The sum of the first n terms of a geometric sequence is 341. If A1=1/3 and A2=8/3. What is the value of n?

Question

The sum of the first n terms of a geometric sequence is 341. If A1=1/3 and A2=8/3. What is the value of n?

Answer 1

A1 = 1/3

A2 = 8/3 = 8A1 r=common ratio = 8

A3 =64/3

A4 = 512/3

A5 = 4,096/3

S2 =3

S3 =24 1/3

S4 = 195.04

S5 = 1,560 1/3

there must be a typo in the problem

N is between 4 and 5, which doesn't makes sense as N is supposed to be an integer

AN=A1(8)^N

SN = A1(r^N-1)/(r-1) = (1/3)(8^N-1)/(8-1)

S2 = (1/3)(8^2 -1)/7) = 63/21 = 3

S3 = (1/3)(8^3 -1)/7 = 512/21 = 24 1/3

S4 = (1/3)(8^4 -1)/7 = 195.04

S5 = (1/3)(8^5-1)/7 = 1,560.38