Chapter 8 Sequence and Series

NCERT Solutions Class 11 Maths Chapter 8 – Sequence and Series

NCERT Solutions for Class 11 Maths Chapter 8 (Sequence and Series) is the study of numbers in some order having a particular rule. If you have studied about basic patterns in your class 10, then in this chapter you will study about Geometric Progressions (GP) and the relationship between Arithmetic Mean (AM) and Geometric Mean (GM). These concepts are the building blocks for understanding limits, calculus and financial mathematics.

ALLEN’s NCERT Solutions for Class 11 Maths Chapter 8 have been prepared by expert faculty with logical derivations and clear examples to make the transition from basic sequences to complex series easier. The answers concentrate on finding the “general term” of a series, which is the answer to the sum of any series.

It is very important to master the summation of Geometric Progressions and Special Series for Competitive Exams. Such solutions provide a rigorous framework for solving problems about infinite GP and properties of means, which are often the subject of high-level mathematics papers.

1.0Class 11 Maths Chapter 8 : Key Concepts

This chapter focuses on different types of progressions and the mathematical techniques used to sum their terms. Key lessons include:

Sequences and Series: Understanding a sequence as a function whose domain is the set of natural numbers, and a series as the sum of the terms of a sequence.
Arithmetic Progression (AP): Reviewing the sequence where the difference between consecutive terms is constant.
- General term: $a_{n} = a + (n - 1) d$
- Sum of n terms: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Geometric Progression (GP): Learning sequences where the ratio of consecutive terms is constant (common ratio 'r').
- General term: $a_{n} = a r^{n - 1}$
- Sum of n terms: $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} f orr \neq = 1.$
General Term and Sum of Infinite GP: Understanding why an infinite GP converges when |r| < 1: $S_{\infty} = \frac{a}{1 - r}$
Arithmetic Mean (AM) and Geometric Mean (GM):
- $A M = \frac{a + b}{2}$
- $GM = ab$
Relationship between AM and GM: Proving that for any two positive real numbers, $A M \geq GM$ .

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 8

EXERCISE - 8.1 Write the first five terms of the sequences in Q. 1 to 6 whose $n^{th}$ term are :

1. $a_{n} = n (n + 2)$ . Sol. Given, $a_{n} = n (n + 2)$ Putting $n = 1, 2, 3, 4$ , and 5, we get

$a_{1} = 1 (1 + 2) = 3; a_{3} = 3 (3 + 2) = 15; a_{5} = 5 (5 + 2) = 35 a_{2} = 2 (2 + 2) = 8; a_{4} = 4 (4 + 2) = 24$

Therefore, the first five terms are $3, 8, 15, 24$ , and 35. 2. $a_{n} = \frac{n}{n + 1}$ Sol. Given, $a_{n} = \frac{n}{n + 1}$ Putting, $n = 1, 2, 3, 4, 5$ , we get

$a_{1} = \frac{1}{1 + 1} = \frac{1}{2}; a_{3} = \frac{3}{3 + 1} = \frac{3}{4}; a_{5} = \frac{5}{5 + 1} = \frac{5}{6} a_{2} = \frac{2}{2 + 1} = \frac{2}{3}; a_{4} = \frac{4}{4 + 1} = \frac{4}{5};$

Therefore, the first five terms are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}$ , $\frac{4}{5}$ and $\frac{5}{6}$ 3. $a_{n} = 2^{n}$ Sol. Given, $a_{n} = 2^{n}$ Putting $n = 1, 2, 3, 4, 5$ , we get

$a_{1} = 2^{1} = 2; a_{3} = 2^{3} = 8; a_{5} = 2^{5} = 32 a_{2} = 2^{2} = 4; a_{4} = 2^{4} = 16;$

Therefore, the first five terms are $2, 4, 8, 16$ , and 32. 4. $a_{n} = \frac{2 n - 3}{6}$ . Sol. Given, $a_{n} = \frac{2 n - 3}{6}$

Putting $n = 1, 2, 3, 4, 5$ , we get

$a_{1} = \frac{2 \times 1 - 3}{6} = \frac{- 1}{6}; a_{3} = \frac{2 \times 3 - 3}{6} = \frac{1}{2}; a_{5} = \frac{2 \times 5 - 3}{6} = \frac{7}{6} a_{2} = \frac{2 \times 2 - 3}{6} = \frac{1}{6}; a_{4} = \frac{2 \times 4 - 3}{6} = \frac{5}{6};$

Therefore, the first five terms are $\frac{- 1}{6}, \frac{1}{6}, \frac{1}{2}$ ,

$\frac{5}{6} and \frac{7}{6} .$

5. $a_{n} = (- 1)^{n - 1} 5^{n + 1}$ Sol. Given, $a_{n} = (- 1)^{n - 1} 5^{n + 1}$ Putting $n = 1, 2, 3, 4, 5$ , we get

$a_{1} = (- 1)^{1 - 1} 5^{1 + 1} = (- 1)^{0} \cdot 5^{2} = 25 a_{2} = (- 1)^{2 - 1} 5^{2 + 1} = (- 1)^{1} \cdot 5^{3} = - 125 a_{3} = (- 1)^{3 - 1} 5^{3 + 1} = (- 1)^{2} \cdot 5^{4} = 625 a_{4} = (- 1)^{4 - 1} 5^{4 + 1} = (- 1)^{3} \cdot 5^{5} = - 3125 a_{5} = (- 1)^{5 - 1} 5^{5 + 1} = (- 1)^{4} \cdot 5^{6} = 15625$

Therefore, the five terms are $25, - 125, 625$ , -3125, and 15625. 6. $a_{n} = n \frac{n ^{2} + 5}{4}$ Sol. Given, $a_{n} = n \frac{n ^{2} + 5}{4}$ Putting $n = 1, 2, 3, 4, 5$ , we get

$a_{1} = 1 \cdot \frac{1 ^{2} + 5}{4} = \frac{6}{4} = \frac{3}{2}; a_{2} = 2 \cdot \frac{2 ^{2} + 5}{4} = 2 \cdot \frac{9}{4} = \frac{9}{2}; a_{3} = 3 \cdot \frac{3 ^{2} + 5}{4} = 3 \cdot \frac{14}{4} = \frac{21}{2}; a_{4} = 4 \cdot \frac{4 ^{2} + 5}{4} = 4 \cdot \frac{16 + 5}{4} = 21; a_{5} = 5 \cdot \frac{5 ^{2} + 5}{4} = 5 \cdot \frac{30}{4} = \frac{75}{2};$

Therefore, the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}$ , 21 and $\frac{75}{2}$

Find the indicated terms in each of the sequence in Q. 7 to 10 whose $n^{th}$ term are :

7. $a_{n} = 4 n - 3; a_{17}, a_{24}$ Sol. Given, $a_{n} = 4 n - 3$ Putting $n = 17$ , we get

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

Putting $n = 24$ , we get

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

8. $a_{n} = \frac{n ^{2}}{2 n}; a_{7}$ Sol. Given, $a_{n} = \frac{n ^{2}}{2 n}$ Putting $n = 7$ , we get

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

9. $a_{n} = (- 1)^{n - 1} n^{3}; a_{9}$ Sol. Given, $a_{n} = (- 1)^{n - 1} n^{3}$ Putting, $n = 9$ , we get

$a_{9} = (- 1)^{9 - 1} (9)^{3} = (- 1)^{8} \cdot (9)^{3} = 729$

10. $a_{n} = \frac{n ( n - 2 )}{n + 3}; a_{20}$ Sol. Given, $a_{n} = \frac{n ( n - 2 )}{n + 3}$ Putting, $n = 20$ , we get

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

Write the first five terms of each of the sequences in Q. 11 to 13 and get the corresponding series:

11. $a_{1} = 3, a_{n} = 3 a_{n - 1} + 2$ for all $n > 1$ Sol. Given, $a_{1} = 3, a_{n} = 3 a_{n - 1} + 2$ for all $n > 1$

Putting $n = 2, 3, 4$ and 5, we get

$a_{2} = 3 a_{2 - 1} + 2 = 3 a_{1} + 2 = 3 (3) + 2 = 11; a_{3} = 3 a_{3 - 1} + 2 = 3 a_{2} + 2 = 3 (11) + 2 = 35 a_{4} = 3 a_{4 - 1} + 2 = 3 a_{3} + 2 = 3 (35) + 2 = 107; a_{5} = 3 a_{5 - 1} + 2 = 3 a_{4} + 2 = 3 (107) + 2 = 323$

Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323. The corresponding series is

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

12. $a_{1} = - 1, a_{n} = \frac{a _{n - 1}}{n}, n \geq 2$ Sol. Given, $a_{1} = - 1, a_{n} = \frac{a _{n - 1}}{n}, n \geq 2$ Putting $n = 2, 3, 4$ and 5 we get

$a_{2} = \frac{a _{2 - 1}}{2} = \frac{a _{1}}{2} = \frac{- 1}{2}; a_{4} = \frac{a _{4 - 1}}{4} = \frac{a _{3}}{4} = \frac{- 1}{24}; a_{3} = \frac{a _{3 - 1}}{3} = \frac{a _{2}}{3} = \frac{- 1}{6}; a_{5} = \frac{a _{5 - 1}}{5} = \frac{a _{4}}{5} = \frac{- 1}{120}$

Hence, the first five terms of the sequence are

$- 1, \frac{- 1}{2}, \frac{- 1}{6}, \frac{- 1}{24} and \frac{- 1}{120} .$

The corresponding series is

$(- 1) + (\frac{- 1}{2}) + (\frac{- 1}{6}) + (\frac{- 1}{24}) + (\frac{- 1}{120}) + \dots ..$

13. $a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 1, n > 2$ Sol. Given, $a_{1} = a_{2} = 2, a_{n} = a_{n - 1} - 1, n > 2$ Putting $n = 3, 4$ and 5 we get

$a_{3} = a_{3 - 1} - 1 = a_{2} - 1 = 2 - 1 = 1; a_{4} = a_{4 - 1} - 1 = a_{3} - 1 = 1 - 1 = 0; a_{5} = a_{5 - 1} - 1 = a_{4} - 1 = 0 - 1 = - 1$

Hence, the first five terms of the sequence are 2, 2, 1, 0, and -1. The corresponding series is

$2 + 2 + 1 + 0 + (- 1) + \dots$

14.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 $n = 1, 2, 3, 4, 5$ Sol. Given, $1 = a_{1} = a_{2}$

$a_{n} = a_{n - 1} + a_{n - 2}, n > 2$

Putting $n = 1, 2, 3, 4$ and 5 we get

$a_{3} = a_{3 - 1} + a_{3 - 2} = a_{2} + a_{1} = 1 + 1 = 2 a_{4} = a_{4 - 1} + a_{4 - 2} = a_{3} + a_{2} = 2 + 1 = 3; a_{5} = a_{5 - 1} + a_{5 - 2} = a_{4} + a_{3} = 3 + 2 = 5; a_{6} = a_{6 - 1} + a_{6 - 2} = a_{5} + a_{4} = 5 + 3 = 8$

For $n = 1, \frac{a _{n + 1}}{a _{n}} = \frac{a _{2}}{a _{1}} = \frac{1}{1} = 1$ For $n = 2, \frac{a _{n + 1}}{a _{n}} = \frac{a _{3}}{a _{2}} = \frac{2}{1} = 2$ For $n = 3, \frac{a _{n + 1}}{a _{n}} = \frac{a _{4}}{a _{3}} = \frac{3}{2}$ For $n = 4, \frac{a _{n + 1}}{a _{n}} = \frac{a _{5}}{a _{4}} = \frac{5}{3}$ For $n = 5, \frac{a _{n + 1}}{a _{n}} = \frac{a _{6}}{a _{5}} = \frac{8}{5}$ Hence, the terms are $1, 2, \frac{3}{2}, \frac{5}{3}$ and $\frac{8}{3}$

3.0EXERCISE - 8.2

1.Find the $2 0^{th}$ and $n^{th}$ terms of the G.P. $\frac{5}{2}$ , $\frac{5}{4}$ , $\frac{5}{8}$ , ...... Sol. The given G.P. is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots\dots$ Here, $a =$ First term $= \frac{5}{2}$

$r = common ratio = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{1}{2}$

$a_{20} = ar^{20 - 1} = \frac{5}{2} (\frac{1}{2})^{19} = \frac{5}{( 2 ) ( 2 ) ^{19}} = \frac{5}{( 2 ) ^{20}}$ and $a_{n} = ar^{n - 1} = \frac{5}{2} (\frac{1}{2})^{n - 1} = \frac{5}{( 2 ) ( 2 ) ^{n - 1}} = \frac{5}{( 2 ) ^{n}}$ 2. Find the $1 2^{th}$ term of a G.P. whose $8^{th}$ term is 192 and the common ratio is 2. Sol. Let a be the first term of the G.P. Common ratio, $r = 2$ and $a_{8} = 192$

$∴ \Rightarrow ∴ a_{8} = a r^{8 - 1} = a r^{7} = 192 (∵ a_{r} = a r^{n - 1}) a r^{7} = 192 a (2)^{7} = 192 \Rightarrow a = \frac{192}{128} = \frac{3}{2} a_{12} = a r^{12 - 1} = (\frac{3}{2}) (2)^{11} = (3) (2)^{10} = 3072$

3.The $5^{th}, 8^{th}$ and $11^{th}$ terms of a G.P. are p, q and s, respectively. Show that $q^{2} = ps$ . Sol. Let a be the first term and r be the common ratio of the G.P. According to the question,

$a_{5} = a r^{5 - 1} = a r^{4} = p a_{8} = a r^{8 - 1} = a r^{7} = q a_{11} = a r^{11 - 1} = a r^{10} = s$

Dividing equation (2) by (1), we get

$\frac{ar ^{7}}{ar ^{4}} = \frac{q}{p} \Rightarrow r^{3} = \frac{q}{p}$

Dividing equation (3) by (2), we get

$\frac{ar ^{10}}{ar ^{7}} = \frac{s}{q} \Rightarrow r^{3} = \frac{s}{q}$

Equating the values of $r^{3}$ geted in (4) and (5), we get

$\frac{q}{p} = \frac{s}{q} \Rightarrow q^{2} = ps$

Hence proved. 4. The $4^{th}$ term of a G.P. is square of its second term, and the first term is -3 . Determine its $7^{th}$ term. Sol. Let a be the first term and r be the common ratio of the G.P. Given

$∴ \Rightarrow ∴ a = - 3 and a_{4} = (a_{2})^{2} a r^{3} = (a r)^{2} \Rightarrow a r^{3} = a^{2} r^{2} r = a \Rightarrow r = - 3 (∵ a = - 3) a_{7} = a r^{7 - 1} a r^{6} = (- 3) (- 3)^{6} = - (3)^{7} = - 2187$

Thus, the seventh term of the G.P. is -2187. 5. Which term of the following sequences:

(a) $2, 22, 4, \dots$ . is 128 ? (b) $3, 3, 33, \dots .$ . is 729? (c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots .$ . is $\frac{1}{19683}$ ?

Sol.

(a) The given sequence is $2, 22, 4, \dots$ . is 128 . Here, $a = 2$ and $r = \frac{( 2 2 )}{2} = 2$ and $a_{n} = 128$ $(∵ a_{n} = a r^{n - 1})$ $\Rightarrow (2) (2)^{n - 1} = 128$ $\Rightarrow (2)^{n - 1} = \frac{128}{2} \Rightarrow (2)^{n - 1} = 64$ $\Rightarrow (2)^{\frac{n - 1}{2}} = 2^{6} \Rightarrow \frac{n - 1}{2} = 6$ $\Rightarrow n - 1 = 12 \Rightarrow n = 13$

Thus, the $1 3^{th}$ term of the given sequence is 128.

(b) The given sequence is $3, 3, 33, \dots .. a = 3$ and $r = \frac{3}{3} = 3$ and $a_{n} = 729$ $(∵ a_{n} = a r^{n - 1})$ $\Rightarrow (3) (3)^{n - 1} = 729$ $\Rightarrow (3)^{\frac{1}{2}} (3)^{\frac{n - 1}{2}} = (3)^{6} \Rightarrow (3)^{\frac{1}{2} + \frac{n - 1}{2}} = (3)^{6}$ $∴ \frac{1}{2} + \frac{n - 1}{2} = 6$ $\Rightarrow \frac{1 + n - 1}{2} = 6 \Rightarrow n = 12$

Thus, the $1 2^{th}$ term of the given sequence is 729 .

