Question

Sum of the first 14 terms of and AP is 1505 and its first term is 10. Find its 25^th term. 

Answer 1

Let d be the common difference of the AP.
Here, a = 10 and n = 14

Now,

S₁₄ = 1505                 (Given) 

`⇒ 14/2 [2 xx 10 + (14 - 1) xx d ] = 1505                  { S_n = n/2 [ 2a +(n-1) d]}`

⇒ 7(20 +13d)= 1505

⇒ 20+13d = 215

⇒ 13d = 215 -20 = 195

⇒ d= 15

∴ 25^th term of the AP, a₂₅ 

= 10 + (25-1) × 15                             [ a_n = a+(n-1) d]

= 10+360

=370

Hence, the required term is 370.