the sum of first 15 terms of A.P. is 105 and the sum of next 15 terms is 780. find the 3rd term.

Let a be the first term and d be the common difference of A.P.

Also the sum of next 15 terms is 780 i.e.  S30S15= 780

Solving (1) and (2), we get –

∴ From (1), a = –14

Hence the required A.P. is

∴ The third term of A.P. is –8.

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