Question

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

Answer 1

hey,

here your answer....

Sum of n terms = Sn = n/2 [ 2a + (n-1) d ]

where a = first term d = common difference 

therefore, S14 = 14/2 [2*10 + (14-1) d ] = 1505

solve for d, we get, 215 = 20 + 13d

hence, d = 15. 

now, an = a + (n-1) d

therefore, a25 = 10 + (25-1) 15

a25 = 10 + 24 * 15 = 370. 

your answer is 370

hope this helps.

:)

Answer 2

Given First term = 10.

Given sum of first 14 terms = 1505.

We know that sum of first n terms of an AP sn = (n/2)[2a + (n - 1) * d]

⇒ 1505 = (14/2)[20 + 13 * d]

⇒ 3010 = 14[20 + 13d]

⇒ 215 = 20 + 13d

⇒ 195 = 13d

⇒ d = 15.

We know that nth term of an AP an = a + (n - 1) * d

25th term: a + (25 - 1) * d

⇒ 10 + 24 * (15)

⇒ 10 + 360

⇒ 370.

Therefore, its 25th term = 370.

Hope it helps!