Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series (Exercise 8.1) with step-by-step detailed explanations based on the latest CBSE and NCERT syllabus. These solutions cover important concepts such as arithmetic progression (A.P.), common difference, nth term, and sum of n terms, helping students develop a strong foundation in higher mathematics. Each question is solved systematically to improve problem-solving skills and enhance exam performance in CBSE, JEE Main, JEE Advanced, NDA, and CUET exams. Click the print button to download study material and notes in PDF format.
Question 1. Write the first five terms of a sequence whose nth term is $a_n = n(n + 2)$
Solution:
Given, $a_n = n(n + 2)$
Putting $n = 1, 2, 3, 4, 5$, we get:
$$a_1 = 1(1 + 2) = 1(3) = 3$$
$$a_2 = 2(2 + 2) = 2(4) = 8$$
$$a_3 = 3(3 + 2) = 3(5) = 15$$
$$a_4 = 4(4 + 2) = 4(6) = 24$$
$$a_5 = 5(5 + 2) = 5(7) = 35$$
Therefore, the first 5 terms of the given series are:
$3, 8, 15, 24, 35$
Question 2. Write the first 5 terms of the series whose nth term is $a_n = \dfrac{n}{n + 1}$
Solution:
Given, $a_n = \dfrac{n}{n + 1}$
Putting $n = 1, 2, 3, 4, 5$, we get:
$$a_1 = \frac{1}{1 + 1} = \frac{1}{2}$$
$$a_2 = \frac{2}{2 + 1} = \frac{2}{3}$$
$$a_3 = \frac{3}{3 + 1} = \frac{3}{4}$$
$$a_4 = \frac{4}{4 + 1} = \frac{4}{5}$$
$$a_5 = \frac{5}{5 + 1} = \frac{5}{6}$$
Therefore, the first 5 terms of the given series are:
$$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$$
Question 3. Write the first five terms of series whose nth term is $a_n = 2^n$
Solution:
Given, $a_n = 2^n$
Putting $n = 1, 2, 3, 4, 5$, we get:
$$a_1 = 2^1 = 2$$
$$a_2 = 2^2 = 4$$
$$a_3 = 2^3 = 8$$
$$a_4 = 2^4 = 16$$
$$a_5 = 2^5 = 32$$
Therefore, the first 5 terms of the given series are: $2, 4, 8, 16, 32$
Question 4. Write the first five terms of series whose nth term is $a_n = \dfrac{2n – 3}{6}$
Solution:
Given, $a_n = \dfrac{2n – 3}{6}$
Putting $n = 1, 2, 3, 4, 5$, we get:
$$a_1 = \frac{2(1) – 3}{6} = \frac{-1}{6}$$
$$a_2 = \frac{2(2) – 3}{6} = \frac{1}{6}$$
$$a_3 = \frac{2(3) – 3}{6} = \frac{1}{2}$$
$$a_4 = \frac{2(4) – 3}{6} = \frac{5}{6}$$
$$a_5 = \frac{2(5) – 3}{6} = \frac{7}{6}$$
Therefore, the first 5 terms of the given series are:
$$-\frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}$$
Question 5. Write the first 5 terms of the sequence whose nth term is $a_n = (-1)^{n – 1} 5^{n + 1}$
Solution:
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 first 5 terms of the series are: $25, -125, 625, -3125, 15625$
Question 6. Find the first five terms of the series whose nth term is given as $a_n = \dfrac{n(n^2 + 5)}{4}$
Solution:
Given, $a_n = \dfrac{n(n^2 + 5)}{4}$
Putting $n = 1, 2, 3, 4, 5$, we get:
$$a_1 = \dfrac{1(1^2 + 5)}{4} = \dfrac{1(6)}{4} = \dfrac{3}{2}$$
$$a_2 = \dfrac{2(2^2 + 5)}{4} = \dfrac{2(9)}{4} = \dfrac{9}{2}$$
$$a_3 = \dfrac{3(3^2 + 5)}{4} = \dfrac{3(14)}{4} = \dfrac{21}{2}$$
$$a_4 = \dfrac{4(4^2 + 5)}{4} = \dfrac{4(21)}{4} = 21$$
$$a_5 = \dfrac{5(5^2 + 5)}{4} = \dfrac{5(30)}{4} = \dfrac{75}{2}$$
Therefore, the first 5 terms of the series are:
$\dfrac{3}{2}, \dfrac{9}{2}, \dfrac{21}{2}, 21, \dfrac{75}{2}$
Question 7. Find the 17th and 24th terms of the sequence whose nth term is given as $a_n = 4n – 3$
Solution:
Given, $a_n = 4n – 3$
Putting $n = 17$, we get:
$$a_{17} = 4(17) – 3 = 68 – 3 = 65$$
Therefore, the 17th term is $65$.
For 24th term :
Putting $n = 24$, we get:
$$a_{24} = 4(24) – 3 = 96 – 3 = 93$$
Therefore, the 24th term is $93$.
