The 11^th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.

Given: t₁₁ = 16 and t₂₁ = 29

To find: t₄₁

Using n^th term of an A.P. formula

t_n = a + (n – 1)d

we will find value of “a” and “d”

Let, t₁₁ = a + (11 – 1) d

⇒ 16 = a + 10 d …..(1)

t₂₁ = a + (21 – 1) d

⇒ 29 = a + 20 d …..(2)

Subtracting eq. (1) from eq. (2), we get,

⇒ 29 – 16 = (a – a) + (20 d – 10 d)

⇒ 13 = 10 d

Substitute value of “d” in eq. (1) to get value of “a”

⇒ 16 = a + 13

⇒ a = 16 – 13 = 3

Now, we will find the value of t₄₁ using n^th term of an A.P. formula

⇒ t₄₁ = 3 + 4 × 13 = 3 + 52 = 55

Thus, t₄₁ = 55