Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-6 -

Class 11 Applied Mathematics - Chapter 6 Solutions

Check Your Progress 6.1

Insert '

n

' arithmetic means between 1 and 31 such that the ratio of the 7th mean and the (n-1)th mean is 5 : 9. Find the value of

n

and the resulting A.P.

Solution

Step 1: Set up the terms for the Arithmetic Progression.
Let the $n$ arithmetic means be $A_{1}, A_{2}, A_{3}, \dots, A_{n}$ .
Then the sequence $1, A_{1}, A_{2}, \dots, A_{n}, 31$ forms an A.P. with $(n + 2)$ total terms.
The first term $a = 1$ and the last term $b = 31$ .

Step 2: Find the common difference.
The common difference $d$ when inserting $n$ arithmetic means between $a$ and $b$ is given by:
$d = \frac{b - a}{n + 1} = \frac{31 - 1}{n + 1} = \frac{30}{n + 1}$

Step 3: Express the 7th and (n-1)th means.
The $k$ -th arithmetic mean is given by $A_{k} = a + k d$ .
For the 7th mean: $A_{7} = a + 7 d = 1 + 7 (\frac{30}{n + 1}) = 1 + \frac{210}{n + 1} = \frac{n + 1 + 210}{n + 1} = \frac{n + 211}{n + 1}$ For the (n-1)th mean: $A_{n - 1} = a + (n - 1) d = 1 + (n - 1) (\frac{30}{n + 1}) = \frac{n + 1 + 30 n - 30}{n + 1} = \frac{31 n - 29}{n + 1}$

Step 4: Use the given ratio to solve for $n$ .
We are given that $\frac{A_{7}}{A_{n - 1}} = \frac{5}{9}$ . Substituting the expressions: $\frac{(\frac{n + 211}{n + 1})}{(\frac{31 n - 29}{n + 1})} = \frac{5}{9}$ The denominators cancel out, giving: $\frac{n + 211}{31 n - 29} = \frac{5}{9}$ Cross-multiply to solve for $n$ : $9 (n + 211) = 5 (31 n - 29)$ $9 n + 1899 = 155 n - 145$ $1899 + 145 = 155 n - 9 n$ $2044 = 146 n \Rightarrow n = \frac{2044}{146} = 14$

Step 5: Find the resulting A.P.
Now substitute $n = 14$ back into the expression for $d$ : $d = \frac{30}{14 + 1} = \frac{30}{15} = 2$ The resulting A.P. starts at 1, has a common difference of 2, and ends at 31.

Final Answer:
The value of

n

is 14.
The resulting A.P. is:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31

The arithmetic mean of two numbers is 34. If one number is increased by 6 and the other is decreased by 4, their arithmetic mean becomes 35. If the difference between the two numbers is 8, find the two numbers.

Solution

Step 1: Set up the variables and equations based on the given information.
Let the two numbers be $x$ and $y$ , where $x > y$ .

From the first statement, their arithmetic mean is 34: $\frac{x + y}{2} = 34 \Rightarrow x + y = 68 --- (Equation 1)$

From the third statement, the difference between the two numbers is 8: $x - y = 8 --- (Equation 2)$

Step 2: Solve the simultaneous equations.
Add Equation 1 and Equation 2: $(x + y) + (x - y) = 68 + 8$ $2 x = 76 \Rightarrow x = 38$ Substitute the value of $x$ into Equation 1: $38 + y = 68 \Rightarrow y = 68 - 38 = 30$

Step 3: Verify with the second statement.
The problem states that if one is increased by 6 and the other is decreased by 4, the mean becomes 35. Let's verify our numbers:
New numbers: $38 + 6 = 44$ and $30 - 4 = 26$ .
New Arithmetic Mean: $\frac{44 + 26}{2} = \frac{70}{2} = 35$ The condition holds true.

Final Answer:
The two numbers are 38 and 30.

Three numbers are in A.P. such that their sum is 27 and the sum of their squares is 293. Find the numbers.

Solution

Step 1: Select convenient variables for three numbers in an A.P.
When dealing with three numbers in an Arithmetic Progression, it is easiest to define them symmetrically as: $a - d, a, a + d$ where $a$ is the middle term and $d$ is the common difference.

Step 2: Use the sum condition to find $a$ .
The problem states that their sum is 27. $(a - d) + a + (a + d) = 27$ $3 a = 27 \Rightarrow a = 9$ So, the three numbers are $9 - d, 9,$ and $9 + d$ .

Step 3: Use the sum of squares condition to find $d$ .
The sum of their squares is 293. $(9 - d)^{2} + 9^{2} + (9 + d)^{2} = 293$ Expand the squared binomials: $(81 - 18 d + d^{2}) + 81 + (81 + 18 d + d^{2}) = 293$ Combine like terms (notice that the $- 18 d$ and $+ 18 d$ cancel out): $243 + 2 d^{2} = 293$ $2 d^{2} = 293 - 243$ $2 d^{2} = 50 \Rightarrow d^{2} = 25$ $d = \pm 5$

Step 4: Determine the three numbers.

If $d = 5$ , the numbers are: $9 - 5, 9, 9 + 5$ ⇒ $4, 9, 14$ .
If $d = - 5$ , the numbers are: $9 - (- 5), 9, 9 + (- 5)$ ⇒ $14, 9, 4$ .

Final Answer:
The three numbers are 4, 9, and 14.

Check Your Progress 6.2

Find the indicated terms in each of the Geometric progressions given below:
(i)

4, 12, 36, \dots

; 5th term
(ii)

3, - 1, \frac{1}{3}, - \frac{1}{9}, \dots

; 4th term, nth term

Solution

(i) Finding the 5th term of $4, 12, 36, \dots$
Here, the first term $a = 4$ .
The common ratio $r = \frac{12}{4} = 3$ .
The $n$ -th term of a G.P. is given by $a_{n} = a r^{n - 1}$ .
For the 5th term ( $n = 5$ ): $a_{5} = 4 \times 3^{5 - 1} = 4 \times 3^{4} = 4 \times 81 = 324$

(ii) Finding the 4th term and $n$ -th term of $3, - 1, \frac{1}{3}, - \frac{1}{9}, \dots$
Here, the first term $a = 3$ .
The common ratio $r = \frac{- 1}{3} = - \frac{1}{3}$ .
For the 4th term ( $n = 4$ ): $a_{4} = a r^{3} = 3 {(- \frac{1}{3})}^{3} = 3 (- \frac{1}{27}) = - \frac{1}{9}$ For the $n$ -th term: $a_{n} = a r^{n - 1} = 3 {(- \frac{1}{3})}^{n - 1}$

Answers:
(i) 5th term =

324

(ii) 4th term =

- \frac{1}{9}

n

-th term =

3 {(- \frac{1}{3})}^{n - 1}

Which term of the following sequences:
(i)

5, 10, 20, 40, \dots

is 5120
(ii)

2, 2 \sqrt{2}, 4, \dots

is 128
(iii)

2, 1, \frac{1}{2}, \frac{1}{4}, \dots

\frac{1}{128}

Solution

(i) For the sequence $5, 10, 20, 40, \dots$
First term $a = 5$ , common ratio $r = \frac{10}{5} = 2$ . We need to find $n$ such that $a_{n} = 5120$ . $a r^{n - 1} = 5120 \Rightarrow 5 \times 2^{n - 1} = 5120$ $2^{n - 1} = \frac{5120}{5} = 1024$ Since $1024 = 2^{10}$ , we have: $n - 1 = 10 \Rightarrow n = 11$

(ii) For the sequence $2, 2 \sqrt{2}, 4, \dots$
First term $a = 2$ , common ratio $r = \frac{2 \sqrt{2}}{2} = \sqrt{2}$ . We need to find $n$ such that $a_{n} = 128$ . $a r^{n - 1} = 128 \Rightarrow 2 {(\sqrt{2})}^{n - 1} = 128$ ${(\sqrt{2})}^{n - 1} = 64$ Writing both sides as powers of 2 (note $\sqrt{2} = 2^{\frac{1}{2}}$ and $64 = 2^{6}$ ): ${(2^{\frac{1}{2}})}^{n - 1} = 2^{6} \Rightarrow 2^{\frac{n - 1}{2}} = 2^{6}$ Equating the exponents: $\frac{n - 1}{2} = 6 \Rightarrow n - 1 = 12 \Rightarrow n = 13$

(iii) For the sequence $2, 1, \frac{1}{2}, \frac{1}{4}, \dots$
First term $a = 2$ , common ratio $r = \frac{1}{2}$ . We need to find $n$ such that $a_{n} = \frac{1}{128}$ . $a r^{n - 1} = \frac{1}{128} \Rightarrow 2 {(\frac{1}{2})}^{n - 1} = \frac{1}{128}$ ${(\frac{1}{2})}^{n - 1} = \frac{1}{256}$ Since $256 = 2^{8}$ , we have: ${(\frac{1}{2})}^{n - 1} = {(\frac{1}{2})}^{8}$ $n - 1 = 8 \Rightarrow n = 9$