Question 8. Find the 7th term of the sequence whose nth term is given as $a_n = \dfrac{n^2}{2n}$
Solution:
Substituting $n = 7$, we get:
$$a_7 = \dfrac{7^2}{2 \times 7} = \dfrac{49}{14} = \dfrac{7}{2}$$
Therefore, the 7th term is $\dfrac{7}{2}$.
Question 9. Find the 9th term of the sequence whose nth term is given as $a_n = (-1)^{n – 1} n^3$
Solution:
Substituting $n = 9$, we get:
$$a_9 = (-1)^{9 – 1} 9^3 = (-1)^8 \times 729 = 729$$
Therefore, the 9th term is $729$.
Question 10. Find the 20th term of the sequence whose nth term is given as $a_n = \dfrac{n(n – 2)}{n + 3}$
Solution:
Given, $a_n = \dfrac{n(n – 2)}{n + 3}$
Putting $n = 20$, we get:
$$a_{20} = \dfrac{20(20 – 2)}{20 + 3} = \dfrac{20 \times 18}{23} = \dfrac{360}{23}$$
Therefore, the 20th term is $\dfrac{360}{23}$.
Question 11. Find the first 5 terms of the following sequence:
$a_1 = 3$, $a_n = 3a_{n-1} + 2$ for all $n > 1$
Solution:
$a_1 = 3$
$$a_2 = 3a_{2-1} + 2 = 3a_1 + 2 = 3 \times 3 + 2 = 9 + 2 = 11$$
$$a_3 = 3a_{3-1} + 2 = 3a_2 + 2 = 3 \times 11 + 2 = 33 + 2 = 35$$
$$a_4 = 3a_{4-1} + 2 = 3a_3 + 2 = 3 \times 35 + 2 = 105 + 2 = 107$$
$$a_5 = 3a_{5-1} + 2 = 3a_4 + 2 = 3 \times 107 + 2 = 321 + 2 = 323$$
Therefore, the first 5 terms are:
$3,\ 11,\ 35,\ 107,\ 323$
Question 12. Write the first 5 terms of the following sequence:
$a_1 = -1$, $a_n = \dfrac{a_{n-1}}{n}$ for all $n > 1$
Solution:
$a_1 = -1$
$$a_2 = \dfrac{a_{2-1}}{2} = \dfrac{a_1}{2} = \dfrac{-1}{2} = -\dfrac{1}{2}$$
$$a_3 = \dfrac{a_{3-1}}{3} = \dfrac{a_2}{3} = \dfrac{-\frac{1}{2}}{3} = -\dfrac{1}{6}$$
$$a_4 = \dfrac{a_{4-1}}{4} = \dfrac{a_3}{4} = \dfrac{-\frac{1}{6}}{4} = -\dfrac{1}{24}$$
$$a_5 = \dfrac{a_{5-1}}{5} = \dfrac{a_4}{5} = \dfrac{-\frac{1}{24}}{5} = -\dfrac{1}{120}$$
Therefore, the first 5 terms of the series are:
$-1,\ -\dfrac{1}{2},\ -\dfrac{1}{6},\ -\dfrac{1}{24},\ -\dfrac{1}{120}$
Question 13. Write the first 5 terms of the following sequence:
$a_1 = a_2 = 2$, $a_n = a_{n-1} – 1$ for all $n > 2$
Solution:
$a_1 = 2$
$a_2 = 2$
$$a_3 = a_{3-1} – 1 = a_2 – 1 = 2 – 1 = 1$$
$$a_4 = a_{4-1} – 1 = a_3 – 1 = 1 – 1 = 0$$
$$a_5 = a_{5-1} – 1 = a_4 – 1 = 0 – 1 = -1$$
Therefore, the first 5 terms of the series are:
$2,\ 2,\ 1,\ 0,\ -1$
Question 14. The Fibonacci sequence is given by:
$a_1 = a_2 = 1$, $a_n = a_{n-1} + a_{n-2}$ for $n > 2$
Find $\dfrac{a_{n+1}}{a_n}$ for the first 5 terms.
Solution:
$a_1 = 1,\ a_2 = 1$
$$a_3 = a_1 + a_2 = 1 + 1 = 2$$
$$a_4 = a_2 + a_3 = 1 + 2 = 3$$
$$a_5 = a_3 + a_4 = 2 + 3 = 5$$
$$a_6 = a_4 + a_5 = 3 + 5 = 8$$
Now,
For $n = 1$: $\dfrac{a_{1+1}}{a_1} = \dfrac{a_2}{a_1} = \dfrac{1}{1} = 1$
For $n = 2$: $\dfrac{a_{2+1}}{a_2} = \dfrac{a_3}{a_2} = \dfrac{2}{1} = 2$
For $n = 3$: $\dfrac{a_{3+1}}{a_3} = \dfrac{a_4}{a_3} = \dfrac{3}{2}$
For $n = 4$: $\dfrac{a_{4+1}}{a_4} = \dfrac{a_5}{a_4} = \dfrac{5}{3}$
For $n = 5$: $\dfrac{a_{5+1}}{a_5} = \dfrac{a_6}{a_5} = \dfrac{8}{5}$
Therefore, the required ratios are:
$1,\ 2,\ \dfrac{3}{2},\ \dfrac{5}{3},\ \dfrac{8}{5}$
