Question

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

⇒ 13d = 195

⇒ d = `195/13`

⇒ 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.