Answers:
(i) 11th term
(ii) 13th term
(iii) 9th term

Find the sum to indicated number of terms in each of the geometric progressions:
(i)

\sqrt{3}, 3, 3 \sqrt{3}, \dots

6 terms
(ii)

0.15 + 0.015 + 0.0015 + \dots

20 terms

Solution

(i) For the sequence $\sqrt{3}, 3, 3 \sqrt{3}, \dots$ up to 6 terms
First term $a = \sqrt{3}$ .
Common ratio $r = \frac{3}{\sqrt{3}} = \sqrt{3}$ .
The sum of $n$ terms of a G.P. is $S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ (since $r > 1$ ). $S_{6} = \frac{\sqrt{3} ({(\sqrt{3})}^{6} - 1)}{\sqrt{3} - 1}$ Calculate ${(\sqrt{3})}^{6} = 3^{3} = 27$ : $S_{6} = \frac{\sqrt{3} (27 - 1)}{\sqrt{3} - 1} = \frac{26 \sqrt{3}}{\sqrt{3} - 1}$ Rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{3} + 1)$ : $S_{6} = \frac{26 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \frac{26 (3 + \sqrt{3})}{3 - 1} = \frac{26 (3 + \sqrt{3})}{2}$ $= 13 (3 + \sqrt{3}) = 39 + 13 \sqrt{3}$

(ii) For the sequence $0.15 + 0.015 + 0.0015 + \dots$ up to 20 terms
First term $a = 0.15$ .
Common ratio $r = \frac{0.015}{0.15} = 0.1$ .
Since $r < 1$ , the sum formula is $S_{n} = \frac{a (1 - r^{n})}{1 - r}$ . $S_{20} = \frac{0.15 (1 - 0 . 1^{20})}{1 - 0.1}$ $= \frac{0.15 (1 - 0 . 1^{20})}{0.9} = \frac{15}{90} (1 - 0 . 1^{20})$ $= \frac{1}{6} (1 - 0 . 1^{20})$

Answers:
(i) $39 + 13 \sqrt{3}$
(ii) $\frac{1}{6} [1 - (0.1)^{20}]$

Evaluate:

\sum_{k = 1}^{10} (3 + 2^{k})

Solution

Step 1: Separate the summation into two parts.
Using the properties of summation: $\sum_{k = 1}^{10} (3 + 2^{k}) = \sum_{k = 1}^{10} 3 + \sum_{k = 1}^{10} 2^{k}$

Step 2: Evaluate the first summation.
The sum of a constant $3$ added $10$ times is: $\sum_{k = 1}^{10} 3 = 3 \times 10 = 30$

Step 3: Evaluate the second summation.
The series $2^{1} + 2^{2} + 2^{3} + \dots + 2^{10}$ is a Geometric Progression with 10 terms.
First term $a = 2$ , and common ratio $r = 2$ . $S_{10} = \frac{a (r^{10} - 1)}{r - 1} = \frac{2 (2^{10} - 1)}{2 - 1}$ $= 2 (1024 - 1) = 2 (1023) = 2046$ Alternatively, leaving it in powers of 2 (as matching typical textbook answers): $2 (2^{10} - 1) = 2^{11} - 2$

Step 4: Combine the results.
Total sum = $30 + 2^{11} - 2 = 28 + 2^{11}$ (or $2076$ ).

Final Answer:

28 + 2^{11} (which evaluates to 2076)

Insert 6 geometric means between

27

and

\frac{1}{81}

Solution

Step 1: Set up the terms of the Geometric Progression.
Let the 6 geometric means be $G_{1}, G_{2}, G_{3}, G_{4}, G_{5}, G_{6}$ .
The resulting sequence forms a G.P.: $27, G_{1}, G_{2}, G_{3}, G_{4}, G_{5}, G_{6}, \frac{1}{81}$ This sequence has a total of $n = 8$ terms.
First term, $a = 27$ .
Eighth term, $a_{8} = \frac{1}{81}$ .

Step 2: Find the common ratio ( $r$ ).
Using the formula for the $n$ -th term of a G.P., $a_{n} = a r^{n - 1}$ : $a_{8} = a r^{7}$ $\frac{1}{81} = 27 \times r^{7}$ $r^{7} = \frac{1}{81 \times 27}$ Expressing the denominators as powers of 3: $81 = 3^{4}$ and $27 = 3^{3}$ . $r^{7} = \frac{1}{3^{4} \times 3^{3}} = \frac{1}{3^{7}}$ $r^{7} = {(\frac{1}{3})}^{7} \Rightarrow r = \frac{1}{3}$

Step 3: Calculate the 6 geometric means.
Multiply each term by the common ratio $r = \frac{1}{3}$ to find the next term.
$G_{1} = a r = 27 \times \frac{1}{3} = 9$ $G_{2} = G_{1} r = 9 \times \frac{1}{3} = 3$ $G_{3} = G_{2} r = 3 \times \frac{1}{3} = 1$ $G_{4} = G_{3} r = 1 \times \frac{1}{3} = \frac{1}{3}$ $G_{5} = G_{4} r = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ $G_{6} = G_{5} r = \frac{1}{9} \times \frac{1}{3} = \frac{1}{27}$

Final Answer:
The 6 geometric means are $9, 3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}$ .

Express

0.4 \bar{73}

as a rational number using the concept of infinite G.P.

Solution

Step 1: Write the repeating decimal as an infinite series.
Let $x = 0.4 \bar{73} = 0.4737373 \dots$
We can separate the non-repeating part from the repeating part: $x = 0.4 + 0.073 + 0.00073 + 0.0000073 + \dots$ Convert the decimals into fractions: $x = \frac{4}{10} + (\frac{73}{1000} + \frac{73}{100000} + \frac{73}{10000000} + \dots)$

Step 2: Identify the infinite G.P. and its parameters.
The terms inside the parenthesis form an infinite Geometric Progression. $First term (a) = \frac{73}{1000}$ $Common ratio (r) = \frac{(\frac{73}{100000})}{(\frac{73}{1000})} = \frac{1}{100}$

Step 3: Calculate the sum of the infinite G.P.
Since $| r | < 1$ , the sum of the infinite G.P. is $S_{\infty} = \frac{a}{1 - r}$ . $S_{\infty} = \frac{(\frac{73}{1000})}{1 - \frac{1}{100}} = \frac{(\frac{73}{1000})}{(\frac{99}{100})}$ $= \frac{73}{1000} \times \frac{100}{99} = \frac{73}{990}$

Step 4: Add the sum back to the non-repeating part.
Now substitute this sum back into our expression for $x$ : $x = \frac{4}{10} + \frac{73}{990}$ Find a common denominator, which is 990: $x = \frac{4 \times 99}{990} + \frac{73}{990}$ $x = \frac{396}{990} + \frac{73}{990} = \frac{396 + 73}{990} = \frac{469}{990}$

Final Answer:

0.4 \bar{73} = \frac{469}{990}

Class 11 Applied Mathematics - Chapter 6 Solutions

Practice Exercise (Q1 - Q10)

If the A.M. of two numbers is

17

and one number is

12

, then the other number is:

$5$
$22$
$29$
$34$

Solution

Step 1: Understand the formula for Arithmetic Mean (A.M.).
The Arithmetic Mean of two numbers $a$ and $b$ is given by the formula:

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

Step 2: Substitute the known values.
We are given that the A.M. is $17$ , and one of the numbers (let's say $a$ ) is $12$ . We need to find $b$ .

17 = \frac{12 + b}{2}

Step 3: Solve for the unknown number.
Multiply both sides by 2:

34 = 12 + b

b = 34 - 12 = 22

Therefore, the other number is

22

. The correct option is (b).

7

arithmetic means are inserted between

2

and

34

, then the 4th arithmetic mean is:

$14$
$16$
$18$
$20$

Solution

Step 1: Set up the Arithmetic Progression.
When we insert $n$ arithmetic means between two numbers $a$ and $b$ , the entire sequence forms an A.P. with $(n + 2)$ terms.
Here, $a = 2$ , $b = 34$ , and we are inserting $7$ means. So, the total number of terms is $7 + 2 = 9$ .

Step 2: Find the common difference ( $d$ ).
The last term ( $t_{9}$ ) is 34. Using the formula for the nth term of an A.P., $t_{n} = a + (n - 1) d$ :

34 = 2 + (9 - 1) d

34 = 2 + 8 d

8 d = 32 ⟹ d = 4

Step 3: Calculate the 4th arithmetic mean.
The sequence is: $a, A_{1}, A_{2}, A_{3}, A_{4}, ...$
The 4th arithmetic mean, $A_{4}$ , is actually the 5th term of the A.P. ( $t_{5}$ ). $A_{4} = t_{5} = a + 4 d$ $A_{4} = 2 + 4 (4) = 2 + 16 = 18$

Therefore, the 4th arithmetic mean is

18

. The correct option is (c).