(c) The given sequence is $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ . Here, a $= \frac{1}{3}$ are $r = \frac{1}{9} \div \frac{1}{3} = \frac{1}{3}$ and

$a_{n} = \frac{1}{19683}$

$\Rightarrow (\frac{1}{3}) (\frac{1}{3})^{n - 1} = \frac{1}{19683} (∵ a_{n} = ar^{n - 1})$ $\Rightarrow (\frac{1}{3})^{n} = (\frac{1}{3})^{9} \Rightarrow n = 9$

Thus, the $9^{th}$ term of the given sequence is $\frac{1}{19683}$

6.For what values of x, the numbers $\frac{2}{7}, x, - \frac{7}{2}$ are in G.P? Sol. The given numbers are $\frac{2}{7}, x, - \frac{7}{2}$ in G.P. Common ratio $= \frac{x}{\frac{- 2}{7}} = \frac{- 7 x}{2}$ Also, common ratio $= \frac{\frac{- 7}{2}}{x} = \frac{- 7}{2 x}$ $∴ \frac{- 7 x}{2} = \frac{- 7}{2 x} \Rightarrow x^{2} = \frac{- 2 \times 7}{- 2 \times 7} = 1 \Rightarrow x = \pm 1$ Thus, for $x = \pm 1$ , the given numbers will be in G.P. Find the sum to 20 terms in the geometric progression Q. 7 to 10

7.0.15, 0.015, 0.0015 ... Sol. The given G.P. is 0.15, 0.015, 0.00015 ... Here, $a = 0.15$ and $r = \frac{0.015}{0.15} = 0.1$ $∴ S_{20} = \frac{0.15 [ 1 - ( 0.1 ) ^{20} ]}{1 - 0.1}$ $(∵ S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ , when r < 1) $= \frac{0.15}{0.9} [1 - (0.1)^{20}] = \frac{15}{90} [1 - (0.1)^{20}]$ $= \frac{1}{6} [1 - (0.1)^{20}]$ 8. $7, 21, 37, \dots\dots$ Sol. The given G.P. is $7, 21, 37, \dots\dots$ . Here, $a = 7$ and $r = \frac{21}{7} = 3$ $\Rightarrow S_{n} = \frac{7 [ ( 3 ) ^{n} - 1 ]}{3 - 1}$ $(∵ S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ , when r $> 1)$ $\Rightarrow S_{n} = \frac{7 [ ( 3 ) ^{n} - 1 ]}{3 - 1} \times \frac{3 + 1}{3 + 1}$ $\Rightarrow S_{n} = \frac{7 ( 3 + 1 ) [ ( 3 ) ^{n} - 1 ]}{3 - 1}$ $\Rightarrow S_{n} = \frac{7 ( 3 + 1 ) [ ( 3 ) ^{n} - 1 ]}{2}$

9. $1, - a, a^{2}, - a^{3} \dots .$ . (if $a \neq = - 1$ ) Sol. The given G.P. is $1, - a, a^{2}, - a^{3}$ ..... Here, first term $= a = 1$ Common ratio $= r = - a$

$(∵ S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}, where r < 1)$

$∴ S_{n} = \frac{[ 1 - ( - a ) ^{n} ]}{1 - ( - a )} = \frac{[ 1 - ( - a ) ^{n} ]}{1 + a}$ 10. $x^{3}, x^{5}, x^{7} \dots .$ . (if $x \neq = \pm 1$ ) Sol. The given G.P. is $x^{3}, x^{5}, x^{7} \dots .$ . Here, $a = x^{3}$ and $r = \frac{x ^{5}}{x ^{3}} = x^{2}$

$S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{x ^{3} [ 1 - ( x ^{2} ) ^{n} ]}{1 - x ^{2}} = \frac{x ^{3} ( 1 - x ^{2 n} )}{1 - x ^{2}}$

11.Evaluate $\sum_{k = 1}^{11} (2 + 3^{k})$ Sol. Given, $\sum_{k = 1}^{11} (2 + 3^{k}) = \sum_{k = 1}^{11} (2) + \sum_{k = 1}^{11} 3^{k}$

$= 2 (11) + k = 1 \sum 11 3^{k} = 22 + k = 1 \sum 11 3^{k}$

$\sum_{k = 1}^{11} 3^{k} = 3^{1} + 3^{2} + 3^{3} + \dots .. + 3^{11}$

Here, $a = 3$ and $r = 3$ $\Rightarrow S_{11} = \frac{3 [ ( 3 ) ^{11} - 1 ]}{3 - 1} (∵ S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ , when r > 1 $)$ $\Rightarrow S_{11} = \frac{3}{2} (3^{11} - 1)$ $∴ \sum_{k = 1}^{11} 3^{k} = \frac{3}{2} (3^{11} - 1)$ Putting this value in equation (1), we get

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

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

$\frac{a}{r} + a + a r = \frac{39}{10} (\frac{a}{r}) (a) (a r) = 1 \Rightarrow a^{3} = 1$

$\Rightarrow a = 1$ (Considering real roots only) Putting a = 1 in equation (1), we get

$\frac{1}{r} + 1 + r = \frac{39}{10}$

$\Rightarrow 1 + r + r^{2} = \frac{39}{10} r$ $\Rightarrow 10 + 10 r + 10 r^{2} - 39 r = 0$ $\Rightarrow 10 r^{2} - 29 r + 10 = 0$ $\Rightarrow 10 r^{2} - 25 r - 4 r + 10 = 0$ $\Rightarrow 5 r (2 r - 5) - 2 (2 r - 5) = 0$ $\Rightarrow (5 r - 2) (2 r - 5) = 0$ $\Rightarrow r = \frac{2}{5}$ or $\frac{5}{2}$ when $r = \frac{2}{5}$ , thus the three terms of G.P. are $\frac{5}{2}$ , 1 and $\frac{2}{5}$ when $r = \frac{5}{2}$ , thus the three terms of G.P. are $\frac{2}{5}$ , 1 and $\frac{5}{2}$

13.How many terms of G.P. 3, $3^{2}, 3^{3} \dots$ are needed to give the sum 120? Sol. The given G.P. is $3, 3^{2}, 3^{3} \dots$ and $S_{n} = 120$ . Here, $a = 3$ and $r = 3$ $∴ S_{n} = 120 = \frac{3 ( 3 ^{n} - 1 )}{3 - 1}$

$(∵ S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}, when r > 1)$

$\Rightarrow 120 = \frac{3 ( 3 ^{n} - 1 )}{2} \Rightarrow \frac{120 \times 2}{3} = 3^{n} - 1$ $\Rightarrow 3^{n} - 1 = 80 \Rightarrow 3^{n} = 81$ $\Rightarrow 3^{n} = 3^{4} ∴ n = 4$

Thus, four terms of the given G.P. are required to get the sum as 120.

14.The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. Sol. Let the G.P. be a, ar, $ar^{2}, ar^{3}, \dots$ According to question,

$a + ar + ar^{2} = 16$

$\Rightarrow a (1 + r + r^{2}) = 16$ and $ar^{3} + ar^{4} + ar^{5} = 128$ $\Rightarrow ar^{3} (1 + r + r^{2}) = 128$

Dividing equation (2) by (1), we get

$\frac{ar ^{3} ( 1 + r + r ^{2} )}{a ( 1 + r + r ^{2} )} = \frac{128}{16}$

$\Rightarrow r^{3} = 8 ∴ r = 2$ Putting $r = 2$ in (1), we get

$a (1 + 2 + 4) = 16$

$\Rightarrow a = \frac{16}{7}$ $\Rightarrow S_{n} = \frac{16}{7} \frac{( 2 ^{n} - 1 )}{2 - 1} = \frac{16}{7} (2^{n} - 1)$

$(∵ S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}, when r > 1)$

15.Given a G.P. with $a = 729$ and $7^{th}$ term 64, determine $S_{7}$ . Sol. Given, $a = 729, a_{7} = 64$

