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Question
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
Solution
Given, Sn = `n/2 (3n + 5)`
∴ `S_(n - 1) = (n - 1)/2 [3(n - 1) + 5]`
or `S_(n - 1) = (n - 1)/2 (3n + 2)`
Since, an = Sn – Sn–1
= `n/2(3n + 5) - (n - 1)/2(3n + 2)`
= `(3n^2)/2 + (5n)/2 - (3n(n - 1))/2 - (2(n - 1))/2`
= `(3n^2)/2 + (5n)/2 - (3n^2)/2 + (3n)/2 - n + 1`
= `(8n)/2 - n + 1`
= 4n – n + 1
= 3n + 1
Now, a25 = 3(25) + 1
or, a25 = 75 + 1 = 76
Thus, 25th term of AP. is 76.
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