The 5th term of the G.P.

3, 6, 12, 24, ...

is:

$36$
$48$
$96$
$192$

Solution

Step 1: Identify the components of the Geometric Progression (G.P.).
The given G.P. is $3, 6, 12, 24, ...$
First term ( $a$ ) = $3$
Common ratio ( $r$ ) = $\frac{6}{3} = 2$

Step 2: Use the formula for the nth term of a G.P.
The formula for the nth term is: $t_{n} = a r^{n - 1}$ We need to find the 5th term ( $n = 5$ ):

t_{5} = 3 \times 2^{5 - 1}

t_{5} = 3 \times 2^{4}

t_{5} = 3 \times 16 = 48

Therefore, the 5th term is

48

. The correct option is (b).

Which term of the G.P.

2, 8, 32, 128, ...

131072

$8$ th
$9$ th
$10$ th
$11$ th

Solution

Step 1: Identify the components of the G.P.
First term ( $a$ ) = $2$
Common ratio ( $r$ ) = $\frac{8}{2} = 4$
Given nth term ( $t_{n}$ ) = $131072$

Step 2: Use the nth term formula and solve for $n$ .

t_{n} = a r^{n - 1}

131072 = 2 \times (4)^{n - 1}

Divide both sides by 2:

65536 = 4^{n - 1}

Express $65536$ as a power of 4. We know that $4^{4} = 256$ , and $256 \times 256 = 65536$ . Since $4^{4} \times 4^{4} = 4^{8}$ , we have:

4^{8} = 4^{n - 1}

Equating the exponents:

8 = n - 1 ⟹ n = 9

Therefore,

131072

is the 9th term. The correct option is (b).

If the 3rd term of a G.P. is

6

, then the product of its first 5 terms is:

$56$
$6^{5}$
$5^{2}$
$6^{2}$

Solution

Step 1: Write the given information mathematically.
Let the first term of the G.P. be $a$ and the common ratio be $r$ .
The 3rd term is given as $6$ :

t_{3} = a r^{2} = 6

Step 2: Express the product of the first 5 terms.
The first 5 terms are: $a, a r, a r^{2}, a r^{3}, a r^{4}$ .
Their product ( $P$ ) is:

P = a \times a r \times a r^{2} \times a r^{3} \times a r^{4}

P = a^{5} r^{(1 + 2 + 3 + 4)} = a^{5} r^{10}

Step 3: Substitute the value of the 3rd term into the product.
Notice that $a^{5} r^{10}$ can be rewritten as:

P = (a r^{2})^{5}

Since $a r^{2} = 6$ , we have:

P = 6^{5}

Therefore, the product of its first 5 terms is

6^{5}

. The correct option is (b).

The sum of the series

\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ...

up to 6 terms is:

$\frac{364}{243}$
$\frac{364}{729}$
$\frac{728}{729}$
$\frac{728}{243}$

Solution

Step 1: Identify the G.P. parameters.
The series is a Geometric Progression.
First term ( $a$ ) = $\frac{1}{3}$
Common ratio ( $r$ ) = $\frac{1 / 9}{1 / 3} = \frac{1}{3}$
Number of terms ( $n$ ) = $6$

Step 2: Apply the sum formula for a G.P.
Since $r < 1$ , the sum of $n$ terms is:

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

S_{6} = \frac{\frac{1}{3} (1 - (\frac{1}{3})^{6})}{1 - \frac{1}{3}}

Step 3: Simplify the expression.

S_{6} = \frac{\frac{1}{3} (1 - \frac{1}{729})}{\frac{2}{3}}

The $\frac{1}{3}$ in the numerator and denominator simplifies:

S_{6} = \frac{1}{2} \times (\frac{729 - 1}{729})

S_{6} = \frac{1}{2} \times \frac{728}{729} = \frac{364}{729}

Therefore, the sum is

\frac{364}{729}

. The correct option is (a).

How many terms of the G.P.

3, 3^{2}, 3^{3}, ...

are needed to give the sum

120

Solution

Step 1: Identify the G.P. components.
First term ( $a$ ) = $3$
Common ratio ( $r$ ) = $\frac{3^{2}}{3} = 3$
Sum to $n$ terms ( $S_{n}$ ) = $120$

Step 2: Use the formula for the sum of $n$ terms.
Since $r > 1$ , the formula is:

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

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

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

Step 3: Solve for $n$ .

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

80 = 3^{n} - 1

3^{n} = 81

Since $3^{4} = 81$ , we have $n = 4$ .

Therefore,

4

terms are needed. The correct option is (b).

Three geometric means between

1

and

256

are:

$2, 8, 32$
$8, 32, 128$
$4, 32, 128$
$4, 16, 64$

Solution

Step 1: Set up the sequence.
Let the 3 geometric means be $G_{1}, G_{2}, G_{3}$ . The sequence forms a G.P.: $1, G_{1}, G_{2}, G_{3}, 256$ .

Step 2: Find the common ratio ( $r$ ).
The total number of terms is $5$ . The first term $a = 1$ and the 5th term is $256$ .

t_{5} = a r^{4}

256 = 1 \times r^{4}

r^{4} = 256

Taking the 4th root, we get $r = 4$ (assuming real positive terms for standard geometric means).

Step 3: Calculate the means.

G_{1} = a r = 1 \times 4 = 4

G_{2} = a r^{2} = 1 \times 4^{2} = 16

G_{3} = a r^{3} = 1 \times 4^{3} = 64

Therefore, the three geometric means are

4, 16, 64

. The correct option is (d).

The sum of an infinite G.P. is

3

and the sum of the squares of its terms is also

3

, then its first term and common ratio are:

$1, \frac{1}{2}$
$\frac{3}{2}, \frac{1}{2}$
$\frac{1}{3}, \frac{1}{2}$
$1, \frac{1}{4}$

Solution

Step 1: Set up the equations based on the formulas.
Let the first term be $a$ and the common ratio be $r$ , where $| r | < 1$ .
The sum of an infinite G.P. is:

\frac{a}{1 - r} = 3

From this, we get:

a = 3 (1 - r) — (Equation 1)

Step 2: Formulate the equation for the sum of squares.
The squares of the terms of the G.P. form a new infinite G.P.: $a^{2}, a^{2} r^{2}, a^{2} r^{4}, ...$
The first term is $a^{2}$ and the common ratio is $r^{2}$ . Its sum is also given as 3:

\frac{a^{2}}{1 - r^{2}} = 3 — (Equation 2)

Step 3: Solve the system of equations.
Substitute $a = 3 (1 - r)$ from Eq 1 into Eq 2:

\frac{[3 (1 - r)]^{2}}{1 - r^{2}} = 3

\frac{9 (1 - r)^{2}}{(1 - r) (1 + r)} = 3

Cancel out $(1 - r)$ assuming $r \neq 1$ :

\frac{9 (1 - r)}{1 + r} = 3

Divide both sides by 3:

\frac{3 (1 - r)}{1 + r} = 1

3 - 3 r = 1 + r

4 r = 2 ⟹ r = \frac{1}{2}

Step 4: Find the first term ( $a$ ).
Substitute $r = \frac{1}{2}$ back into Equation 1:

a = 3 (1 - \frac{1}{2}) = 3 (\frac{1}{2}) = \frac{3}{2}

Therefore, the first term is

\frac{3}{2}

and the common ratio is

\frac{1}{2}

. The correct option is (c).

Q10

If A and G are the A.M. and G.M. between two positive numbers respectively, then the relation between them is:

$A < G$
$A = G$
$A \geq G$
$A > G$ for unequal numbers

Solution

Step 1: Write the formulas for A.M. and G.M.
Let the two positive numbers be $a$ and $b$ .
Their Arithmetic Mean (A.M.) is $A = \frac{a + b}{2}$
Their Geometric Mean (G.M.) is $G = \sqrt{a b}$

Step 2: Compare A and G by finding their difference.

A - G = \frac{a + b}{2} - \sqrt{a b}

A - G = \frac{a + b - 2 \sqrt{a b}}{2}

Notice that the numerator is a perfect square:

A - G = \frac{(\sqrt{a} - \sqrt{b})^{2}}{2}

Step 3: Analyze the result.
Since the square of any real number is always non-negative, $(\sqrt{a} - \sqrt{b})^{2} \geq 0$ .
Therefore, $A - G \geq 0$ , which implies $A \geq G$ .

Note: $A = G$ strictly when $a = b$ . If $a$ and $b$ are unequal, then $A > G$ . Both option (c) and (d) express valid mathematical truths, but generally, the fundamental relationship established is $A \geq G$ without specific constraints, and explicitly $A > G$ for unequal numbers as accurately stated in option (d).

