a

_{n}= a_{1}+ (n - 1)d, where a_{n}= nth term, a_{1}= first term, n = number of terms, d = common differencea

_{15}= 3 + (15 - 1)4a

_{15}= 3 + 14(4)a

_{15}= 3 + 56a

_{15}= 59S

_{n}= (n/2)(2a_{1}+ (n - 1)d)S

_{11}= (11/2)(2(3) + (11 - 1)4)S

_{11}= (11/2)(6 + 40)S

_{11}= (11/2)(46)S

_{11}= 253Mark M.

tutor

If each term is multiplied by a "common ratio," you have a geometric sequence/series. It has its own set of formulas.

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06/21/15

John Y.

06/18/15