$a_{7} = a r^{7 - 1} = (729) r^{6} (∵ a_{n} = a r^{n - 1}) \Rightarrow 64 = 729 r^{6} \Rightarrow r^{6} = \frac{64}{729} = (\frac{2}{3})^{6} \Rightarrow r = \frac{2}{3} ∴ S_{7} = \frac{729 [ 1 - ( \frac{2}{3} ) ^{7} ]}{1 - \frac{2}{3}} = 3 \times 729 [1 - (\frac{2}{3})^{7}] (∵ S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}, when r < 1) = (3)^{7} [\frac{( 3 ) ^{7} - ( 2 ) ^{7}}{( 3 ) ^{7}}] = (3)^{7} - (2)^{7} = 2187 - 128 = 2059$

16.Find a G.P. for which sum of the first two terms is - 4 and the fifth term is 4 times the third term. Sol. Let a be the first term and r be the common ratio of the G.P. According to question,

$S_{2} = - 4 = \frac{a ( 1 - r ^{2} )}{1 - r} a_{5} = 4 \times a_{3} \Rightarrow ar^{4} = 4 ar^{2}$

$\Rightarrow r^{2} = 4$ $∴ r = \pm 2$ From (1), we get

$- 4 = \frac{a [ 1 - ( 2 ) ^{2} ]}{1 - 2} for r = 2$

$\Rightarrow - 4 = \frac{a ( 1 - 4 )}{- 1} \Rightarrow - 4 = a (3)$ $\Rightarrow a = \frac{- 4}{3}$

Also, $- 4 = \frac{a [ 1 - ( - 2 ) ^{2} ]}{1 - ( - 2 )}$ for $r = - 2$ $\Rightarrow - 4 = \frac{a ( 1 - 4 )}{1 + 2} \Rightarrow - 4 = \frac{a ( - 3 )}{3}$ $\Rightarrow a = 4$ Thus, the required G.P. is $\frac{- 4}{3}, \frac{- 8}{3}, \frac{- 16}{3}$ or 4, -8, 16, -32 17. If the $4^{th}, 1 0^{th}$ and $1 6^{th}$ terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. Sol. Let a be the first term and r be the common ratio of the G.P. According to question,

$a_{4} = a r^{3} = x a_{10} = a r^{9} = y a_{16} = a r^{15} = z$

Dividing (2) by (1), we get

$\frac{y}{x} = \frac{ar ^{9}}{ar ^{3}} \Rightarrow \frac{y}{x} = r^{6}$

Dividing (3) by (2), we get

$\frac{z}{y} = \frac{ar ^{15}}{ar ^{9}} \Rightarrow \frac{z}{y} = r^{6}$

$∴ \frac{y}{x} = \frac{z}{y}$ Thus, x, y, z are in G. P. 18. Find the sum to n terms of the sequence, 8, 88, 888, 8888... Sol. The given sequence is $8, 88, 888, 8888 \dots$

$S_{n} = 8 + 88 + 888 + 8888 + \dots\dots . to n terms = \frac{8}{9} [9 + 99 + 999 + 9999 + \dots .. to n terms] = \frac{8}{9} [(10 - 1) + (1 0^{2} - 1) + (1 0^{3} - 1) + (1 0^{4} - 1) + \dots .. to n terms] = \frac{8}{9} [(10 + 1 0^{2} + \dots .. n terms) - (1 + 1 + 1 + \dots .. n terms)] = \frac{8}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{10 - 1} - n] = \frac{8}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{80}{81} (1 0^{n} - 1) - \frac{8}{9} n$

19.Find the sum of the products of the corresponding terms of the sequences

$2, 4, 8, 16, 32 and 128, 32, 8, 2, \frac{1}{2} .$

Sol. Required sum

$= 2 \times 128 + 4 \times 32 + 8 \times 8 + 16 \times 2 + 32 \times \frac{1}{2} = 256 + 128 + 64 + 32 + 16 are in G.P.$

First term, $a = 256$

$Common ratio, r = \frac{128}{256} = \frac{1}{2} and n = 5 ∴ S_{5} = \frac{4 [ 1 - ( \frac{1}{2} ) ^{5} ]}{1 - \frac{1}{2}} = \frac{4 [ 1 - \frac{1}{32} ]}{\frac{1}{2}} (∵ S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}, where r < 1) = 8 (\frac{32 - 1}{32}) = \frac{31}{4}$

$∴$ Required sum $= 64 (\frac{31}{4}) = (16) (31) = 496$ 20. Show that the products of the corresponding terms of the sequences form $a, ar, ar^{2}, \dots . ar^{n - 1}$ and A, AR, AR $^{2}$ , ..... AR $^{n - 1}$ a G.P, and find the common ratio. Sol. It has to be proved that the sequence: aA, arAR, $ar^{2} AR^{2}, \dots ar^{n - 1} AR^{n - 1}$ , forms a G.P. Now we have $\frac{Second term}{First term} = \frac{arAR}{aA} = rR$ Similiary, $\frac{Third term}{Second term} = \frac{ar ^{2} AR ^{2}}{arAR} = rR$ Thus, the above sequence forms a G.P. and the common ratio is rR. 21. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the $4^{th}$ by 18 . Sol. Let a be the first term and r be the common ratio of the G.P.

$a_{1} = a, a_{2} = a r, a_{3} = a r^{2}, a_{4} = a r^{3}$

According to question

$\Rightarrow \Rightarrow a_{3} = a_{1} + 9 a^{2} = a + 9 a_{2} = a_{4} + 18 a r = a r^{3} + 18$

From (1) and (2), we get

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

Dividing (4) by (3), we get

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

Putting the value of r in (1), we get

$4 a = a + 9 \Rightarrow 3 a = 9$

$∴ a = 3$ Thus, the first four numbers of the G.P. are 3, $3 (- 2), 3 (- 2)^{2}$ , and $3 (- 2)^{3}$ i.e., $3, - 6, 12$ , and -24. 22. If $p^{th}, q^{th}$ and $r^{th}$ terms of a G.P. are a, b and c, respectively. Prove that $a^{q - r} . b^{r - p} . c^{p - q} = 1$ . Sol. Let A be the first term and R be the common ratio of the G.P. According to question

$A R^{p - 1} = a, A R^{q - 1} = b, A R^{r - 1} = c a^{q - r} \cdot b^{r - p} \cdot c^{p - q} = A^{q - r} \times R^{(p - 1) (q - r)} \times A^{r - p} \times R^{(q - 1) (r - p)} \times A^{p - q} \times R^{(r - 1) (p - q)} = A^{q - r + r - p + p - q} \times R^{(p r - p r - q + r) + (r q - r + p - pq) + (p r - p - q r + q)} = A^{0} \times R^{0} = 1$

Thus, the given result is proved. 23. If the first and the $n^{th}$ term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that $P^{2} = (ab)^{n}$ . Sol. The first term of the G.P is a and the last term is $b$ . Therefore, the G.P. is a, ar, $ar^{2}, ar^{3} \dots ar^{n - 1}$ , where $r$ is the common ratio.

$b = ar^{n - 1}$

$P = Product of n terms = (a) (ar) (ar^{2}) \dots (ar^{n - 1}) = (a \times a \times \dots a) (r \times r^{2} \times \dots r^{n - 1}) = an r 1 + 2 + \dots (n - 1) \dots (2)$

Here, $1, 2, \dots (n - 1)$ is an A.P.

$∴ 1 P + 2 + \dots\dots\dots + (n - 1) = \frac{n - 1}{2} [2 + (n - 1 - 1) \times 1] = \frac{n - 1}{2} [2 + n - 2] = \frac{n ( n - 1 )}{2} = a^{n} r^{\frac{n ( n - 1 )}{2}}$

$∴ P^{2} = a^{2 n} r^{n (n - 1)} = [a^{2} r^{(n - 1)}]^{n} = [a \times a r^{n - 1}]^{n} = (ab)^{n} [Using (1)]$

Thus, the given result is proved. 24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n + 1)^{th}$ to $(2 n)^{th}$ term is $\frac{1}{r ^{n}}$ . Sol. Let a be the first term and r be the common ratio of the G.P. Sum of the first n terms $= \frac{a ( 1 - r ^{n} )}{( 1 - r )}$ Since there are n terms from ( $n + 1)^{th}$ to $(2 n)^{th}$ term. Sum of terms from $(n + 1)^{th}$ to $(2 n)^{th}$ term

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

Thus, Required ratio $= \frac{a ( 1 - r ^{n} )}{( 1 - r )} \times \frac{( 1 - r )}{ar ^{n} ( 1 - r ^{n} )} = \frac{1}{r ^{n}}$ Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n + 1)^{th}$ to (2n) $^{th}$ term is $\frac{1}{r ^{n}}$ .

