Find the geometric sum: 12 + 36 + 108 + … + 78,732. 39,360 118,092 1,062,876 354,288

This is a geometric sequence with an initial term of 12 and a common ratio of 3.

The rule for a geometric sequence is:

a(n)=ar^(n-1)  we are told that the last term is 78732 so we can find the number of terms in the sequence...

78732=12(3)^(n-1)  divide both sides by 12

6561=3^(n-1)  take the natural log of both sides

ln6561=(n-1)ln3  divide both sides by ln3

ln6561/ln3=n-1  add 1 to both sides

ln6561/ln3 +1=n

9=n  so there are 9 terms in the sequence.

The sum of any geometric sequence is:

s(n)=a(1-r^n)/(1-r), in this case a=12, r=3, and n=9 so

s(9)=12(1-3^9)/(1-3)

s(9)=12(1-19683)/(-2)

s(9)=-6(-19682)

s(9)=118092

The correct answer is $90