NCERT Class 11 Mathematics — Sequences and Series, Exercise 8.1. All 14 questions solved.

A sequence is a list of numbers in a definite order; the corresponding series is what you get by adding them with $+$ signs instead of commas.

A sequence can be specified in two ways, and this exercise practises both:

• Explicitly, by a formula for $a_n$ in terms of $n$ — substitute and read off any term directly (questions 1 to 10).
• Recursively, by the first term or two plus a rule linking each term to its predecessors — here every term must be built up in order (questions 11 to 14).

Key insight. A recursive definition cannot be jumped into. To reach $a_5$ you must have $a_4$, which needs $a_3$, and so on — so work strictly downward from the given starting value and write each term out before using it. An explicit formula, by contrast, lets you compute $a_{20}$ without knowing any other term, which is exactly what questions 7 to 10 exploit.

Write the first five terms of each of the sequences in questions 1 to 6 whose $n$th terms are:

Question 1

$a_n = n(n+2)$

Solution. Substitute $n = 1, 2, 3, 4, 5$ in turn:

$$a_1 = 1(3) = 3, \quad a_2 = 2(4) = 8, \quad a_3 = 3(5) = 15, \quad a_4 = 4(6) = 24, \quad a_5 = 5(7) = 35$$

Question 2

$a_n = \dfrac{n}{n+1}$

Solution.

$$a_1 = \frac12, \quad a_2 = \frac23, \quad a_3 = \frac34, \quad a_4 = \frac45, \quad a_5 = \frac56$$

Each term is a little larger than the last, creeping towards $1$ without reaching it.

Question 3

$a_n = 2^n$

Solution.

$$2,\ 4,\ 8,\ 16,\ 32$$

Question 4

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

Solution. The numerator runs $-1, 1, 3, 5, 7$:

$$a_1 = -\frac16, \quad a_2 = \frac16, \quad a_3 = \frac36 = \frac12, \quad a_4 = \frac56, \quad a_5 = \frac76$$

Note that $a_1$ is negative — the formula does not promise positive terms, and $2(1) – 3 = -1$.

Question 5

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

Solution. The factor $(-1)^{n-1}$ is $+1$ for odd $n$ and $-1$ for even $n$, so the signs alternate starting with a plus. The magnitudes are $5^2, 5^3, 5^4, 5^5, 5^6$:

$$a_1 = 25, \quad a_2 = -125, \quad a_3 = 625, \quad a_4 = -3125, \quad a_5 = 15625$$

Question 6

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

Solution.

$$a_1 = 1 \cdot \frac{6}{4} = \frac32, \qquad a_2 = 2 \cdot \frac{9}{4} = \frac92, \qquad a_3 = 3 \cdot \frac{14}{4} = \frac{21}{2}$$

$$a_4 = 4 \cdot \frac{21}{4} = 21, \qquad a_5 = 5 \cdot \frac{30}{4} = \frac{75}{2}$$

Find the indicated terms in each of the sequences in questions 7 to 10 whose $n$th terms are:

Question 7

$a_n = 4n – 3$; find $a_{17}$ and $a_{24}$.

Solution. No need to compute any intermediate term — substitute directly:

$$a_{17} = 4(17) – 3 = 68 – 3 = 65$$

$$a_{24} = 4(24) – 3 = 96 – 3 = 93$$

Question 8

$a_n = \dfrac{n^2}{2^n}$; find $a_7$.

Solution.

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

Question 9

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

Solution. For $n = 9$ the exponent is $n – 1 = 8$, which is even, so the sign is positive:

$$a_9 = (-1)^{8} \times 9^3 = 729$$

Question 10

$a_n = \dfrac{n(n-2)}{n+3}$; find $a_{20}$.

Solution.

$$a_{20} = \frac{20 \times 18}{23} = \frac{360}{23}$$

The fraction does not simplify — $23$ is prime and divides neither $20$ nor $18$ — so this is the final answer.