25.If a, b, c and d are in G.P. show that:

$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + b c - c d)^{2}$

Sol. a, b, c, d are in G.P. Therefore,

$bc = ad b^{2} = ac c^{2} = bd$

It has to be proved that,

$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + b c - c d)^{2}$

R.H.S.

$= (ab + b c + c d)^{2} = (ab + a d + c d)^{2} = [ab + d (a + c)]^{2} = a^{2} b^{2} + 2 ab d (a + c) + d^{2} (a + c)^{2} = a^{2} b^{2} + 2 a^{2} b d + 2 a c b d + d^{2} (a^{2} + 2 a c + c^{2}) = a^{2} b^{2} + 2 a^{2} c^{2} + 2 b^{2} c^{2} + d^{2} a^{2} + 2 d^{2} b^{2} + d^{2} c^{2}$

[Using (1) and (2)]

$= a^{2} b^{2} = a^{2} b^{2} + a^{2} c^{2} + a^{2} c^{2} + b^{2} c^{2} + b^{2} c^{2} + d^{2} a^{2} + d^{2} b^{2} + d^{2} b^{2} + d^{2} c^{2} + a^{2} c^{2} + a^{2} d^{2} + b^{2} \times b^{2} + b^{2} c^{2} + b^{2} d^{2} + c^{2} b^{2} + c^{2} \times c^{2} + c^{2} d^{2}$

[Using (2) and (3) and rearranging terms]

$= a^{2} (b^{2} + c^{2} + d^{2}) = (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) + b^{2} (b^{2} + c^{2} + d^{2}) + c^{2} (b^{2} + c^{2} + d^{2}) = L.H.S.$

$∴$ L.H.S. = R.H.S. $∴ (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc - cd)^{2}$ 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P. Sol. Let $G_{1}$ and $G_{2}$ be two numbers between 3 and 81 such that the series, $3, G_{1}, G_{2}, 81$ , forms a G.P. Let a be the first term and r be the common ratio of the G.P. $∴ 81 = (3) (r) 3$ $\Rightarrow r^{3} = 27$ $∴ r = 3$ (Taking real roots only)

For $r = 3$ ,

$G_{1} = ar = (3) (3) = 9 G_{2} = ar^{2} = (3) (3)^{2} = 27$

Thus, the required two numbers are 9 and 27.

27.Find the value of n so that $\frac{a ^{n + 1} + b ^{n + 1}}{a ^{n} + b ^{n}}$ may be the geometric mean between a and b . Sol. M of a and b is $ab$ . By the given condition : $\frac{a ^{n + 1} + b ^{n + 1}}{a ^{n} + b ^{n}}$ Squaring both sides, we get

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

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

$\Rightarrow a^{2 n + 2} + 2 a^{n + 1} b^{n + 1} + b^{2 n + 2} = a^{2 n + 1} b + 2 a^{n + 1} b^{n + 1} + a b^{2 n + 1}$

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

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

$\Rightarrow 2 n + 1 = 0$

$\Rightarrow n = \frac{- 1}{2}$

28.The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $(3 + 22) : (3 - 22)$ . Sol. Let the two numbers be a and b.

$G.M. = ab$

According to the given condition,

$a + b = 6 ab$

$\Rightarrow (a + b)^{2} = 36 (ab)$

$Also, (a - b)^{2} = (a + b)^{2} - 4 ab = 36 ab - 4 ab = 32 ab$

$\Rightarrow a - b = 32 ab = 42 ab$

Adding (1) and (2), we get

$2 a = (6 + 42) ab$

$\Rightarrow a = (3 + 22) ab$

Putting the value of a in (1), we get

$b = 6 ab - (3 + 22) ab$

⇒

$b = (3 - 22) ab \frac{a}{b} = \frac{( 3 + 2 2 ) ab}{( 3 - 2 2 ) ab} = \frac{3 + 2 2}{3 - 2 2}$

Thus, the required ratio is

$(3 + 22) : (3 - 22)$

29.If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A \pm (A + G) (A - G)$ Sol. It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

$∴ AM = A = \frac{a + b}{2} GM = G = ab$

Putting the value of a and b from (3) and (4) in the identity

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

we get

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

From (3) and (5), we get

$2 a = 2 A + 2 (A + G) (A - G)$

$\Rightarrow a = A + (A + G) (A - G)$

Putting the value of a in (3), we get

$b = 2 A - A - (A + G) (A - G) = A - (A + G) (A - G)$

Thus, the two numbers are

$A \pm (A + G) (A - G)$

30.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 $2^{nd}$ hour, $4^{th}$ hour and $n^{th}$ hour? Sol. It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P. Here, $a = 30$ and $r = 2$ $∴ a_{3} = ar^{2} = (30) (2)^{2} = 120$

Therefore, the number of bacteria at the end of $2^{nd}$ hour will be 120 .

$a_{5} = a r^{4} = (30) (2)^{4} = 480$

The number of bacteria at the end of $4^{th}$ hour will be 480 .

$a_{n + 1} = a r^{n} = (30) 2^{n}$

Thus, number of bacteria at the end of $n^{th}$ hour will be 30(2) $^{n}$ .

31.What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually? Sol. The amount deposited in the bank is ₹500. At the end of first year, amount

$= ₹500 (1 + \frac{1}{10}) = ₹500 (1.1)$

At the end of $2^{nd}$ year, amount = ₹500 (1.1) (1.1) At the end of $3^{rd}$ year, amount = ₹500 (1.1) (1.1) (1.1) and so on ∴ Amount at the end of 10 years

$= ₹500 (1.1) (1.1) \dots (10 times) = ₹500 (1.1)^{10}$

32.If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then get the quadratic equation. Sol. Let the root of the quadratic equation be a and b.

According to the given condition,

$A.M. = \frac{a + b}{2} = 8 \Rightarrow a + b = 16$

$G.M. = ab = 5 \Rightarrow ab = 25$

The quadratic equation is given by,

$x^{2} - x (Sum of roots) + (Product of roots) = 0 x^{2} - x (a + b) + (ab) = 0 x^{2} - 16 x + 25 = 0 [Using (1) and (2)]$

Thus, the required quadratic equation is

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

4.0EXERCISE - 8.3 (SUPPLEMENTARY)

Find the sum to infinity in each of the following Geometric Progression.

1. $1, \frac{1}{3}, \frac{1}{9}, \dots .$ . Sol. Let $S = 1, \frac{1}{3}, \frac{1}{9}, \dots$ ...

$Here, a = 1 and r = \frac{1}{3} (∵ r = \frac{\frac{1}{3}}{1} = \frac{1}{3})$

So,

$S = \frac{1}{1 - \frac{1}{3}} [Using S_{\infty} = \frac{a}{1 - r}] = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5$

2.6, 1. 2, 0.24 ..... Sol. Let $S = 6, 1.2, 0.24 + \dots$ ..

$Here, a = 6 and r = 0.2 (∵ r = \frac{1.2}{6} = 0.2)$

So,

$S = \frac{6}{1 - 0.2} [Using S_{\infty} = \frac{a}{1 - r}] = \frac{6}{0.8} = 7.5$

3. $5, \frac{20}{7}, \frac{80}{49}, \dots .$ . Sol. Let $S = 5, \frac{20}{7}, \frac{80}{49}, \dots .$ .

$Here, a = 5 and r = \frac{4}{7} (∵ r = \frac{20}{\frac{7}{5}} = \frac{4}{7})$

So, $S = \frac{5}{1 - \frac{4}{7}}$

$[Using S_{\infty} = \frac{a}{1 - r}]$

$= \frac{35}{3}$

4. $\frac{- 3}{4}, \frac{3}{16}, \frac{- 3}{64}, \dots .$ . Sol. Let $S = \frac{- 3}{4}, \frac{3}{16}, \frac{- 3}{64}, \dots .$ .

