+766 The sum of the first three terms of a geometric sequence of integers is equal to seven times the first term, and the sum of the first four terms is $30$. What is the first term of the sequence?
+130458 We need to work with three equations
7a1 = a1 + a1r + a1r^2 → 6a1 = a1r + a1r^2 → 6 = r + r^2 (1)
a1 + a1r + a1r^2 + a1r^3 = 30 → a1 ( 1 + r + r^2 + r^3) = 30 → 1 +( r + r^2) + r^3 = 30/a1 →
1 + 6 + r^3 = 30/a1 → 7 + r^3 = 30/a1 (2)
a1 ( 1 - r^4) / (1 - r) = 30 → (1-r^4) / (1 - r) = 30/a1 (3)
Set (3) = (2) and solve for r
(1 - r^4) / (1 - r) = 7 + r^3
1 -r^4 = (1 -r) (7 + r^3)
1 -r^4 = 7 + r^3 - 7r - r^4
1 = 7 + r^3 -7r
r^3 - 7r + 6 = 0
Possible solutions for r
r =1 (reject...it make equation 3 undefined)
r = 2 , -3
If r = 2
Then
a1 ( 1 - 2^4) / (1 - 2) = 30
a1 (-15)/(-1) = 30
a1 * 15 = 30
a1 = 30 / 15 = 2
If r = -3
a1 [1 - (-3)^4 ] / ( 1 + 3) =30
a1 [ -80] / 4 = 30
a1 * -20 = 30
a1 = 30 / -20 = -3/2
