NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.2#
For an arithmetic progression (AP) with first term $a$ and common difference $d$:
$$t_n = a + (n - 1)d$$$$\text{Recursive rule: } t_1 = a, \quad t_{n+1} = t_n + d$$
$$S_n = \dfrac{n}{2}\bigl(2a + (n - 1)d\bigr) = \dfrac{n}{2}(\text{first term} + \text{last term})$$
1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …. #
Solution
Here $a = 3$ and $d = 8 - 3 = 5$.
$$t_{10} = 3 + (10 - 1) \times 5$$$$t_{10} = 3 + 45$$
$$t_{10} = 48$$$$t_{26} = 3 + (26 - 1) \times 5$$
$$t_{26} = 3 + 125$$
$$t_{26} = 128$$
∴ $t_{10} = $ 48 and $t_{26} = $ 128.
2. Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer. #
Which term is −81?
Here $a = 21$ and $d = 18 - 21 = -3$, so:
$$t_n = 21 + (n - 1)(-3)$$$$t_n = 24 - 3n$$
We set it equal to $-81$:
$$24 - 3n = -81$$$$\Rightarrow 3n = 105$$
$$\Rightarrow n = 35$$
∴ −81 is the 35th term.
Is 0 a term?
$$\Rightarrow 3n = 24$$
$$\Rightarrow n = 8$$
$n = 8$ is a natural number, so yes, 0 is a term — the 8th term. (The AP is 21, 18, 15, 12, 9, 6, 3, 0, …)
3. Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP. #
Solution
Here $a = 11$ and $d = 8 - 11 = -3$. We find $t_n$:
$$= 11 + (n - 1)(-3)$$$$= 11 - 3n + 3$$
$$= 14 - 3n$$
∴ $t_n = 14 - 3n$.
Recursive rule: $t_1 = 11$, $t_{n+1} = t_n - 3$ for $n \geq 1$.
4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term. #
Solution
Let the first term be $a$ and the common difference $d$.
The 3rd term is 12, so:
$$a + 2d = 12 \quad \cdots (1)$$The last (50th) term is 106, so:
$$a + 49d = 106 \quad \cdots (2)$$Subtracting equation (1) from equation (2):
$$\begin{array}{rcl} a + 49d &=& 106 \\ a + 2d &=& 12 \\ \hline 47d &=& 94 \end{array}$$$$\Rightarrow d = 2$$
Substituting $d = 2$ in equation (1):
$$a + 2d = 12$$$$a + 2(2) = 12$$
$$\Rightarrow a = 8$$
So:
$$t_{29} = a + 28d$$$$t_{29} = 8 + 28 \times 2$$
$$t_{29} = 64$$
∴ The 29th term is 64.
5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers? #
Solution
The 2-digit numbers divisible by 3 are 12, 15, 18, …, 99 — an AP with $a = 12$, $d = 3$ and last term 99.
$$12 + (n - 1) \times 3 = 99$$$$\Rightarrow 3(n - 1) = 87$$
$$\Rightarrow n - 1 = 29$$
$$\Rightarrow n = 30$$
So there are 30 such numbers. Their sum is:
$$S_{30} = \dfrac{30}{2}(12 + 99)$$$$S_{30} = 15 \times 111$$
$$S_{30} = 1665$$
∴ There are 30 two-digit numbers divisible by 3, and their sum is 1665.
6. Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000? #
Solution
His salaries form an AP with $a = 500000$ and $d = 20000$. Let the salary in year $n$ be ₹7,00,000:
$$500000 + (n - 1) \times 20000 = 700000$$$$\Rightarrow (n - 1) \times 20000 = 200000$$
$$\Rightarrow n - 1 = 10$$
$$\Rightarrow n = 11$$
So ₹7,00,000 is his salary in the 11th year, that is, after 10 years of increments.
7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all? #
Solution
The total is $1 + 2 + 3 + \cdots + 25$, the sum of the first 25 natural numbers:
$$S_{25} = \dfrac{25 \times 26}{2}$$$$S_{25} = 325$$
∴ The child uses 325 marbles.