac{r}{r-1} = \frac{r^n - 1}{a r^{n-1} (r-1)}$$ Now, consider $P^2 R^n$: $$P^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 = a^{2n} r^{n(n-1)}$$ $$R^n = \left(\frac{r^n - 1}{a r^{n-1} (r-1)}\right)^n = \frac{(r^n - 1)^n}{a^n r^{n(n-1)} (r-1)^n}$$ Multiplying $P^2$ and $R^n$: $$P^2 R^n = a^{2n} r^{n(n-1)} \cdot \frac{(r^n - 1)^n}{a^n r^{n(n-1)} (r-1)^n} = a^n \frac{(r^n - 1)^n}{(r-1)^n} = \left[ \frac{a(r^n - 1)}{r - 1} \right]^n = S^n$$ Hence proved: $P^2 R^n = S^n$.

Exercise - Question 8

Question: If $a, b, c, d$ are in G.P., prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P.

Solution: Since $a, b, c, d$ are in G.P., let the common ratio be $r$. Then: $$b = ar, \quad c = ar^2, \quad d = ar^3$$ We need to prove that $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P., which means we must show: $$(b^n + c^n)^2 = (a^n + b^n)(c^n + d^n)$$ Let's substitute $b, c, d$ in terms of $a$ and $r$: LHS: $$(b^n + c^n)^2 = (ar^n + ar^{2n})^2 = [a r^n (1 + r^n)]^2 = a^2 r^{2n} (1 + r^n)^2$$ RHS: $$(a^n + b^n)(c^n + d^n) = (a^n + a^n r^n)(a^n r^{2n} + a^n r^{3n})$$ $$= a^n(1 + r^n) \cdot a^n r^{2n}(1 + r^n)$$ $$= a^{2n} r^{2n} (1 + r^n)^2$$ Since $\text{LHS} = \text{RHS}$, the terms $(a^n + b^n), (b^n + c^n), (c^n + d^n)$ are in G.P. Hence proved.

Exercise - Question 9

Question: If $a$ and $b$ are the roots of $x^2 - 3x + p = 0$ and $c, d$ are roots of $x^2 - 12x + q = 0$, where $a, b, c, d$ form a G.P. Prove that $(q + p) : (q - p) = 17 : 15$.

Solution: Given equations: 1. $x^2 - 3x + p = 0$ has roots $a$ and $b$. Sum of roots: $a + b = 3$ Product of roots: $ab = p$ 2. $x^2 - 12x + q = 0$ has roots $c$ and $d$. Sum of roots: $c + d = 12$ Product of roots: $cd = q$ Let $a, b, c, d$ be in G.P. Let the first term be $a$ and common ratio be $r$. Thus, $b = ar$, $c = ar^2$, $d = ar^3$. From $a + b = 3$: $$a(1 + r) = 3 \implies a = \frac{3}{1 + r}$$ From $c + d = 12$: $$c + d = ar^2 + ar^3 = ar^2(1 + r) = 12$$ Substitute $a = \frac{3}{1+r}$: $$\left(\frac{3}{1+r}\right) r^2 (1 + r) = 12 \implies 3r^2 = 12 \implies r^2 = 4$$ Since $r^2 = 4$, $r = \pm 2$. Case 1: If $r = 2$ $$a = \frac{3}{1 + 2} = 1$$ $$b = ar = 2$$ $$c = ar^2 = 4$$ $$d = ar^3 = 8$$ Now, find $p$ and $q$: $$p = ab = 1 \times 2 = 2$$ $$q = cd = 4 \times 8 = 32$$ Check ratio $(q + p) : (q - p)$: $$\frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15}$$ Thus, $(q + p) : (q - p) = 17 : 15$. Hence proved.

Exercise - Question 10

Question: The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$, is $m : n$. Show that $a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)$.

Solution:

Let the two positive numbers be $a$ and $b$.

