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. S30– S15= 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.