Question

If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively, then which term of this A.P is zero.

Answer 1

Given that, the 3rd and 9th term of an AP are 4 and – 8.

⟹ a₃ = 4

⟹ a₉ = – 8

From the formula a + (n – 1)d = an, we get :

⟹ a + (3 – 1)d = an

⟹ a + 2d = 4

⟹ a = 4 – 2d . . . . . eq ⑴

Also :

⟹ a + (9 – 1)d = an

⟹ a + 8d = – 8

⟹ a = – 8 – 8d . . . . . . eq ⑵

The first term ‘a’ is common for both of them, so let’s equate ⑴, ⑵ and solve :

⟹ – 8 – 8d = 4 – 2d

⟹ – 8d + 2d = 4 + 8

⟹ – 6d = 12

⟹ d = 12/– 6

⟹ d = – 2

⋆ Therefore, the common difference ‘d’ of this AP is – 2.

Let’s now put this value of ‘d’ in ⑵ to get ‘a’ :

⟹ a = – 8 – 8d

⟹ a = – 8 – 8 (– 2)

⟹ a = – 8 + 16

⟹ a = 8

⋆ Therefore, the first term ‘a’ of this AP is 8.

Let the nth term of the AP be 0. From, an = a + (n – 1)d, we shall find the value of n :

⟹ an = 0

⟹ d = – 2

⟹ a = 8

Substituting and solving for n :

⟹ 0 = 8 + (n – 1) (– 2)

⟹ 0 = 8 + 2 – 2n

⟹ 0 = 10 – 2n

⟹ – 10 = – 2n

⟹ – 10/– 2 = n

⟹ 5 = n

⋆ Therefore, 5th term of the AP is 0.