Therefore, the most precise specific relation given the options is

A > G

for unequal numbers. However, the universal relation is

A \geq G

. (Standard answer key usually accepts (d) for "unequal numbers" condition or (c) as the general rule).

Class 11 Applied Mathematics - Chapter 6 Solutions

Practice Exercise (Q11 to Q30)

Q11

An antique's present worth is ₹9000. If its value appreciates at the rate of 10% per year. Its worth 3 years from now is:

₹ 6,561
₹ 10,890
₹ 11,979
₹ 12,000

Solution

Step 1: Identify the given values.
The appreciation of an asset's value follows the compound interest formula.
Present value ( $P$ ) = $9000$
Rate of appreciation ( $r$ ) = $10 %$ per year
Time period ( $n$ ) = $3$ years

Step 2: Apply the formula for appreciated value.
$A = P {(1 + \frac{r}{100})}^{n}$ Substituting the values: $A = 9000 {(1 + \frac{10}{100})}^{3}$ $A = 9000 {(1 + 0.1)}^{3}$ $A = 9000 {(1.1)}^{3}$

Step 3: Calculate the final amount.
We know that ${(1.1)}^{3} = 1.331$ . $A = 9000 \times 1.331 = 11979$

Therefore, the correct option is (c) ₹ 11,979.

Q12

Veena invests ₹ 5000 in a bond that pays 6% interest p.a compounded semi-annually. The value of the bond in rupees after 5 years is:

$5000 {(1.06)}^{5}$
$5000 {(1.03)}^{5}$
$5000 {(1.06)}^{10}$
$5000 {(1.03)}^{10}$

Solution

Step 1: Identify the given values and adjusting for semi-annual compounding.
Principal ( $P$ ) = $5000$
Annual interest rate ( $R$ ) = $6 %$
Time period in years ( $T$ ) = $5$

Since the interest is compounded semi-annually, we must halve the annual rate and double the number of years to find the number of compounding periods:
Rate per half-year ( $i$ ) = $\frac{6 %}{2} = 3 % = 0.03$
Number of half-years ( $n$ ) = $5 \times 2 = 10$

Step 2: Apply the compound interest formula.
$A = P {(1 + i)}^{n}$ $A = 5000 {(1 + 0.03)}^{10}$ $A = 5000 {(1.03)}^{10}$

Therefore, the correct option is (d) $5000 {(1.03)}^{10}$ .

Q13

a, b

and

c

are in A.P. as well as in G.P., then which of the following is true:

$a = b \neq c$
$a \neq b \neq c$
$a = b = c$
$a \neq b = c$

Solution

Step 1: Use the condition for A.P.
If $a, b, c$ are in Arithmetic Progression, the common difference between consecutive terms is equal. Thus: $2 b = a + c$ This implies $b = \frac{a + c}{2}$ .

Step 2: Use the condition for G.P.
If $a, b, c$ are also in Geometric Progression, the square of the middle term is the product of the extremes: $b^{2} = a c$

Step 3: Solve the equations simultaneously.
Substitute the expression for $b$ from the A.P. condition into the G.P. condition: ${(\frac{a + c}{2})}^{2} = a c$ $\frac{a^{2} + 2 a c + c^{2}}{4} = a c$ $a^{2} + 2 a c + c^{2} = 4 a c$ $a^{2} - 2 a c + c^{2} = 0$ This can be factored as a perfect square: ${(a - c)}^{2} = 0 ⟹ a = c$

Step 4: Find $b$ .
Substitute $a = c$ back into the A.P. equation: $2 b = a + a = 2 a ⟹ b = a$ Hence, $a = b = c$ .

Therefore, the correct option is (c) $a = b = c$ .

Q14

Which number should be added to the numbers 3, 7, 15 to make the resulting numbers in G.P.?

Solution

Step 1: Set up the equation.
Let the number to be added be $x$ . The new numbers will be $(3 + x)$ , $(7 + x)$ , and $(15 + x)$ . Since these three numbers are in Geometric Progression (G.P.), the square of the middle term is equal to the product of the first and third terms: ${(7 + x)}^{2} = (3 + x) (15 + x)$

Step 2: Expand and solve for $x$ .
$49 + 14 x + x^{2} = 45 + 3 x + 15 x + x^{2}$ $49 + 14 x + x^{2} = 45 + 18 x + x^{2}$ Subtract $x^{2}$ from both sides: $49 + 14 x = 45 + 18 x$ Bring like terms to one side: $49 - 45 = 18 x - 14 x$ $4 = 4 x ⟹ x = 1$

Check: If $x = 1$ , the numbers become $4, 8, 16$ . Since $8 / 4 = 2$ and $16 / 8 = 2$ , they form a G.P. with a common ratio of $2$ .

Therefore, the correct option is (a) 1.

Q15

The geometric mean between 3 and 12 is:

Solution

Step 1: Recall the formula for Geometric Mean.
The geometric mean (G.M.) of two positive numbers $a$ and $b$ is given by the square root of their product: $G . M . = \sqrt{a \times b}$

Step 2: Calculate the value.
Here, $a = 3$ and $b = 12$ . $G . M . = \sqrt{3 \times 12}$ $G . M . = \sqrt{36} = 6$

Therefore, the correct option is (b) 6.

Q16

The sum of '

n

' terms of two A.P.s are in the ratio

5 n + 4 : 9 n + 6

. Find the ratio of their 18th terms.

Solution

Step 1: Write the sum formulas for the two A.P.s.
Let the first term and common difference of the first A.P. be $a$ and $d$ respectively. Let the first term and common difference of the second A.P. be $A$ and $D$ respectively. $\frac{S_{n}}{S_{n}^{'}} = \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 A + (n - 1) D]} = \frac{5 n + 4}{9 n + 6}$ Cancel out $\frac{n}{2}$ : $\frac{2 a + (n - 1) d}{2 A + (n - 1) D} = \frac{5 n + 4}{9 n + 6}$

Step 2: Relate the sum to the nth term.
Divide the numerator and denominator of the left side by $2$ : $\frac{a + (\frac{n - 1}{2}) d}{A + (\frac{n - 1}{2}) D} = \frac{5 n + 4}{9 n + 6}$ We need to find the ratio of their 18th terms, which is given by: $\frac{t_{18}}{t_{18}^{'}} = \frac{a + 17 d}{A + 17 D}$

Step 3: Solve for $n$ .
Comparing the two expressions, we set the coefficient of $d$ and $D$ to match: $\frac{n - 1}{2} = 17$ $n - 1 = 34 ⟹ n = 35$

Step 4: Substitute $n = 35$ into the right hand side ratio.
$\frac{t_{18}}{t_{18}^{'}} = \frac{5 (35) + 4}{9 (35) + 6}$ $\frac{t_{18}}{t_{18}^{'}} = \frac{175 + 4}{315 + 6} = \frac{179}{321}$

The ratio of their 18th terms is $179 : 321$ .

Q17

The sum of the first two terms of a G.P. is 36 and the product of first term and third term is 9 times the second term. Find the sum of first 8 terms.

Solution

Step 1: Set up the equations for the G.P.
Let the first term of the G.P. be $a$ and the common ratio be $r$ .
The terms are $a, a r, a r^{2}, \dots$
Given, the sum of the first two terms is 36: $a + a r = 36 ⟹ a (1 + r) = 36 --- (Eq. 1)$ Given, the product of the 1st and 3rd term is 9 times the 2nd term: $a \times a r^{2} = 9 (a r)$

Step 2: Solve for $a$ and $r$ .
$a^{2} r^{2} = 9 a r$ Assuming $a \neq 0$ and $r \neq 0$ , divide both sides by $a r$ : $a r = 9$ This means the second term is 9. Substitute $a r = 9$ into Eq. 1: $a + 9 = 36 ⟹ a = 27$ Now find $r$ : $a r = 9 ⟹ 27 r = 9 ⟹ r = \frac{9}{27} = \frac{1}{3}$

Step 3: Calculate the sum of the first 8 terms.
The sum of $n$ terms of a G.P. is $S_{n} = a \frac{1 - r^{n}}{1 - r}$ for $r < 1$ . $S_{8} = 27 \frac{1 - {(\frac{1}{3})}^{8}}{1 - \frac{1}{3}}$ $S_{8} = 27 \frac{1 - \frac{1}{6561}}{\frac{2}{3}}$ $S_{8} = 27 \times \frac{3}{2} \times \frac{6561 - 1}{6561}$ $S_{8} = \frac{81}{2} \times \frac{6560}{6561}$ Notice that $81 \times 81 = 6561$ , so dividing 6561 by 81 leaves 81 in the denominator: $S_{8} = \frac{1}{2} \times \frac{6560}{81} = \frac{3280}{81}$

The sum of the first 8 terms is $\frac{3280}{81}$ .

