The 5 th and 9 th term of an Arithmetic Progression are 4 and -12 respectively. Find :

ICSE Class 10 Mathematics PYQ · 2023 Paper · Section B

6(ii)Written · 2023 · Section B · 2023 Paper

The 5th and 9th term of an Arithmetic Progression are 4 and -12 respectively. Find :

(a) the first term

(b) common difference

(c) sum of 16 terms of the A.P.

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If first term is a and common difference is d of the A.P.

By formula,

nth term = an = a + (n - 1)d

Given,

⇒ 5th term = 4

⇒ a5 = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9th term = -12

⇒ a9 = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\frac{16}{4}$

⇒ d = -4.

Substituting value of d in equation (1), we get :

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\frac{n}{2}(a + l)$

Sum of 16 terms of A.P. = $\frac{n}{2}(a + a_{16})$

$$\begin{aligned}= \frac{16}{2}[a + a + (16 - 1)d] \\[1em] = 8(2a + 15d) \\[1em] = 8 \times (2 \times 20 + 15 \times -4) \\[1em] = 8 \times (40 - 60) \\[1em] = 8 \times -20 \\[1em] = -160.\end{aligned}$$

Hence, sum of 16 terms of A.P. = -160.

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