The 15th term of an A.P. is 3 more than twice its 7th term. If the 10th term of the A.P. is 41, find its nth term.

Given,

a₁₀ = 41,          (Eq 1)
a₁₅ = 2a₇ + 3.  (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,
⇒ a₁₀ = a + (10 - 1)d = 41
⇒ a + 9d = 41
⇒ a = 41 - 9d     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,
⇒ a₁₅ = 2a₇ + 3
⇒ a + (15 - 1)d = 2(a + (7 - 1)d) + 3
⇒ a + 14d = 2(a + 6d) + 3
⇒ a + 14d = 2a + 12d + 3

Putting value of a from Eq 3 in above equation
⇒ 41 - 9d + 14d = 2(41- 9d) + 12d + 3
⇒ 41 + 5d = 82 - 18d + 12d + 3
⇒ 41 + 5d = 82 - 6d + 3
⇒ 5d + 6d = 85 - 41
⇒ 11d = 44
⇒ d = 4.

∴ a = 41 - 9d = 41 - 9(4) = 41 - 36 = 5.

nth term = a_n = a + (n - 1)d = 5 + 4(n - 1) = 5 + 4n - 4 = 4n + 1.

Hence, the nth term of the A.P. is 4n + 1.