Q18

Find the sum to

n

terms of the sequence:

7, 77, 777, 7777, \dots

Solution

Step 1: Write the sum as a series.
$S_{n} = 7 + 77 + 777 + 7777 + \dots to n terms$

Step 2: Factor out the common digit.
$S_{n} = 7 (1 + 11 + 111 + 1111 + \dots to n terms)$

Step 3: Multiply and divide by 9.
$S_{n} = \frac{7}{9} (9 + 99 + 999 + 9999 + \dots to n terms)$

Step 4: Rewrite the terms as powers of 10 minus 1.
$S_{n} = \frac{7}{9} [(10 - 1) + (10^{2} - 1) + (10^{3} - 1) + \dots + (10^{n} - 1)]$ Separate the powers of 10 and the ones: $S_{n} = \frac{7}{9} [(10 + 10^{2} + 10^{3} + \dots + 10^{n}) - (1 + 1 + 1 + \dots to n terms)]$

Step 5: Apply the geometric progression sum formula.
The series $10 + 10^{2} + 10^{3} + \dots$ is a G.P. with $a = 10$ and $r = 10$ . Its sum is $10 \frac{10^{n} - 1}{10 - 1} = \frac{10}{9} (10^{n} - 1)$ . Also, adding 1 $n$ times is $n$ . $S_{n} = \frac{7}{9} [\frac{10 (10^{n} - 1)}{9} - n]$ Multiply out the constant: $S_{n} = \frac{70}{81} (10^{n} - 1) - \frac{7 n}{9}$

The sum to

n

terms is $\frac{70}{81} (10^{n} - 1) - \frac{7 n}{9}$ .

Q19

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

Step 1: Choose suitable terms for the G.P.
When the product of three terms of a G.P. is given, it is mathematically convenient to assume the three terms are: $\frac{a}{r}, a, a r$ where $a$ is the middle term and $r$ is the common ratio.

Step 2: Use the product condition to find $a$ .
Given, product = 1 $(\frac{a}{r}) \times a \times (a r) = 1$ $a^{3} = 1 ⟹ a = 1$ So, the three terms are $\frac{1}{r}, 1, r$ .

Step 3: Use the sum condition to find $r$ .
Given, sum = $\frac{39}{10}$ $\frac{1}{r} + 1 + r = \frac{39}{10}$ $\frac{1 + r + r^{2}}{r} = \frac{39}{10}$ Cross-multiply to get a quadratic equation: $10 (1 + r + r^{2}) = 39 r$ $10 + 10 r + 10 r^{2} = 39 r$ $10 r^{2} - 29 r + 10 = 0$ Factorize the quadratic equation: $10 r^{2} - 25 r - 4 r + 10 = 0$ $5 r (2 r - 5) - 2 (2 r - 5) = 0$ $(5 r - 2) (2 r - 5) = 0$ Therefore, $r = \frac{2}{5}$ or $r = \frac{5}{2}$ .

Step 4: Find the terms.
If $r = \frac{5}{2}$ , the terms are $\frac{1}{5 / 2}, 1, \frac{5}{2}$ , which is $\frac{2}{5}, 1, \frac{5}{2}$ .
If $r = \frac{2}{5}$ , the terms are $\frac{1}{2 / 5}, 1, \frac{2}{5}$ , which is $\frac{5}{2}, 1, \frac{2}{5}$ .

Common ratio

r = \frac{5}{2}

\frac{2}{5}

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

Q20

Find four numbers forming a G.P. in which the third term is greater than the first term by 9, and the second term is greater than the 4th term by 18.

Solution

Step 1: Set up the terms and equations.
Let the four numbers in G.P. be $a, a r, a r^{2}, a r^{3}$ .

Condition 1: The 3rd term is greater than the 1st term by 9. $a r^{2} = a + 9$ $a r^{2} - a = 9 ⟹ a (r^{2} - 1) = 9 --- (Eq. 1)$
Condition 2: The 2nd term is greater than the 4th term by 18. $a r = a r^{3} + 18$ $a r - a r^{3} = 18 ⟹ a r (1 - r^{2}) = 18$ Factoring out a negative sign to match Eq. 1: $- a r (r^{2} - 1) = 18 --- (Eq. 2)$

Step 2: Solve for $r$ .
Divide Eq. 2 by Eq. 1: $\frac{- a r (r^{2} - 1)}{a (r^{2} - 1)} = \frac{18}{9}$ $- r = 2 ⟹ r = - 2$

Step 3: Solve for $a$ .
Substitute $r = - 2$ into Eq. 1: $a ({(- 2)}^{2} - 1) = 9$ $a (4 - 1) = 9$ $3 a = 9 ⟹ a = 3$

Step 4: Find the four numbers.
The terms are $a, a r, a r^{2}, a r^{3}$ : $3, 3 (- 2), 3 {(- 2)}^{2}, 3 {(- 2)}^{3}$ $3, - 6, 12, - 24$

The four numbers forming the G.P. are $3, - 6, 12, - 24$ .

Q21

If the AM of two unequal positive real numbers

a

and

b

(

a > b

), be twice as much as their G.M., show that

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

Solution

Step 1: Write the expressions for A.M. and G.M.
For two positive real numbers $a$ and $b$ , the Arithmetic Mean (A.M.) and Geometric Mean (G.M.) are: $A . M . = \frac{a + b}{2}$ $G . M . = \sqrt{a b}$

Step 2: Use the given condition.
It is given that $A . M . = 2 \times G . M .$ $\frac{a + b}{2} = 2 \sqrt{a b}$ $a + b = 4 \sqrt{a b}$

Step 3: Square both sides to form a quadratic equation in terms of ratio.
${(a + b)}^{2} = 16 a b$ $a^{2} + b^{2} + 2 a b = 16 a b$ $a^{2} - 14 a b + b^{2} = 0$ Divide the entire equation by $b^{2}$ : ${(\frac{a}{b})}^{2} - 14 (\frac{a}{b}) + 1 = 0$

Step 4: Solve for $\frac{a}{b}$ using the quadratic formula.
$\frac{a}{b} = \frac{14 \pm \sqrt{{(- 14)}^{2} - 4 (1) (1)}}{2}$ $\frac{a}{b} = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2}$ $\frac{a}{b} = \frac{14 \pm 8 \sqrt{3}}{2} = 7 \pm 4 \sqrt{3}$ Since we are given $a > b$ , the ratio $\frac{a}{b}$ must be greater than 1. Thus, we take the positive root: $\frac{a}{b} = 7 + 4 \sqrt{3}$

Step 5: Verify the required format.
We need to show the ratio is $\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$ . Let's rationalize this target ratio: $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{(2 + \sqrt{3}) (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}$ $= \frac{{(2 + \sqrt{3})}^{2}}{2^{2} - {(\sqrt{3})}^{2}} = \frac{4 + 3 + 4 \sqrt{3}}{4 - 3} = 7 + 4 \sqrt{3}$

Since both expressions equal

7 + 4 \sqrt{3}

, we have successfully proven that $a : b = (2 + \sqrt{3}) : (2 - \sqrt{3})$ .

Q22

Let

S

be the sum,

P

the product and

R

the sum of reciprocals of

n

terms of a G.P. Prove that

P^{2} R^{n} = S^{n}

Solution

Step 1: Express $S$ , $P$ , and $R$ mathematically.
Let the $n$ terms of the G.P. be $a, a r, a r^{2}, \dots, a r^{n - 1}$ .
The sum $S$ of these $n$ terms is: $S = a \frac{(r^{n} - 1)}{r - 1}$ The product $P$ is: $P = a \times a r \times a r^{2} \times \dots \times a r^{n - 1}$ $P = a^{n} r^{1 + 2 + \dots + (n - 1)} = a^{n} r^{\frac{n (n - 1)}{2}}$ The sum of the reciprocals $R$ is: $R = \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} + \dots + \frac{1}{a r^{n - 1}}$ This is a G.P. with the first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$ : $R = \frac{1}{a} \frac{1 - {(\frac{1}{r})}^{n}}{1 - \frac{1}{r}} = \frac{1}{a} \frac{\frac{r^{n} - 1}{r^{n}}}{\frac{r - 1}{r}} = \frac{r^{n} - 1}{a r^{n - 1} (r - 1)}$

Step 2: Compare $S$ and $R$ to find a relation.
Divide $S$ by $R$ : $\frac{S}{R} = \frac{a \frac{(r^{n} - 1)}{r - 1}}{\frac{r^{n} - 1}{a r^{n - 1} (r - 1)}}$ $\frac{S}{R} = a \frac{r^{n} - 1}{r - 1} \times \frac{a r^{n - 1} (r - 1)}{r^{n} - 1} = a^{2} r^{n - 1}$

Step 3: Prove the identity.
Raise the resulting equation $\frac{S}{R} = a^{2} r^{n - 1}$ to the power of $n$ : ${(\frac{S}{R})}^{n} = {(a^{2} r^{n - 1})}^{n}$ $\frac{S^{n}}{R^{n}} = a^{2 n} r^{n (n - 1)}$ Recall our expression for $P$ , and square it: $P^{2} = {(a^{n} r^{\frac{n (n - 1)}{2}})}^{2} = a^{2 n} r^{n (n - 1)}$ Since both $\frac{S^{n}}{R^{n}}$ and $P^{2}$ equal the same expression, we have: $\frac{S^{n}}{R^{n}} = P^{2}$

Cross-multiplying yields the required proof:
$P^{2} R^{n} = S^{n}$

Q23

What will ₹ 5000 amount to in 10 years after it is deposited in a bank which pays annual interest of 8% compounded annually?