The Arithmetic Mean (A.M.) of $a$ and $b$ is given by:

$$\text{A.M.} = \frac{a + b}{2}$$

The Geometric Mean (G.M.) of $a$ and $b$ is given by:

$$\text{G.M.} = \sqrt{ab}$$

According to the given condition, the ratio of A.M. to G.M. is $m : n$:

$$\frac{\frac{a + b}{2}}{\sqrt{ab}} = \frac{m}{n} \implies \frac{a + b}{2\sqrt{ab}} = \frac{m}{n}$$

Using componendo and dividendo, we get:

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

Notice that the numerator and denominator can be written as complete squares:

$$\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{m + n}{m - n}$$

Taking square roots on both sides:

$$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{m + n}}{\sqrt{m - n}}$$

Again, applying componendo and dividendo:

$$\frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$$ $$\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} \implies \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$$

Squaring both sides:

$$\frac{a}{b} = \frac{(m + n) + (m - n) + 2\sqrt{(m + n)(m - n)}}{(m + n) + (m - n) - 2\sqrt{(m + n)(m - n)}}$$ $$\frac{a}{b} = \frac{2m + 2\sqrt{m^2 - n^2}}{2m - 2\sqrt{m^2 - n^2}} = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}}$$

Thus, $a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)$. Hence proved.

Exercise - Question 11

Question: Find the sum of the following series up to $n$ terms:
(i) $5 + 55 + 555 + \dots$
(ii) $.6 + .66 + .666 + \dots$

Solution:

(i) $S_n = 5 + 55 + 555 + \dots$ up to $n$ terms

$$S_n = 5(1 + 11 + 111 + \dots \text{ to } n \text{ terms})$$

Multiply and divide by $9$:

$$S_n = \frac{5}{9}(9 + 99 + 999 + \dots \text{ to } n \text{ terms})$$ $$S_n = \frac{5}{9}\left[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)\right]$$ $$S_n = \frac{5}{9}\left[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + \dots + n \text{ times})\right]$$

The first part is a geometric series with first term $10$ and common ratio $10$:

$$S_n = \frac{5}{9}\left[\frac{10(10^n - 1)}{10 - 1} - n\right] = \frac{5}{81}\left[10(10^n - 1) - 9n\right]$$

(ii) $S_n = 0.6 + 0.66 + 0.666 + \dots$ up to $n$ terms

$$S_n = 6(0.1 + 0.11 + 0.111 + \dots \text{ to } n \text{ terms})$$

Multiply and divide by $9$:

$$S_n = \frac{6}{9}(0.9 + 0.99 + 0.999 + \dots \text{ to } n \text{ terms})$$ $$S_n = \frac{2}{3}\left[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots \text{ to } n \text{ terms}\right]$$ $$S_n = \frac{2}{3}\left[(1 + 1 + \dots + n \text{ times}) - (0.1 + 0.01 + 0.001 + \dots \text{ to } n \text{ terms})\right]$$ $$S_n = \frac{2}{3}\left[n - \frac{0.1(1 - (0.1)^n)}{1 - 0.1}\right] = \frac{2}{3}\left[n - \frac{1}{9}(1 - 10^{-n})\right] = \frac{2}{27}\left[9n - 1 + 10^{-n}\right]$$

Exercise - Question 12

Question: Find the $20^{\text{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots + n$ terms.

Solution:

Let the given series be $S_n = (2 \times 4) + (4 \times 6) + (6 \times 8) + \dots$ up to $n$ terms.

Let us write the $n^{\text{th}}$ term ($T_n$) of the series by observing the patterns of the first and second factors:

First factors: $2, 4, 6, \dots$ (an A.P. with $a = 2, d = 2$) $\implies a_n = 2 + (n-1)2 = 2n$.

Second factors: $4, 6, 8, \dots$ (an A.P. with $a = 4, d = 2$) $\implies b_n = 4 + (n-1)2 = 2n + 2$.

