Question

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.

Answer 1

Let a be the first term and d be a common difference.
We have,
a₁₀ = 41
⇒ a + 9d = 41                     ...(i)
and 
a₁₅ = 2a + 3
⇒ a + 14d = 2(a + 6d) + 3
⇒ a + 14d = 2a + 12d + 3
⇒ a – 2d = 3                        ...(ii)
Subtracting (ii) from (i), we get
9d + 2d = 41 + 3
⇒ 11d = 44
⇒ d = 4
Now, from (i), we get
a + 9 x 4 = 41
⇒ a + 36 = 41
⇒ a = 5
Now,
nth term
= a_n
= a + (n – 1)d
= 5 + (n – 1)4
= 4n + 1.