$Here, a = \frac{- 3}{4} and r = \frac{- 1}{4} (∵ r = \frac{\frac{3}{16}}{- \frac{3}{4}} = - \frac{1}{4}) So, S = \frac{\frac{- 3}{4}}{1 - ( - \frac{1}{4} )} [Using S_{\infty} = \frac{a}{1 - r}] = \frac{\frac{- 3}{4}}{1 + \frac{1}{4}} = \frac{- 3}{4} \times \frac{4}{5} = \frac{- 3}{5}$

5.Prove that : $3^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \dots$ .. Sol. LHS $= 3^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} \dots$ .

$= 3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots\dots}$

[Power of 3 is in the form of a GP with

$a = \frac{1}{2} and r = \frac{1}{2}] = 3^{(\frac{\frac{1}{2}}{1 - \frac{1}{2}})} [Using S_{\infty} = \frac{a}{1 - r}] = 3^{(\frac{\frac{1}{2}}{\frac{1}{2}})} = 3^{1} = 3 (∵ r = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2})$

6.Let $x = 1 + a + a^{2} +$ ..... and $y = 1 + b + b^{2}$

$+ \dots .., where ∣ a ∣ < 1 and ∣ b ∣ < 1.$

Prove that: $1 + ab + a^{2} b^{2} + \dots\dots = \frac{xy}{x + y - 1}$ Sol. Here $x = 1 + a + a^{2} +$ .....

$= \frac{1}{1 - a} [Using S_{\infty} = \frac{a}{1 - r}]$

Ans

$y = 1 + b + b^{2} + \dots . = \frac{1}{1 - b} [Using S_{\infty} = \frac{a}{1 - r}]$

$RHS = \frac{x y}{x + y - 1} = \frac{( \frac{1}{1 - a} ) ( \frac{1}{1 - b} )}{( \frac{1}{1 - a} ) + ( \frac{1}{1 - b} ) - 1} = \frac{\frac{1}{( 1 - a ) ( 1 - b )}}{\frac{1 - b + 1 - a - ( 1 - a ) ( 1 - b )}{( 1 - a ) ( 1 - b )}} = \frac{1}{( 1 - a ) ( 1 - b )} \times \frac{( 1 - a ) ( 1 - b )}{2 - a - b - 1 + a + b - ab} = \frac{1}{1 - ab}$

From (1) and (2)

$1 + ab + a^{2} b^{2} + \dots . = \frac{x y}{x + y - 1}$

Hence Proved

5.0MISCELLANEOUS EXERCISE

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

$for all x f (1) = 3$

Taking $x = y = 1$ in (1), we get $f (1 + 1) = f (2) = f (1) f (1) = 3 \times 3 = 9$ Similarly,

$f (1 + 2) = f (3) = f (1) f (2) = 3 \times 9 = 27 f (1 + 3) = f (4) = f (1) f (3) = 3 \times 27 = 81$

$∴ f (1), f (2), f (3), \dots .$ . , that is $3, 9, 27, \dots .$ . , forms a G.P. with both the first term and common ratio equal to 3 . It is known that, $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ It is given that, $\sum_{x = 1}^{n} f (x) = 120$

$∴ \Rightarrow \Rightarrow \Rightarrow ∴ f (1) + f (2) + f (3) + \dots\dots .. + f (n) = 120 3 + 9 + 27 + 81 + \dots\dots .. + n terms = 120 120 = \frac{3 ( 3 ^{n} - 1 )}{3 - 1} 120 = \frac{3}{2} (3^{n} - 1) 3^{n} - 1 = 80 3^{n} = 81 = 3^{4} n = 4$

Thus, the value of n is 4. 2. 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. Sol. Given, $a = 5, r = 2$ and $S_{n} = 315$ It is given that the first term a is 5 and common ratio r is 2.

$∴ 315 = \frac{5 ( 2 ^{n} - 1 )}{2 - 1} [∵ S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}]$

$\Rightarrow 2^{n} - 1 = 63$ $\Rightarrow 2^{n} = 64 = (2)^{6}$ $\Rightarrow n = 6$ ∴ Last term of the $G . P = 6^{th}$ term

$= ar^{6 - 1} = (5) (2)^{5} = (5) (32) = 160$

Thus, the last term of the G.P. is 160. 3. 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. Sol. Let a and r be the first term and the common ratio of the G.P. respectively.

$∴ a = 1, a_{3} = ar^{2} = r^{2}, a_{5} = ar^{4} = r^{4}$ $∴ r^{2} + r^{4} = 90$ $\Rightarrow r^{4} + r^{2} - 90 = 0$ Thus, the common ratio of the G.P. is ±3. $\Rightarrow r^{2} = \frac{- 1 + 1 + 360}{2} = \frac{- 1 \pm 361}{2}$

$= \frac{- 1 \pm 19}{2} = - 10 or 9$

$∴ r = \pm 3$ (Taking real roots) Thus, the common ratio of the the G.P. is ± 3.

4.The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we get an arithmetic progression. Find the numbers. Sol. Let the three numbers in G.P. be a, ar, and $ar^{2}$ . From the given condition,

$a + ar + ar^{2} = 56$

$\Rightarrow a (1 + r + r^{2}) = 56$ $a - 1, ar - 7, ar^{2} - 21$ forms an A.P. $∴ (ar - 7) - (a - 1) = (a r^{2} - 21) - (a r - 7)$ $\Rightarrow a r - a - 6 = a r^{2} - a r - 14$ $\Rightarrow a r^{2} - 2 a r + a = 8$ $\Rightarrow a r^{2} - a r - a r + a = 8$ $\Rightarrow a (r^{2} + 1 - 2 r) = 8$ $\Rightarrow a (r - 1)^{2} = 8$ From (1) and (2), we get $\Rightarrow 7 (r^{2} - 2 r + 1) = 1 + r + r^{2}$ $\Rightarrow 7 r^{2} - 14 r + 7 - 1 - r - r^{2} = 0$ $\Rightarrow 6 r^{2} - 15 r + 6 = 0$ $\Rightarrow 6 r^{2} - 12 r - 3 r + 6 = 0$ $\Rightarrow 6 r (r - 2) - 3 (r - 2) = 0$ $\Rightarrow (6 r - 3) (r - 2) = 0$ When $r = 2, a = 8$ When Therefore, when $r = 2$ , the three numbers in G.P. are 8, 16, and 32. When, $r = 1/2$ , the three numbers in G.P. are 32, 16, and 8. Thus, in either case, the three required numbers are 8, 16, and 32. 5. 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. Sol. Let the G.P. be $T_{1}, T_{2}, T_{3}, T_{4} \dots T_{2 n}$ . Number of terms $= 2 n$ According to the given condition,

$\Rightarrow \Rightarrow T_{1} + T_{2} + T_{3} + \dots + T_{2 n} = 5 [T_{1} + T_{3} + \dots + T_{2 n - 1}] T_{1} + T_{2} + T_{3} + \dots + T_{2 n} - 5 [T_{1} + T_{3} + \dots + T_{2 n - 1}] = 0 T_{2} + T_{4} + \dots + T_{2 n} = 4 [T_{1} + T_{3} + \dots + T_{2 n - 1}]$

Let the G.P. be a, ar, ar $^{2}, ar^{3} \dots$ $∴ \frac{ar ( r ^{n} - 1 )}{r - 1} = \frac{4 \times a ( r ^{n} - 1 )}{r - 1}$ $\Rightarrow ar = 4 a \Rightarrow r = 4$ Thus, the common ratio of the G.P. is 4. 6. $\frac{If - a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - ds} (x \neq = 0)$ then show that a, b, c and d are in G.P. Sol. It is given that, $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx}$

$\Rightarrow \Rightarrow (a + b x) (b - c x) = (b + c x) (a - b x) ab - a c x + b^{2} x - b c x^{2} = ab - b^{2} x + a c x - b c x^{2}$

$\Rightarrow 2 b^{2} x = 2 acx$

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

Also, $\frac{b + cx}{b - cx} = \frac{c + dx}{c - ds}$

$\Rightarrow \Rightarrow (b + cx) (c - dx) = (b - cx) (c + dx) bc - bdx + c^{2} x - cdx^{2} = bc + bdx - c^{2} x - cdx^{2}$

$\Rightarrow 2 c^{2} x = 2 bdx$

$\Rightarrow c^{2} = bd \Rightarrow \frac{c}{d} = \frac{d}{c}$

From (1) and (2), we get

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

Thus, a, b, c and d are in G.P. 7. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $P^{2} R^{n} = S^{n}$