Thus, the $n^{\text{th}}$ term of the series is:

$$T_n = (2n)(2n + 2) = 4n^2 + 4n$$

To find the $20^{\text{th}}$ term ($T_{20}$), substitute $n = 20$:

$$T_{20} = 4(20)^2 + 4(20) = 4(400) + 80 = 1600 + 80 = 1680$$

Answer: The $20^{\text{th}}$ term of the series is $1680$.

Exercise - Question 13

Question: A farmer buys a used tractor for Rs $12000$. He pays Rs $6000$ cash and agrees to pay the balance in annual instalments of Rs $500$ plus $12\%$ interest on the unpaid amount. How much will the tractor cost him?

Solution:

Total cost of the tractor = Rs $12000$.

Cash payment made = Rs $6000$.

Unpaid balance (principal amount to be paid in installments) = $\text{Rs } 12000 - \text{Rs } 6000 = \text{Rs } 6000$.

The balance is paid in annual installments of Rs $500$ each, along with $12\%$ interest on the unpaid amount each year.

• 1st Year: Unpaid amount = Rs $6000$.
  Interest for 1st year = $12\% \text{ of } 6000 = \frac{12}{100} \times 6000 = \text{Rs } 720$.
  Payment made at the end of 1st year = $\text{Rs } 500 + \text{Rs } 720$.
• 2nd Year: Unpaid amount = $6000 - 500 = \text{Rs } 5500$.
  Interest for 2nd year = $12\% \text{ of } 5500 = \frac{12}{100} \times 5500 = \text{Rs } 660$.
  Payment made at the end of 2nd year = $\text{Rs } 500 + \text{Rs } 660$.
• 3rd Year: Unpaid amount = $5500 - 500 = \text{Rs } 5000$.
  Interest for 3rd year = $12\% \text{ of } 5000 = \frac{12}{100} \times 5000 = \text{Rs } 600$.

$\dots$ continuing this pattern for $12$ years ($\frac{6000}{500} = 12$ installments).

Total interest paid is the sum of interests for all 12 years:

$$\text{Total Interest} = 720 + 660 + 600 + \dots + \text{to } 12 \text{ terms}$$

This is an arithmetic progression with first term $a = 720$, common difference $d = 660 - 720 = -60$, and number of terms $n = 12$:

$$S_{12} = \frac{n}{2}[2a + (n-1)d] = \frac{12}{2}[2(720) + (12-1)(-60)]$$ $$S_{12} = 6[1440 + 11(-60)] = 6[1440 - 660] = 6[780] = \text{Rs } 4680$$

Total cost of the tractor = Cash down payment + Unpaid balance + Total interest

$$\text{Total Cost} = 6000 + 6000 + 4680 = \text{Rs } 16680$$

Answer: The tractor will cost the farmer Rs $16680$.

Exercise - Question 14

Question: Shamshad Ali buys a scooter for Rs $22000$. He pays Rs $4000$ cash and agrees to pay the balance in annual instalment of Rs $1000$ plus $10\%$ interest on the unpaid amount. How much will the scooter cost him?

Solution:

Total cost of the scooter = $\text{Rs } 22000$.

Cash payment made initially = $\text{Rs } 4000$.

Unpaid amount (Principal) = $22000 - 4000 = \text{Rs } 18000$.

The balance of $\text{Rs } 18000$ is to be paid in annual instalments of $\text{Rs } 1000$ each along with $10\%$ interest on the unpaid amount.

