Solved 2023 Question Paper ICSE Class 10 Mathematics — Question 8

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

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th 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 = $-\dfrac{16}{4}$

⇒ 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. = $\dfrac{n}{2}(a + l)$

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

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

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