Write the first five terms of each of the sequences in questions 11 to 13 and obtain the corresponding series:

Question 11

$a_1 = 3$, $a_n = 3a_{n-1} + 2$ for all $n > 1$

Solution. Each term is three times the previous plus $2$:

$$a_1 = 3$$ $$a_2 = 3(3) + 2 = 11$$ $$a_3 = 3(11) + 2 = 35$$ $$a_4 = 3(35) + 2 = 107$$ $$a_5 = 3(107) + 2 = 323$$

Question 12

$a_1 = -1$, $a_n = \dfrac{a_{n-1}}{n}$ for $n \ge 2$

Solution. Each term is the previous divided by the current index:

$$a_1 = -1$$ $$a_2 = \frac{-1}{2} = -\frac12$$ $$a_3 = \frac{-1/2}{3} = -\frac16$$ $$a_4 = \frac{-1/6}{4} = -\frac{1}{24}$$ $$a_5 = \frac{-1/24}{5} = -\frac{1}{120}$$

The denominators are $1, 2, 6, 24, 120$ — the factorials — so in fact $a_n = -\dfrac{1}{n!}$.

Question 13

$a_1 = a_2 = 2$, $a_n = a_{n-1} – 1$ for $n > 2$

Solution. The first two terms are given, and the rule applies only from the third onwards:

$$a_1 = 2, \qquad a_2 = 2$$ $$a_3 = a_2 – 1 = 1$$ $$a_4 = a_3 – 1 = 0$$ $$a_5 = a_4 – 1 = -1$$

The repeated $2$ at the start is deliberate — applying the rule too early would give $a_2 = 1$ and change everything after it.

Question 14

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

Solution. First generate the terms:

$$a_1 = 1, \quad a_2 = 1, \quad a_3 = 1 + 1 = 2, \quad a_4 = 2 + 1 = 3, \quad a_5 = 3 + 2 = 5, \quad a_6 = 5 + 3 = 8$$

Now form the consecutive ratios:

$$\frac{a_2}{a_1} = \frac11 = 1, \qquad \frac{a_3}{a_2} = \frac21 = 2, \qquad \frac{a_4}{a_3} = \frac32$$

$$\frac{a_5}{a_4} = \frac53, \qquad \frac{a_6}{a_5} = \frac85$$

The ratios oscillate — above, below, above — and close in on the golden ratio $\dfrac{1+\sqrt5}{2} \approx 1.618$. As a decimal the five values are $1$, $2$, $1.5$, $1.667$, $1.6$, tightening at each step.

Common mistakes

• Question 4, expecting all terms to be positive. $a_1 = -\tfrac16$, because $2(1) – 3 = -1$. Substituting mechanically rather than assuming a pattern avoids this.
• Question 5, getting the sign phase wrong. $(-1)^{n-1}$ starts at $+1$ for $n = 1$; $(-1)^n$ would start at $-1$. Checking $n = 1$ explicitly fixes the phase once and for all.
• Question 5, misreading the exponent. The power is $5^{n+1}$, so the first term is $5^2 = 25$, not $5$.
• Questions 7 to 10, computing every term up to the one asked for. These are explicit formulae; substitute the index directly.
• Question 9, taking $(-1)^{n-1}$ as negative for $n = 9$. The exponent is $8$, which is even, so the sign is positive.
• Question 12, dividing by $n – 1$ instead of $n$. The rule is $a_n = \tfrac{a_{n-1}}{n}$, so $a_3 = \tfrac{a_2}{3}$, not $\tfrac{a_2}{2}$.
• Question 13, applying the recursion from $n = 2$. The condition is $n > 2$, and $a_2$ is given as $2$. Starting the rule a step early makes the sequence $2, 1, 0, -1, -2$ — every term after the first is wrong.
• Question 14, computing $\tfrac{a_n}{a_{n+1}}$ instead. The question asks for $\tfrac{a_{n+1}}{a_n}$, so the later term goes on top and the ratios exceed $1$.

Practise next

• Exercise 8.2 — geometric progressions, where the recursive rule $a_n = r\,a_{n-1}$ has an explicit solution $a_n = ar^{n-1}$.
• Miscellaneous Exercise on Chapter 8 — sums of special series and mixed AP/GP problems.