If the nth term of AP – 9, 7, 5, .... is the same as the nth term of the AP – 15, 12, 9...then find the value of n.

We know, nth term of an AP is

a_n = a + (n – 1)d

First AP:

9, 7, 5, …,

First term, a = 9

Common difference, d = a₂ – a = 7 – 9 = –2

Let the nth term be a_n

Second AP:

15, 12, 9, …,

First term, A = 15

Common difference, D = A₂ – A = 12 – 15 = –3

Let the nth term be A_n

Given,

nth term of first AP = nth term of second AP

⇒ a_n = A_n

⇒ a + (n – 1)d = A + (n – 1)D

⇒ 9 + (n – 1)(–2) = 15 + (n – 1)(–3)

⇒ 9 – 2n + 2 = 15 – 3n + 3

⇒ n = 7