Solution

Step 1: Identify the given variables.
Principal amount ( $P$ ) = ₹ 5000
Rate of interest ( $r$ ) = 8% per annum
Number of years ( $n$ ) = 10

Step 2: Apply the compound interest formula.
The amount $A$ after $n$ years compounded annually is: $A = P {(1 + \frac{r}{100})}^{n}$ Substitute the given values into the formula: $A = 5000 {(1 + \frac{8}{100})}^{10}$ $A = 5000 {(1 + 0.08)}^{10}$ $A = 5000 {(1.08)}^{10}$

The amount after 10 years will be $5000 {(1.08)}^{10}$ .

Q24

If the first and the nth term of a G.P. are

a

and

b

, respectively, and if

P

is the product of

n

terms, prove that

P^{2} = {(a b)}^{n}

Solution

Step 1: Write down the known terms of the G.P.
Let the common ratio be $r$ . The G.P. sequence is $a, a r, a r^{2}, \dots, a r^{n - 1}$ .
We are given the $n$ th term is $b$ . $t_{n} = a r^{n - 1} = b$ This gives us the relationship: $r^{n - 1} = \frac{b}{a}$

Step 2: Express the product $P$ .
$P = a \times a r \times a r^{2} \times \dots \times a r^{n - 1}$ $P = a^{n} r^{1 + 2 + \dots + (n - 1)}$ Using the formula for the sum of an arithmetic progression, the exponent of $r$ is: $1 + 2 + \dots + (n - 1) = \frac{(n - 1) n}{2}$ So, $P = a^{n} r^{\frac{n (n - 1)}{2}}$ .

Step 3: Square $P$ and substitute $r^{n - 1}$ .
$P^{2} = {(a^{n} r^{\frac{n (n - 1)}{2}})}^{2}$ $P^{2} = a^{2 n} r^{n (n - 1)}$ This can be rewritten as: $P^{2} = a^{2 n} {(r^{n - 1})}^{n}$ Now substitute $r^{n - 1} = \frac{b}{a}$ : $P^{2} = a^{2 n} {(\frac{b}{a})}^{n}$ $P^{2} = a^{2 n} \frac{b^{n}}{a^{n}}$ $P^{2} = a^{n} b^{n} = {(a b)}^{n}$

Hence proved, $P^{2} = {(a b)}^{n}$ .

Q25

A certain type of bacteria doubles its population every 20 minutes. Assuming no bacteria die, how many bacteria will there be after 3 hours if there are 1 million bacteria at present?

Solution

Step 1: Determine the parameters of the geometric progression.
The initial population of bacteria is $a = 1, 000, 000$ .
The population doubles every 20 minutes, so the common ratio $r = 2$ .
We need to find the population after 3 hours. Let's convert 3 hours to minutes: $3 hours = 3 \times 60 = 180 minutes$

Step 2: Calculate the number of doubling intervals.
The number of 20-minute intervals in 180 minutes is: $n = \frac{180}{20} = 9$ Since the population doubles 9 times, we need to find the term after 9 doublings. The initial population is the 1st term ( $t_{1}$ ), so the population after 9 doublings is the 10th term ( $t_{10}$ ).

Step 3: Calculate the population.
$t_{10} = a r^{9}$ $t_{10} = 1, 000, 000 \times 2^{9}$ Since $2^{9} = 512$ : $t_{10} = 1, 000, 000 \times 512 = 512, 000, 000$

There will be $512$ million bacteria after 3 hours.

Q26

One side of an equilateral triangle is 24 cm. The mid points of its sides are joined to form another triangle whose mid points are joined to form yet another triangle and so on. This process continues indefinitely. Find the sum of the perimeters of all the triangles.

Solution

Step 1: Find the perimeter of the first triangle.
The side of the first equilateral triangle is $a_{1} = 24$ cm.
Its perimeter is $P_{1} = 3 \times 24 = 72$ cm.

Step 2: Find the perimeters of the subsequent triangles.
When the midpoints of the sides of an equilateral triangle are joined, the side of the new triangle formed is half the side of the original triangle. Therefore, the perimeter is also halved.
The side of the second triangle $a_{2} = \frac{24}{2} = 12$ cm.
Its perimeter is $P_{2} = 3 \times 12 = 36$ cm.
Similarly, $P_{3} = 18$ cm, and so on.

Step 3: Sum the infinite geometric progression.
The perimeters form an infinite G.P.: $72, 36, 18, \dots$
The first term $a = 72$ and the common ratio $r = \frac{1}{2}$ .
The sum to infinity is given by $S_{\infty} = \frac{a}{1 - r}$ . $S_{\infty} = \frac{72}{1 - 0.5}$ $S_{\infty} = \frac{72}{0.5} = 144$

The sum of the perimeters of all the triangles is $144 cm$ .

Q27

On a certain day in a hospital, during covid crisis, the patients in the OPD were 1000. Due to efforts of the doctors and health care warriors and precautions taken by general public numbers declined by 50 per day. As per the decline in the number of patients, do you think that there would be a day with no patients in the OPD? If yes, which day would it be from the day when there were 1000 patients?

Solution

Step 1: Identify the progression.
The number of patients on the first day is $1000$ .
The number declines by a constant $50$ each day. This forms an Arithmetic Progression (A.P.).
First term ( $a$ ) = $1000$
Common difference ( $d$ ) = $- 50$

Step 2: Solve for the day with 0 patients.
We need to find the day $n$ when the number of patients ( $a_{n}$ ) becomes $0$ . $a_{n} = a + (n - 1) d$ $0 = 1000 + (n - 1) (- 50)$ $50 (n - 1) = 1000$ $n - 1 = \frac{1000}{50}$ $n - 1 = 20 ⟹ n = 21$

Yes, there will be a day with no patients. It will be the 21st day.

Q28

After striking a floor a certain ball rebounds a fraction of the height from which it has fallen. If the ball is dropped from a height of 240 cm, find the total distance the ball travels before coming to rest.
(Note: The fraction is missing in the original textbook question. Based on the correct answer of 21.6 m, we assume the ball rebounds four-fifths (4/5) of its previous height.)

Solution

Step 1: Set up the geometric progression for total distance.
The ball is dropped from an initial height $h = 240$ cm.
It rebounds to a fraction $f = \frac{4}{5}$ of its height.
The total distance $D$ traveled includes the initial drop and the up-and-down distance of every bounce: $D = h + 2 (h f) + 2 (h f^{2}) + 2 (h f^{3}) + \dots$

Step 2: Simplify the expression.
Factor out $2 h f$ from the subsequent terms: $D = h + 2 h f (1 + f + f^{2} + \dots)$ The series in parentheses is an infinite G.P. with sum $\frac{1}{1 - f}$ . $D = h + \frac{2 h f}{1 - f}$ This can be factored further as: $D = h (1 + \frac{2 f}{1 - f}) = h (\frac{1 - f + 2 f}{1 - f}) = h (\frac{1 + f}{1 - f})$

Step 3: Calculate the value.
Substitute $h = 240$ and $f = \frac{4}{5}$ : $D = 240 (\frac{1 + 4 / 5}{1 - 4 / 5})$ $D = 240 (\frac{9 / 5}{1 / 5})$ $D = 240 \times 9 = 2160 cm$ $D = 21.6 meters$

The total distance the ball travels before coming to rest is $21.6 m$ .

Q29

Suppose a person mails a letter to five of his friends. He asks each one of them to mail it further to five additional friends with instruction that they move the chain further. Assuming the chain is not broken and no person receives the mail more than once, determine the amount spent on postage when the 8th set of letters is mailed, if cost of postage of each letter is 50 paisa.

Solution

Step 1: Write down the geometric progression.
In the 1st set, letters are sent to 5 friends.
In the 2nd set, each of the 5 friends sends it to 5 more, so $5 \times 5 = 25$ letters are sent.
In the 3rd set, $25 \times 5 = 125$ letters are sent.
This forms a Geometric Progression: $5, 25, 125, \dots$
Here, the first term $a = 5$ and the common ratio $r = 5$ .

Step 2: Find the total number of letters mailed up to the 8th set.
We need the sum of the first 8 terms ( $n = 8$ ). $S_{8} = \frac{a (r^{n} - 1)}{r - 1}$ $S_{8} = \frac{5 (5^{8} - 1)}{5 - 1}$ $S_{8} = \frac{5 (390625 - 1)}{4}$ $S_{8} = \frac{5 (390624)}{4}$ $S_{8} = 5 \times 97656 = 488280$ Total number of letters sent is 488,280.

