The 10^th term of an A.P. is (–4) and its 22^nd term is (–16). Find its 38^th term.

Given:

T₁₀ = –4

T₂₂ = –16

We know that,

n^th term = T_n = a + (n – 1)d

where,

a = first term

d = common difference

∴

T₁₀ = a + (10 – 1)d = -4

⇒ a + 9d = -4 …(1)

And,

T₂₂ = a + (22 – 1)d = -16

⇒ a + 21d = -16 …(2)

Subtract (1) from (2),

(a + 21d) – (a + 9d) = -16 – (-4)

⇒ a + 21d – a – 9d = -16 + 4

⇒ 12d = -12

⇒ d = -1

Put value of d in (1),

a + 9d = -4

⇒ a + 9(-1) = -4

⇒ a – 9 = -4

⇒ a = -4 + 9

⇒ a = 5

Now,

T₃₈ = 5 + (38 – 1)(-1)

T₃₈ = 5 + 37(-1)

T₃₈ = 5 – 37

T₃₈ = – 32

Hence, the 38^th term is – 32