Sol. Let the G.P. be a, ar, $ar^{2}, ar^{3} \dots ar^{n - 1}$ According to the given information,

$P = a^{n} \times r^{1 + 2 + \dots\dots + n - 1} = a^{n} r^{\frac{n ( 1 - 1 )}{2}}$

$[∵$ Sum of first n natural numbers is $\frac{n ( n - 1 )}{2}]$

$R = \frac{1}{a} + \frac{1}{a r} + \dots . + \frac{1}{a r ^{n - 1}} = \frac{r ^{n - 1} + r ^{n - 2} + \dots .. r + 1}{a r ^{n - 1}} [∵ S = \frac{a ( r ^{n} - 1 )}{r - 1}] = \frac{1 ( r ^{n} - 1 )}{( r - 1 )} \times \frac{1}{a r ^{n - 1}} [∵ 1, r, .. r^{n - 1} forms a G.P.] = \frac{r ^{n} - 1}{a ^{n - 1} ( r - 1 )}$

$∴ P^{2} R^{n} = a^{2 n} r^{n (n - 1)} \frac{( r ^{n} - 1 ) ^{n}}{a ^{n} r ^{n (n - 1)} ( r - 1 ) ^{n}}$

$= \frac{a ^{n} ( r ^{n} - 1 ) ^{n}}{( r - 1 ) ^{n}} = [\frac{a ( r ^{n} - 1 )}{( r - 1 )}]^{n} = S^{n}$

Hence proved, $P^{2} R^{n} = S^{n}$ 8. 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. Sol. It is given that $a, b, c$ , and d are in G.P.

$∴ b^{2} = a c c^{2} = b d a d = b c$

It has to be proved that $(a^{n} + b^{n}), (b^{n} + c^{n})$ , $(c^{n} + d^{n})$ are in G.P. i.e.,

$(b^{n} + c^{n})^{2} = (a^{n} + b^{n}) (c^{n} + d^{n})$

Consider L.H.S.

$(b^{n} + c^{n})^{2} = b^{2 n} + 2 b^{n} c^{n} + c^{2 n} = (b^{2})^{n} + 2 b^{n} c^{n} + (c^{2})^{n} = (a c)^{n} + 2 b^{n} c^{n} + (b d)^{n}$

[Using (1) and (2)]

$= a^{n} c^{n} + b^{n} c^{n} + b^{n} c^{n} + b^{n} d^{n} = a^{n} c^{n} + b^{n} c^{n} + a^{n} d^{n} + b^{n} d^{n} [Using (3)] = c^{n} (a^{n} + b^{n}) + d^{n} (a^{n} + b^{n}) = (a^{n} + b^{n}) (c^{n} + d^{n})$

= R.H.S. $∴ (b^{n} + c^{n})^{2} = (a^{n} + b^{n}) (c^{n} + d^{n})$ Thus, $(a^{n} + b^{n}), (b^{n} + c^{n})$ , and $(c^{n} + d^{n})$ are in G.P. 9. If a and b are the roots of $x^{2} - 3 x + p = 0$ and c, d are roots of $x^{2} - 12 x + q = 0$ , where a, b, c, d, form a G.P. Prove that $(q + p) : (q - p) = 17 : 15$ . Sol. It is given that a and b are the roots of $x^{2} - 3 x + p = 0$ $∴ a + b = 3$ and $ab = p$

Also, c and d are the roots of

$∴ c + d = 12$ and $cd = q$

It is given that $a, b, c, d$ are in G.P. Let $a = x, b = xr, c = xr^{2}, d = xr^{3}$ From (1) and (2), we get

$x + x r = 3$

$\Rightarrow x (1 + r) = 3$ $xr^{2} + xr^{3} = 12$ $\Rightarrow xr^{2} (1 + r) = 12$ On dividing, we get

$\frac{xr ^{2} ( 1 + r )}{x ( 1 + r )} = \frac{12}{3}$

$\Rightarrow r^{2} = 4$ $\Rightarrow r = \pm 2$ When $r = 2, x = \frac{3}{1 + 2} = \frac{3}{3} = 1$ When $r = - 2, x = \frac{3}{1 - 2} = \frac{3}{- 1} = - 3$ Case I: When $r = 2$ and $x = 1$ ,

$p = ab = x^{2} r = 2 q = c d = x^{2} r^{5} = 32$

$∴ \frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15}$ i.e. $(q + p) : (q - p) = 17 : 15$

Case II: When $r = - 2, x = - 3, ab = x^{2} r = - 18$

$cd = x^{2} r^{5} = - 288$

$∴ \frac{q + p}{q - p} = \frac{- 288 - 18}{- 288 + 18} = \frac{- 306}{- 270} = \frac{17}{15}$

i.e. $(q + p) : (q - p) = 17 : 15$ Thus, in both the cases, we get

$(q + p) : (q - p) = 17 : 15$

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

Sol. Let the two numbers be a and b.

$A.M = \frac{a + b}{2} and G.M. = ab$

According to question,

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

$\Rightarrow \frac{( a + b ) ^{2}}{4 ( ab )} = \frac{m ^{2}}{n ^{2}} \Rightarrow (a + b)^{2} = \frac{4 ab m ^{2}}{n ^{2}}$

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

Using this in the identity $(a - b)^{2} = (a + b)^{2} - 4 ab$ , we get

$(a - b)^{2} = \frac{4 ab m ^{2}}{n ^{2}} - 4 ab = \frac{4 ab ( m ^{2} - n ^{2} )}{n ^{2}}$

$\Rightarrow (a - b) = \frac{2 ab m ^{2} - n ^{2}}{n}$

Adding (1) and (2), we get

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

$\Rightarrow a = \frac{ab}{n} (m + m^{2} - n^{2})$

Putting the value of a in (1), we get

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

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

Thus, $a : b = (m + m^{2} - n^{2}) : (m - m^{2} - n^{2})$ 11. Find the sum of the following series up to n terms:

(i) $5 + 55 + 555 + \dots$ (ii) $0.6 + .66 + .666 + \dots$

Sol.

(i) $5 + 55 + 555 + \dots$ Let $S_{n} = 5 + 55 + 555 + \dots .$ . to n terms

$= \frac{5}{9} [9 + 99 + 999 + \dots .. to n terms] = \frac{5}{9} [(10 - 1) + (1 0^{2} - 1) + (1 0^{3} - 1) +$

$= \frac{5}{9} [(10 + 1 0^{2} + 1 0^{3} \dots .. to n terms) - (1 + 1 + \dots .. n terms)] = \frac{5}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{10 - 1} - n] [∵ S = \frac{a ( r ^{n} - 1 )}{r - 1}, r > 1] = \frac{5}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{50}{81} (1 0^{n} - 1) - \frac{5 n}{9}$

(ii) . $6 + .66 + .666 + \dots$ Let $S_{n} = 0.6 + 0.66 + 0.666 + \dots$ .to n terms

$= \frac{6}{9} [0.1 + 0.11 + 0.111 + \dots .. to terms] = \frac{6}{9} [0.9 + 0.99 + 0.999 + \dots .. to terms] = \frac{6}{9} [(1 - \frac{1}{10}) + (1 - \frac{1}{1 0 ^{2}}) + (1 - \frac{1}{1 0 ^{3}}) + \dots . to n terms] = \frac{2}{3} [(1 + 1 + \dots . n terms) - \frac{1}{10} (1 + \frac{1}{10} + \frac{1}{1 0 ^{2}} + \dots . to n terms)]$

$= \frac{2}{3} [n - \frac{1}{10} (\frac{1 - ( \frac{1}{10} ) ^{n}}{1 - \frac{1}{10}})] [∵ S = \frac{a ( 1 - r ^{n} )}{1 - r}, r < 1]$

$= \frac{2}{3} n - \frac{2}{30} \times \frac{10}{9} (1 - 1 0^{- n}) = \frac{2}{3} n - \frac{2}{27} (1 - 1 0^{- n})$

12.Find the $2 0^{th}$ term of the series $2 \times 4 + 4 \times 6$

$+ 6 \times 8 + \dots + n terms.$

Sol. The given series is $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots n$ terms

$n^{th} term = a_{n} = 2 n \times (2 n + 2) = 4 n^{2} + 4 n a_{20} = 4 (20)^{2} + 4 (20) = 4 (400) + 80 = 1600 + 80 = 1680$

Thus, the $2 0^{th}$ term of the series is 1680. 13. A farmer buys a used tractor for ₹12000. He pays ₹6000 cash and agrees to pay the balance in annual installments of ₹500 plus 12% interest on the unpaid amount. How much will be the tractor cost him? Sol. It is given that the farmer pays ₹6000 in cash. Therefore, unpaid amount

$= ₹ 12000 - ₹ 6000 = Rs 6000$

According to the given condition, the interest paid annually is

$12% of 6000, 12% of 5500, 12% of 5000 \dots 12% of 500$

Thus, total interest to be paid

$= 12% of 6000 + 12% of 5500 + 12% of 5000 + \dots + 12% of 500 = 12% of (6000 + 5500 + 5000 + \dots + 500) = 12% of (500 + 1000 + 1500 + \dots + 6000)$

Now, the series $500, 1000, 1500 \dots 6000$ is an A.P. with both the first term and common difference equal to 500.