Step 3: Calculate the amount spent on postage.
Cost per letter is 50 paisa, which is ₹ 0.50. $Total Cost = 488280 \times 0.50 = 244140$

The total amount spent on postage is ₹ 244,140.

Q30

Due to reduced taxes an individual has an extra ₹ 30,000 in spendable income. If we assume that an individual spends 70% of this on consumer goods and the producers of these goods in turn spends 70% on consumer goods and this process continues indefinitely. What is the total amount spent on consumer goods.

Solution

Step 1: Identify the amounts spent.
The individual gets an extra ₹ 30,000. He spends 70% of this amount.
First amount spent ( $a$ ) = $30000 \times 0.70 = 21000$ .
The producers receive ₹ 21,000 and spend 70% of it.
Second amount spent = $21000 \times 0.70 = 14700$ .
This forms an infinite Geometric Progression where the first term $a = 21000$ and the common ratio $r = 0.70$ .

Step 2: Calculate the total sum to infinity.
The sum to infinity of a G.P. is given by: $S_{\infty} = \frac{a}{1 - r}$ Substitute the values: $S_{\infty} = \frac{21000}{1 - 0.70}$ $S_{\infty} = \frac{21000}{0.30}$ $S_{\infty} = \frac{210000}{3} = 70000$

The total amount spent on consumer goods is ₹ 70,000.

Q31

A machine depreciates in value by one-fifth each year. If the machine is now worth ₹ 51,000, how much will it be worth 3 years from now?

Solution

Step 1: Understand the depreciation rate.
The machine depreciates by $\frac{1}{5}$ of its value each year. This means it retains $1 - \frac{1}{5} = \frac{4}{5}$ of its value each year.
Present value ( $P$ ) = ₹ 51,000.
Time period ( $n$ ) = 3 years.

Step 2: Apply the depreciation formula.
Value after $n$ years is given by: $V = P {(\frac{4}{5})}^{n}$ $V = 51000 {(\frac{4}{5})}^{3}$ $V = 51000 \times \frac{64}{125}$ $V = 408 \times 64 = 26112$

The machine will be worth ₹ 26,112 three years from now.

Case Studies

Q32

Case Study: An architect is designing a stepped garden for a new eco-friendly corporate building. The garden is designed as a series of terraces. The lowest terrace (Terrace 1) is 20 meters long. Due to the shape of the building, each subsequent upper terrace is 1.5 meters shorter than the one immediately below it. The architect plans to build 12 such terraces.

Based on the above information, answer the following questions:
i. Write the sequence representing the length of the terraces and identify the common difference.
ii. What will be the length of the topmost terrace?
iii. The architect decides to place solar panels along the edge of every terrace. If 1 meter of solar panelling costs ₹500, calculate the total cost of panelling all 12 terraces.
iv. The architect checks the inventory and finds they have materials sufficient to build exactly 118 meters of total terrace length. How many terraces can be constructed using this exact total length?

Solution

i. Sequence and common difference:
The length of the first terrace is $a = 20$ m. Each subsequent terrace is 1.5 m shorter, so the common difference is $d = - 1.5$ m. The sequence is: $20, 18.5, 17, \dots$
Common difference ( $d$ ) = $- 1.5$

ii. Length of the topmost (12th) terrace:
We need to find $a_{12}$ for the A.P. $a_{12} = a + 11 d$ $a_{12} = 20 + 11 (- 1.5) = 20 - 16.5 = 3.5$ Length of topmost terrace = 3.5 m

iii. Total cost of panelling 12 terraces:
First, find the total length of 12 terraces using the sum formula: $S_{12} = \frac{12}{2} (a + a_{12})$ $S_{12} = 6 (20 + 3.5) = 6 \times 23.5 = 141 m$ Total cost = Length × Rate = $141 \times 500$ = ₹ 70,500

iv. Number of terraces for exactly 118 meters:
We set $S_{n} = 118$ and solve for $n$ : $\frac{n}{2} [2 (20) + (n - 1) (- 1.5)] = 118$ $n [40 - 1.5 n + 1.5] = 236$ $41.5 n - 1.5 n^{2} = 236$ Multiply by 2 to clear decimals: $83 n - 3 n^{2} = 472$ $3 n^{2} - 83 n + 472 = 0$ Solving this quadratic equation by factorization or formula: $n = \frac{83 \pm \sqrt{83^{2} - 4 (3) (472)}}{2 (3)} = \frac{83 \pm \sqrt{6889 - 5664}}{6}$ $n = \frac{83 \pm \sqrt{1225}}{6} = \frac{83 \pm 35}{6}$ Valid integer solution: $n = \frac{83 - 35}{6} = \frac{48}{6} = 8$ . Number of terraces = 8

Q33

Case Study: A tech startup launches a new gaming app. On the first day of the launch, they record 10 direct downloads. The game becomes an instant hit, and the number of new daily downloads triples (becomes 3 times) every subsequent day compared to the previous day.

Based on the above information, answer the following questions:
i. Write the geometric progression representing the downloads for the first three days and identify the common ratio (r).
ii. How many new downloads will happen specifically on the 5th day?
iii. The startup management sets a target to achieve a total cumulative download count (sum of all days) of at least 3,500 by the end of the 6th day. Will they achieve this target?
iv. If the trend continues, on which specific day will the daily new downloads cross 5,000 for the first time?

Solution

i. Sequence and common ratio:
First day downloads $a = 10$ .
Since they triple every day, the common ratio $r = 3$ .
The G.P. for the first three days is: $10, 30, 90, \dots$ with $r = 3$ .

ii. New downloads on the 5th day:
We need to find the 5th term ( $t_{5}$ ): $t_{5} = a r^{4}$ $t_{5} = 10 \times 3^{4} = 10 \times 81 = 810$ 810 downloads

iii. Cumulative downloads by the 6th day:
We need to find the sum of the first 6 terms ( $S_{6}$ ): $S_{6} = \frac{a (r^{6} - 1)}{r - 1}$ $S_{6} = \frac{10 (3^{6} - 1)}{3 - 1} = \frac{10 (729 - 1)}{2} = 5 \times 728 = 3640$ Since $3640 \geq 3500$ , Yes, they will achieve the target.

iv. Day when daily downloads cross 5,000:
We set $t_{n} > 5000$ : $10 \times 3^{n - 1} > 5000 ⟹ 3^{n - 1} > 500$ Checking powers of 3: $3^{5} = 243$ and $3^{6} = 729$ . Thus, $n - 1 = 6 ⟹ n = 7$ . 7th day

Q34

Case Study: The Rebounding Ball Experiment
A physics student drops a highly elastic superball from the roof of a building 80 meters high. Every time the ball hits the ground, it rebounds to 3/4 (or 75%) of the height from which it fell. The ball continues to bounce until it eventually comes to rest.

Based on the above information, answer the following questions:
i. What is the height reached by the ball after the first rebound?
ii. Calculate the specific height the ball reaches after the 2nd rebound.
iii. Calculate the total vertical distance the ball travels before coming to rest.
iv. The student repeats the experiment with a different ball (a tennis ball) dropped from the same 80m height. This tennis ball is less elastic and only rebounds to half of its previous height. Calculate the total distance this new ball travels before coming to rest.

Solution

i. Height after first rebound:
Initial height $h = 80$ m. Rebound fraction $f = \frac{3}{4}$ .
Height $h_{1} = 80 \times \frac{3}{4} =$ 60 m

ii. Height after 2nd rebound:
$h_{2} = h_{1} \times \frac{3}{4} = 60 \times \frac{3}{4} =$ 45 m

iii. Total vertical distance for superball:
The total distance $D$ includes the initial drop and the up-and-down of all rebounds. $D = h + 2 (h_{1} + h_{2} + \dots)$ This forms an infinite G.P. with sum $\frac{h_{1}}{1 - f}$ . $D = 80 + 2 (\frac{60}{1 - 0.75})$ $D = 80 + 2 (\frac{60}{0.25}) = 80 + 2 (240) = 80 + 480 =$ 560 m

iv. Total vertical distance for tennis ball:
Here, $h = 80$ m and $f = \frac{1}{2}$ . The first rebound $h_{1} = 40$ m. $D = 80 + 2 (\frac{40}{1 - 0.5})$ $D = 80 + 2 (\frac{40}{0.5}) = 80 + 2 (80) = 80 + 160 =$ 240 m

Q35

Case Study: A financial analyst is comparing the growth of two different stocks over two years to determine their volatility and average performance.
Stock A: Started at ₹100, went to ₹150 in Year 1, and ₹225 in Year 2.
Stock B: Two specific growth values, a and b, are analyzed. The analyst notes that the Arithmetic Mean (A.M.) of these two values is 25, and their Geometric Mean (G.M.) is 20.

