NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.1#
An explicit rule gives the $n^{th}$ term directly from $n$ (for example $t_n = 5n - 3$). A recursive rule gives each term from the terms before it (for example $t_{n+1} = t_n + 3$), together with the starting term(s).
To check whether a number is a term: set $t_n$ equal to it and solve for $n$. It is a term only if $n$ comes out as a natural number.
1. Find the first five terms of the sequence in which the nth term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \geq 1$. #
(i) $t_n = 3n - 4$
$$t_2 = 3(2) - 4 = 2$$
$$t_3 = 3(3) - 4 = 5$$
$$t_4 = 3(4) - 4 = 8$$
$$t_5 = 3(5) - 4 = 11$$
∴ The first five terms are −1, 2, 5, 8, 11.
(ii) $t_n = 2 - 5n$
$$t_2 = 2 - 5(2) = -8$$
$$t_3 = 2 - 5(3) = -13$$
$$t_4 = 2 - 5(4) = -18$$
$$t_5 = 2 - 5(5) = -23$$
∴ The first five terms are −3, −8, −13, −18, −23.
(iii) $t_n = n^2 - 2n + 3$
$$t_2 = 4 - 4 + 3 = 3$$
$$t_3 = 9 - 6 + 3 = 6$$
$$t_4 = 16 - 8 + 3 = 11$$
$$t_5 = 25 - 10 + 3 = 18$$
∴ The first five terms are 2, 3, 6, 11, 18.
2. Find the 10th and 15th terms of the sequence $t_n = 5n - 3$ for $n \geq 1$. #
Solution
$$t_{15} = 5(15) - 3 = 72$$
∴ $t_{10} = $ 47 and $t_{15} = $ 72.
3. Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \geq 1$. #
Is 97 a term?
$$\Rightarrow 5n = 100$$
$$\Rightarrow n = 20$$
$n = 20$ is a natural number, so 97 is a term — the 20th term.
Is 172 a term?
$$\Rightarrow 5n = 175$$
$$\Rightarrow n = 35$$
$n = 35$ is a natural number, so 172 is a term — the 35th term.
4. Which term of the sequence $t_n = 5n - 3$ for $n \geq 1$ is 607? #
Solution
$$\Rightarrow 5n = 610$$
$$\Rightarrow n = 122$$
∴ 607 is the 122nd term.
5. A sequence is given by the recursive rule $t_1 = -5$, $t_{n+1} = t_n + 3$ for $n \geq 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it? #
First five terms
$$t_2 = t_1 + 3 = -2$$
$$t_3 = t_2 + 3 = 1$$
$$t_4 = t_3 + 3 = 4$$
$$t_5 = t_4 + 3 = 7$$
∴ The first five terms are −5, −2, 1, 4, 7.
Is 52 a term?
Each step adds 3, so after starting at $-5$, the $n^{th}$ term has had $(n - 1)$ threes added:
$$t_n = -5 + 3(n - 1)$$We set it equal to 52:
$$-5 + 3(n - 1) = 52$$$$\Rightarrow 3(n - 1) = 57$$
$$\Rightarrow n - 1 = 19$$
$$\Rightarrow n = 20$$
∴ Yes, 52 is the 20th term.
6. Let $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$. Find $T_4$, $T_5$, $T_6$, $T_7$, and $T_8$. #
Solution
Each term is the sum of the three terms before it:
$$T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7$$$$T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13$$
$$T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24$$
$$T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44$$
$$T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81$$
∴ $T_4, \ldots, T_8$ are 7, 13, 24, 44, 81.