The 7th term of an A.P. is – 4 and its 13th term is – 16. Find the A.P. [CBSE 20041

We have

a₇ = a + (7 – 1)d = a + 6d = –4 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = –16 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = –4 – (–16)

⇒ – 6d = –4 + 16

⇒ – 6d = 12

⇒ d = –2

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

a + 6(–2) = –4

⇒ a – 12 = –4

⇒ a = 8

Hence, the required AP is 8, 6, 4, 2,…