Based on the above information, answer the following questions:
i. For Stock A, verify if the prices form a G.P. If so, find the value of r.
ii. Write the relationship inequality that always holds true between A.M. and G.M. for distinct positive numbers.
iii. For Stock B, find the two specific values a and b given that their A.M. is 25 and G.M. is 20.
iv. Using the two values found in Q3, the analyst wants to create a linear growth projection. Insert 2 Arithmetic Means between these two values to find the intermediate price targets.

Solution

i. Verify G.P. for Stock A:
The prices are $100, 150, 225$ .
$\frac{150}{100} = 1.5$ and $\frac{225}{150} = 1.5$ .
Since the ratios are constant, Yes, they form a G.P. with $r = 1.5$ .

ii. Inequality relationship:
For any two distinct positive numbers, the Arithmetic Mean is strictly greater than the Geometric Mean.
$A . M . > G . M .$

iii. Values of $a$ and $b$ for Stock B:
Given $A . M . = \frac{a + b}{2} = 25 ⟹ a + b = 50$ .
Given $G . M . = \sqrt{a b} = 20 ⟹ a b = 400$ .
We form the quadratic equation $x^{2} - (a + b) x + a b = 0$ : $x^{2} - 50 x + 400 = 0$ $(x - 10) (x - 40) = 0$ The two values are 10 and 40.

iv. Inserting 2 Arithmetic Means:
Let the two means be $A_{1}$ and $A_{2}$ between 10 and 40. The sequence $10, A_{1}, A_{2}, 40$ forms an A.P.
Here, the 4th term $a_{4} = 40$ and first term $a = 10$ . $a + 3 d = 40 ⟹ 10 + 3 d = 40 ⟹ 3 d = 30 ⟹ d = 10$ $A_{1} = 10 + 10 = 20$ $A_{2} = 20 + 10 = 30$ The intermediate targets are 20 and 30.

Q36

Case Study: A construction company buys a large crane for ₹5,000,000 (50 Lakhs). The value of the machine depreciates at a rate of 20% per annum on the diminishing value.

Based on the given information, answer the following questions:
i. What will be the value of the machine after 1 year?
ii. Write the expression for calculating the value of the machine after n years.
iii. Calculate the estimated value of the machine at the end of the 4th year.
iv. The company plans to sell the machine as scrap when its value drops below ₹1,500,000. Will they sell it after the 5th year or the 6th year?

Solution

i. Value after 1 year:
Initial value $P = 50, 00, 000$ . Depreciation rate = 20%, so it retains 80% (or $0.8$ ) of its value. $V_{1} = 5000000 \times 0.8 =$ ₹ 40,00,000

ii. Expression after n years:
Using the formula for compound depreciation: $V = P {(1 - \frac{r}{100})}^{n}$ $V_{n} = 5000000 {(0.8)}^{n}$

iii. Value at the end of 4th year:
$V_{4} = 5000000 {(0.8)}^{4}$ Since ${0.8}^{4} = 0.4096$ : $V_{4} = 5000000 \times 0.4096 =$ ₹ 20,48,000

iv. When to sell the machine:
We need to check when the value falls below ₹ 15,00,000.
Value after 5th year: $V_{5} = V_{4} \times 0.8 = 2048000 \times 0.8 = 16, 38, 400$ .
Value after 6th year: $V_{6} = V_{5} \times 0.8 = 1638400 \times 0.8 = 13, 10, 720$ .
The value drops below 15 Lakhs after the 6th year. They will sell it after the 6th year.

Assertion Reason Questions

Q37

Assertion (A): If three positive real numbers

x, y, z

are in Geometric Progression (G.P.), then

2 \log y = \log x + \log z

.
Reason (R): If three numbers

a, b, c

are in Geometric progression, then

2 b = a + c

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Solution

Analyze Assertion (A):
If $x, y, z$ are in G.P., then $y^{2} = x z$ . Taking logarithms on both sides yields: $\log (y^{2}) = \log (x z) ⟹ 2 \log y = \log x + \log z$ Thus, Assertion (A) is True.

Analyze Reason (R):
If $a, b, c$ are in G.P., the correct relation is $b^{2} = a c$ . The relation $2 b = a + c$ is the condition for numbers to be in an Arithmetic Progression (A.P.), not a G.P.
Thus, Reason (R) is False.

Therefore, the correct option is (c) A is true but R is false.

Q38

Assertion (A): The sum of the infinite series

3 + \frac{9}{2} + \frac{27}{4} + \frac{81}{8} + \dots

- 2

.
Reason (R): The sum to infinity of a geometric series is defined if and only if the absolute value of the common ratio

| r | < 1

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Solution

Analyze Assertion (A):
The given series has a first term $a = 3$ . The common ratio $r = \frac{9 / 2}{3} = \frac{3}{2} = 1.5$ .
Since the absolute value of the common ratio $| r | > 1$ , the sum to infinity diverges and is not defined. The formula $\frac{a}{1 - r}$ cannot be applied here.
Thus, Assertion (A) is False.

Analyze Reason (R):
The sum to infinity for a geometric progression converges and has a defined limit only when $| r | < 1$ . This statement is mathematically sound.
Thus, Reason (R) is True.

Therefore, the correct option is (d) A is false but R is true.

Q39

Assertion (A): If the first term of an A.P. is 5 and the common difference is 2, the sum of the first 10 terms is 140.
Reason (R): The sum of the first

n

odd natural numbers is given by

n^{2}

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Solution

Analyze Assertion (A):
We use the A.P. sum formula: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
Given $a = 5$ , $d = 2$ , and $n = 10$ : $S_{10} = \frac{10}{2} [2 (5) + (10 - 1) 2] = 5 [10 + 18] = 5 \times 28 = 140$ Thus, Assertion (A) is True.

Analyze Reason (R):
The sum of the first $n$ odd natural numbers ( $1 + 3 + 5 + \dots$ ) is indeed $n^{2}$ . This is a known mathematical fact.
Thus, Reason (R) is True.

Relationship:
While both statements are true, Reason (R) applies specifically to the sequence of odd numbers starting from 1. The sequence in Assertion (A) is $5, 7, 9, \dots$ . You can evaluate A using the general A.P. sum formula, but R is not its direct general explanation (even if it's tangentially related by summing odds offset by 1 and 3).

Therefore, the correct option is (b) Both A and R are true but R is not the correct explanation of A.

Q40

Assertion (A): If

a, b, c

are in G.P., then

a^{2}, b^{2}, c^{2}

are also in G.P.
Reason (R): If three numbers are in G.P., their squares, cubes, or any equal powers are also in G.P.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Solution

Analyze Assertion (A):
If $a, b, c$ are in G.P., then $b^{2} = a c$ . Squaring this relation gives ${(b^{2})}^{2} = a^{2} c^{2}$ , which means $a^{2}, b^{2}, c^{2}$ form a G.P.
Thus, Assertion (A) is True.

Analyze Reason (R):
For any sequence in G.P. with a common ratio $r$ , raising the terms to a power $k$ produces a new sequence with a common ratio $r^{k}$ . This means that any equal powers of a G.P. will also be in a G.P.
Thus, Reason (R) is True, and it provides the exact general property that explains Assertion (A).

Therefore, the correct option is (a) Both A and R are true and R is the correct explanation of A.

Q41

Assertion (A): The sum of the series

5 + 55 + 555 + \dots

n

terms cannot be calculated directly using the formula of sum of n terms of G.P.
Reason (R): The terms of the series

5, 55, 555, \dots

do not have a constant common ratio between consecutive terms.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Solution

Analyze Reason (R):
Let's evaluate the ratio between consecutive terms of the given series:
Ratio 1: $\frac{55}{5} = 11$
Ratio 2: $\frac{555}{55} \approx 10.09$
Since $11 \neq 10.09$ , there is no constant common ratio. The series is not a direct Geometric Progression.
Thus, Reason (R) is True.

Analyze Assertion (A):
Because the series is not a G.P., the direct formula for the sum of $n$ terms of a G.P. ( $S_{n} = a \frac{r^{n} - 1}{r - 1}$ ) cannot be applied straight away. (We must manipulate it first by factoring out the digit and converting it into a difference of a G.P. and an A.P.).
Thus, Assertion (A) is True.

Relationship:
The exact reason we cannot directly apply the G.P. sum formula is that the terms fail to maintain a constant common ratio, as explained in Reason (R).

Therefore, the correct option is (a) Both A and R are true and R is the correct explanation of A.

Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-6

Chapter 6: Sequence and Series

Check Your Progress 6.1

Check Your Progress 6.2

Chapter 6: Sequence and Series

Practice Exercise (Q1 - Q10)

Chapter 6: Sequence and Series

Practice Exercise (Q11 to Q30)

Case Studies

Assertion Reason Questions