What is #s_n# of the geometric series with #a_1=4#, #a_n=256#, and #n=4#?

1 Answer
Jul 30, 2014

The sum is #340#.

Since the #nth# term is given as #256# but #n# is given as #4#, that means #256# is the #4th# term. But the #4th# term of a GP equals #ar^3#, where a is the first term and #r# is the common ratio of the GP. Dividing #256# by the first term (which is given as #4#) shows us that #r^3 = 256/4 = 64#.

If #r^3 = 64#, then the common ratio #r# must equal #4# as well. This gives us all the information we need to use the formula for the sum of a GP, #S = (a(r^n - 1))/(r - 1)#.

In this case, #S = (4(4^4 - 1))/(4 - 1) = (4(256 - 1))/3 = 1020/3 = 340#.