Let the number of terms of the A.P. be n.

$∴ \Rightarrow \Rightarrow 6000 = 500 + (n - 1) 500 1 + (n - 1) = 12 n = 12$

∴ Sum of the A.P

$= \frac{12}{2} [2 (500) + (12 + 1) (500)] = 6 [1000 + 5500] = 6 (6500) = 39000$

Thus, total interest to be paid

$= 12% of (500 + 1000 + 1500 + \dots + 6000) = 12% of 39000 = ₹4680$

Thus, cost of tractor $= (₹12000 + ₹4680)$ = ₹16680

14.Shamshad Ali buys a scooter for ₹22000. He pays ₹4000 cash and agrees to pay the balance in annual installment of ₹1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? Sol. It is given that Shamshad Ali buys a scooter for ₹22000 and pays ₹4000 in cash.

$∴ Unpaid amount = ₹22000 - ₹4000 = Rs 18000$

According to the given condition, the interest paid annually is

$10% of 18000, 10% of 17000, 10% of 16000 \dots 10% of 1000$

Thus, total interest to be paid

$= = = 10% of 18000 + 10% of 17000 + 10% of 16000 + \dots + 10% of 1000 10% of (18000 + 17000 + 16000 + \dots + 1000) 10% of (1000 + 2000 + 3000 + \dots + 18000)$

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.

Let the number of terms be n .

$∴ \Rightarrow ∴ 18000 = 1000 + (n - 1) (1000) n = 18 1000 + 2000 + \dots .. + 18000 = \frac{18}{2} [2 (1000) + (18 - 1) (1000)] = 9 [2000 + 17000] = 171000$

Total interest paid

$= 10% of (18000 + 17000 + 16000 + \dots + 1000) = 10% of ₹ 171000 = ₹ 17100$

Cost of scooter

$= ₹22000 + ₹17100 = ₹39100$

15.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 $8^{th}$ set of letter is mailed.

Sol. The numbers of letters mailed forms a G.P. 4, $4^{2}, \dots 4^{8}$ First term = 4 Common ratio = 4 Number of terms $= 8$ It is known that the sum of n terms of a G.P. is given by

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

It is given that the cost to mail one letter is 50 paisa.

$∴ Cost of mailing 87380 letters = ₹87380 \times \frac{50}{100} = ₹43690$

Thus, the amount spent when $8^{th}$ set of letter is mailed is ₹43690. 16. A man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years. Sol. It is given that the man deposited ₹10000 in a bank at the rate of 5% simple interest annually.

$= \frac{5}{100} \times ₹10000 = ₹500$

∴ Interest in first year

$= 10000 + 14 times 500 + 500 + \dots . + 500$

Amount in 15th year $= ₹$

$= ₹10000 + 14 \times ₹500 = ₹10000 + ₹7000 = ₹17000$

Amount after 20 years

$= 10000 + 20 times 500 + 500 + \dots . + 500 = ₹10000 + 20 \times ₹500 = ₹10000 + ₹10000 = ₹20000$

17.A manufacturer reckons that the value of a machine, which costs him ₹15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. Sol. Cost of machine $= ₹15625$ Machine depreciates by 20% every year. Therefore, its value after every year is 80% of the original cost i.e., $\frac{4}{5}$ of the original cost. $∴$ Value at the end of 5 years

$= 15625 \times 5 times \frac{4}{5} \times \frac{4}{5} \times \dots\dots \times \frac{4}{5} = 5 \times 1024 = 5120$

Thus, the value of the machine at the end of 5 years is ₹5120.

18.150 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 32 day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. Sol. Let x be the number of days in which 150 workers finish the work. According to the given information, $150 x = 150 + 146 + 142 + \dots . (x + 8)$ terms The series $150 + 146 + 142 + \dots . (x + 8)$ terms is an A.P. with first term 146, common difference - 4 and number of terms as ( $x + 8$ ) $\Rightarrow 150 x = \frac{( x + 8 )}{2} [2 (150) + (x + 8 - 1) (- 4)]$ $\Rightarrow 150 x = (x + 8) [150 + (x + 7) (- 2)]$ $\Rightarrow 150 x = (x + 8) (150 - 2 x - 14)$ $\Rightarrow 150 x = (x + 8) (136 - 2 x)$ $\Rightarrow 75 x = (x + 8) (68 - x)$ $\Rightarrow 75 x = 68 x - x^{2} + 544 - 8 x$ $\Rightarrow x^{2} + 75 x - 60 x - 544 = 0$ $\Rightarrow x^{2} + 15 x - 544 = 0$ $\Rightarrow x^{2} + 32 x - 17 x - 544 = 0$ $\Rightarrow x (x + 32) - 17 (x + 32) = 0$ $\Rightarrow x (x - 17) (x + 32) = 0$ $\Rightarrow x = 17$ or $x = - 32$ However, x cannot be negative. $x = 17$ Therefore, originally, the number of days in which the work was completed is 17. Thus, required number of days $= (17 + 8) = 25$

6.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Here, you can find chapter-wise NCERT solutions along with the answers for NCERT Class 11 Maths. You will also find key formulas and easy-to-understand explanations to enhance your understanding and improve your problem-solving skills.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

7.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 8

Exercises	Number of Questions	Important Topics Covered
Exercise 8.1	14 Questions and Solutions	Types of sequences and finding the nth term, including AP, GP and HP
Exercise 8.2	32 Questions and Solutions	Geometric Progression (GP), nth term and sum of n terms
Miscellaneous Exercise	18 Questions and Solutions	Mixed questions covering the key concepts of Sequences and Series

8.0Key Features of NCERT Solutions for Class 11 Maths Chapter 8 (Sequences and Series)

Clear Understanding of Sequences and Series:
Learn how sequences and series are formed and understand important concepts such as arithmetic progressions (AP) and geometric progressions (GP) with simple explanations.
Step-by-Step Formula Applications:
The solutions explain how to find the (n^{th}) term and the sum of terms in an arithmetic or geometric progression using the correct formulas and clear steps.
Arithmetic and Geometric Mean:
Understand how to insert arithmetic means and geometric means between two given numbers and solve related questions step by step.
Sum of a Geometric Progression:
Learn how to derive and apply the formula for the sum of the first (n) terms of a GP, making it easier to solve progression-based questions.
Sum to Infinity of a GP:
Understand when the sum of an infinite geometric progression exists and apply the condition (|r| < 1) to find its sum.
Prepared by ALLEN Subject Experts:
These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage:
All questions from the NCERT exercises, including the miscellaneous exercise, are covered with clear, step-by-step solutions to help students understand Sequences and Series.

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Why are NCERT Solutions for Class 11 Maths Chapter 8 Sequence and Series important for students?

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NCERT Solutions Class 11 Maths Chapter 8 – Sequence and Series

1.0Class 11 Maths Chapter 8 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 8

3.0EXERCISE - 8.2

4.0EXERCISE - 8.3 (SUPPLEMENTARY)

5.0MISCELLANEOUS EXERCISE

6.0Class 11 Maths NCERT Solutions – Chapter-wise Links

7.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 8

8.0Key Features of NCERT Solutions for Class 11 Maths Chapter 8 (Sequences and Series)

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