Question

What is the sum of an 8-term geometric series if the first term is 12 and the last term is -3,359,232?

Answer 1

\[a=12 \]\[ar^8 \]\[12r^8=-3359,232 \]\[r=(-3359,232/12)^1/8 =-4.7960350032986424\]\[Use S(n)=a(1-r^n)/(1-r)\]\[S(8)=12(1-(-4.79603)^8/1--4.79603=-579572.06\]

Answer 2

umm that isnt in any of the mult choice
-3,599,175
-3,359,230
-3,119,285
-2,879,340

Answer 3

The 8th term is \(\large ar^7\) and not \(\large ar^8\)

Answer 4

i just realized my mistake thank @aum

Answer 5

np

Answer 6

hmm how would you rework it

Answer 7

\[12r^7=-3359232\]
\[r=(-3359232/12)^{1/7}=5.9999=6\]
\[S(8)=12(1-(-6)^8)/(1-(-6)=-2879340\]