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Question
Sum of the first 14 terms of and AP is 1505 and its first term is 10. Find its 25th term.
Solution
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
∴ 25th term of the AP, a25
= 10 + (25-1) × 15 [ an = a+(n-1) d]
= 10+360
=370
Hence, the required term is 370.
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