The 7th term of an A.P. is -4 and its 13th term is -16. Find its (i) first term (ii) common difference (iii) nth term.

Let first term be a and common difference be d.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 7th term is -4.

∴ a₇ = a + (7 - 1)d

⇒ -4 = a + 6d

⇒ a + 6d = -4 ….(1)

Given,

13th term is -16

∴ a₁₃ = a + (13 - 1)d

⇒ -16 = a + 12d

⇒ a + 12d = -16 ….(2)

Subtracting (1) from (2) we get,

⇒ a + 12d - (a + 6d) = -16 - (-4)

⇒ a + 12d - a - 6d = -16 + 4

⇒ 6d = -12

⇒ d = $\dfrac{-12}{6}$

⇒ d = -2.

Substituting value of d in (1) we get,

⇒ a + 6(-2) = -4

⇒ a - 12 = -4

⇒ a = -4 + 12

⇒ a = 8.

∴ a_n = 8 + (n - 1)(-2)

= 8 - 2n + 2

= 10 - 2n.

Hence, a = 8, d = -2 and a_n = 10 - 2n.