Since the annual principal payment is $\text{Rs } 1000$, the number of instalments required to pay off the principal is:

$$\text{Number of instalments } (n) = \frac{18000}{1000} = 18$$

Let us calculate the interest paid each year:

• 1st year unpaid amount: $\text{Rs } 18000$, Interest = $10\% \text{ of } 18000 = 1800$
• 2nd year unpaid amount: $\text{Rs } 17000$, Interest = $10\% \text{ of } 17000 = 1700$
• 3rd year unpaid amount: $\text{Rs } 16000$, Interest = $10\% \text{ of } 16000 = 1600$
• ... and so on, until the last year
• 18th year unpaid amount: $\text{Rs } 1000$, Interest = $10\% \text{ of } 1000 = 100$

Thus, the total interest paid is the sum of an arithmetic progression:

$$\text{Total Interest} = 1800 + 1700 + 1600 + \dots + 100$$

This is an Arithmetic Progression with:

• First term, $a = 1800$
• Last term, $l = 100$
• Number of terms, $n = 18$

Using the sum formula for an AP, $S_n = \frac{n}{2}(a + l)$:

$$\text{Total Interest} = \frac{18}{2}(1800 + 100) = 9 \times 1900 = \text{Rs } 17100$$

Therefore, the total cost of the scooter is:

$$\text{Total Cost} = \text{Cash Down Payment} + \text{Principal Amount} + \text{Total Interest}$$ $$\text{Total Cost} = 4000 + 18000 + 17100 = \text{Rs } 39100$$

Answer: The scooter will cost Shamshad Ali Rs $39100$.

Question 15

Question: A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs $50$ paise to mail one letter. Find the amount spent on the postage when $8^{\text{th}}$ set of letter is mailed.

Solution:

Let us find the number of letters mailed in each set:

• 1st set: $4$ letters
• 2nd set: Each of the $4$ persons writes to $4$ friends $\Rightarrow 4 \times 4 = 16 = 4^2$ letters
• 3rd set: $16 \times 4 = 64 = 4^3$ letters

Thus, the number of letters in each set forms a geometric progression:

$$4, 4^2, 4^3, \dots$$

Here, the first term $a = 4$, the common ratio $r = 4$, and the number of terms is $n = 8$.

The total number of letters mailed up to the $8^{\text{th}}$ set is the sum of the first $8$ terms of this G.P. ($S_8$):

$$S_n = \frac{a(r^n - 1)}{r - 1}$$ $$S_8 = \frac{4(4^8 - 1)}{4 - 1} = \frac{4(65536 - 1)}{3} = \frac{4 \times 65535}{3} = 4 \times 21845 = 87380$$

Given that the cost of mailing one letter is $50$ paise (or $\text{Rs } 0.50$), the total amount spent on postage for $87380$ letters is:

$$\text{Total Amount} = 87380 \times 0.50 = \text{Rs } 43690$$

Question 16

Question: A man deposited Rs $10000$ in a bank at the rate of $5\%$ simple interest annually. Find the amount in $15^{\text{th}}$ year since he deposited the amount and also calculate the total amount after $20$ years.

Solution:

Given:

• Principal ($P$) = $\text{Rs } 10000$
• Rate of simple interest ($R$) = $5\%$ per annum

The simple interest for $1$ year is:

$$\text{Interest per year} = \frac{P \times R \times 1}{100} = \frac{10000 \times 5 \times 1}{100} = \text{Rs } 500$$

Part 1: Amount in the $15^{\text{th}}$ year

The amount at the end of $t$ years is given by $A_t = P + \text{Simple Interest for } t \text{ years}$.

For the $15^{\text{th}}$ year (i.e., after $15$ years):

$$\text{Interest for } 15 \text{ years} = 15 \times 500 = \text{Rs } 7500$$ $$\text{Amount after } 15 \text{ years} = 10000 + 7500 = \text{Rs } 17500$$

Part 2: Total amount after $20$ years

Interest for $20$ years:

$$\text{Interest for } 20 \text{ years} = 20 \times 500 = \text{Rs } 10000$$ $$\text{Total Amount after } 20 \text{ years} = 10000 + 10000 = \text{Rs } 20000$$

Question 17

Question: A manufacturer reckons that the value of a machine, which costs him Rs. $15625$, will depreciate each year by $20\%$. Find the estimated value at the end of $5$ years.

