The sum of the first 15 terms of a geometric sequence with 7 as the first term and a common ratio of -3 is

Answer 1

a1=7

a2=-21

a3 =63

a4 =-189

an=a1(r^n-1)

an=7(-3)^(n-1)

a15 = 7(-3)^(15-1)

=7(-3)^(14) =33,480,783 = the 15th term

sum of the geometric sequence = Sn = a1(1-r^n)/(1-r) with n=15, r=-3 a1 = 7

S1 = 7

S2 = -14 7(1-9)/(1--3) = 7(-8)/4=-14

S3 = 49 7(1+27)/4 = 7(7) = 49

S4 = -140 7(1-81)/4 = 7(-80)/4 =-7(20) = -140

Sn = a1[(1-r)^(n-1)]/(1-r)

S15 = 7(1-(-3)^15/(1--3) = 25,110,589 = sum of first 15 terms