Question

The 7th term of an AP is –4 and its 13th term is –16. Find the AP.

Answer 1

We have

T₇ = a + (n – 1)d

⇒ a + 6d = –4   ...(1)

T₁₃ = a + (n – 1)d

⇒ a + 12d = –16   ...(2) 

On solving (1) and (2), we get

(a + 12d) – (a + 6d) = –16 – (–4)

6d = –12

d = `(-12)/6`

d = –2

Putting the value of d in equation (1), we get

a + 6(–2) = –4

a – 12 = –4

a = –4 + 12

a = 8

Thus, first term = 8 and common difference = –2

∴ The term of the AP are 8, 6, 4, 2,.........