Solution:

Given:

• Initial cost of the machine ($P$) = $\text{Rs } 15625$
• Rate of depreciation ($r$) = $20\%$ per year
• Time ($n$) = $5$ years

Since the value depreciates by $20\%$ each year, the value of the machine at the end of each year is $80\%$ (or $\frac{4}{5}$) of its value at the beginning of that year.

The values at the end of successive years form a geometric progression:

• End of 1st year: $15625 \times \left(1 - \frac{20}{100}\right) = 15625 \times \left(\frac{4}{5}\right)$
• End of 2nd year: $15625 \times \left(\frac{4}{5}\right)^2$
• ...
• End of 5th year: $15625 \times \left(\frac{4}{5}\right)^5$

Calculating the value at the end of $5$ years:

$$\text{Value} = 15625 \times \left(\frac{4}{5}\right)^5 = 15625 \times \frac{1024}{3125}$$

Since $15625 = 5^6$ and $3125 = 5^5$:

$$\text{Value} = \frac{15625}{3125} \times 1024 = 5 \times 1024 = \text{Rs } 5120$$

Thus, the estimated value of the machine at the end of $5$ years is Rs $5120$.

Exercise Chapter 5 - Question 18

Question: $150$ workers were engaged to finish a job in a certain number of days. $4$ workers dropped out on second day, $4$ more workers dropped out on third day and so on. It took $8$ more days to finish the work. Find the number of days in which the work was completed.

Solution:

Let the original number of days planned to finish the job be $x$ days.

Since $150$ workers were engaged to finish the job in $x$ days, the total amount of work (measured in worker-days) is:

$$\text{Total Work} = 150x$$

According to the problem, the number of workers working each day forms an Arithmetic Progression (AP):

• 1st day: $150$ workers
• 2nd day: $150 - 4 = 146$ workers
• 3rd day: $146 - 4 = 142$ workers
• and so on.

It took $8$ more days than the scheduled $x$ days to complete the work. Therefore, the actual number of days taken to finish the work is $x + 8$.

The total work done is the sum of workers working on each day from day $1$ to day $x + 8$. This gives the sum of the first $n = x + 8$ terms of the AP where the first term $a = 150$ and common difference $d = -4$:

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

Substitute $n = x + 8$, $a = 150$, and $d = -4$ into the formula:

$$\text{Total Work} = \frac{x + 8}{2} [2(150) + (x + 8 - 1)(-4)]$$

$$\text{Total Work} = \frac{x + 8}{2} [300 - 4(x + 7)]$$

$$\text{Total Work} = \frac{x + 8}{2} [300 - 4x - 28]$$

$$\text{Total Work} = \frac{x + 8}{2} [272 - 4x]$$

Factoring out $2$ from the bracket:

$$\text{Total Work} = (x + 8)(136 - 2x)$$

Since the total amount of work remains constant, we can equate the two expressions for the total work:

$$150x = (x + 8)(136 - 2x)$$

Divide the entire equation by $2$:

$$75x = (x + 8)(68 - x)$$

Expand the right-hand side:

$$75x = 68x - x^2 + 544 - 8x$$

$$75x = 60x - x^2 + 544$$

Rearrange into a standard quadratic equation form ($ax^2 + bx + c = 0$):

$$x^2 + 75x - 60x - 544 = 0$$

$$x^2 + 15x - 544 = 0$$

Now, factorize the quadratic equation by finding two numbers that multiply to $-544$ and add up to $15$. These numbers are $32$ and $-17$:

$$x^2 + 32x - 17x - 544 = 0$$

$$x(x + 32) - 17(x + 32) = 0$$

$$(x - 17)(x + 32) = 0$$

This yields two possible values for $x$:

$$x = 17 \quad \text{or} \quad x = -32$$

Since the number of days cannot be negative, we discard $x = -32$. Thus:

$$x = 17$$

The original number of days planned was $17$. The question asks for the number of days in which the work was actually completed, which is $x + 8$:

$$\text{Actual days} = 17 + 8 = 25$$

Answer: The work was completed in $25$ days.