Question

what is the sum of a 7-term geometric series if the first term is -11, the last term is -45,056, and the common ratio is -4?

Answer 1

Sn=(a1)((1-r^n)/(1-r))
=(-11)(1-(-4)^7)/(1-(-4))
=(-11)(1-(-16384))/5)
=(-11)(3277)
=-36047

Answer 2

Answer:

The sum of a 7-term geometric series = -36047

Step-by-step explanation:

We have 7-term geometric series if the first term is -11, the last term is -45,056, and the common ratio is -4.

So the GP is

        -11, 44, -176,704,-2816,11264,-45056

Adding these terms we will get

      Sum = -36047

Using equation:

 We have

                  $s_n=\frac{a(r^n-1)}{r-1}$

                  a = -11, r = -4 and n= 7

 Substituting

                 $s_7=\frac{-11((-4)^7-1)}{-4-1}=-36047$

The sum of a 7-term geometric series = -36047