Mark M. answered • 06/05/16

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a

_{n}= a_{1}+ (n - 1)d399 = 3 + (n - 1)4

396 = 4n - 4

400 = 4n

100 = n

S

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

_{100}= (100/2)(3 + 399)S

_{100 }= (50)(402)S

_{100}= 20100
Mark M.

_{1}._{0}is used to represent initial amount as in growth and decay models.06/05/16