Question

the sum of the first two terms in a geometric series is 12. the sum of the first three terms of the same series is 62. determine the first four terms of the series.

Answer 1

a + ar = 12
a(1+r) = 12 , #1

a + ar + ar^2 = 62
a(1 + r + r^2) = 62 , #2

divide #2 by #1
(1+r+ r^2)/(1+r) = 62/12 = 31/6
6r^2 + 6r + 6 = 31 + 31r
6r^2 - 25r - 25 = 0
(r-5)(6r + 5) = 0
r = 5 or r = -5/6

if r = 5 , in #1
6a = 12
a = 2

the first 4 terms are
2 , 10, 50, 250

if r = -5/6
a(1-5/6) = 12
(1/6)a = 12
a = 72

the first 4 terms are
72 , -60 , 50 , - 125/3

Answer 2

thank you !