+208 The sum of the first 10 terms of a geometric sequence is 10, and the sum of the first 30 terms of the same sequence is 70. Find the sum of the first 40 terms.
+128570 S10 = a ( 1 - r^10) / ( 1 -r) = 10
S30 = a ( 1 - r^30) / (1- r) = 70
a / (1 -r ) = 10 / (1 -r^10)
a/(1- r) = 70 / (1-r^30)
10 / ( (1 -r^10) = 70 / ( 1 -r^30)
10 ( 1 - r^30) = 70 (1 -r^10)
10 - 10r^30 = 70 - 70r^10
70r^10 - 10r^30 = 60
7r^10 - r^30 = 6
r^30 - 7r^10 + 6 = 0
Let r^30 = x^3 , 7r^10 = 7x
So we have
x^3 - 7x + 6 = 0
1 is a root
Use synthetic division to find the other roots
1 [ 1 0 -7 6]
1 1 -6
____________
1 1 -6 0
The remaining polynomial is
x^2 + x - 6 = 0
(x + 3) ( x -2) = 0
x = 2 is the only feasible solution
So
r^30 = x^3
r ^10 = x
r^10 = 2
r = 2^(1/10)
a ( 1 - r*10) / (1 - r) = 10
a ( 1 - 2) / ( 1 - 2^(1/10) ) = 10
a = - 10 ( 1 - 2^(1/10))
a = 10 [ 2^(1/10) - 1)]
S40 = 10 [ 2^(1/10) - 1 ] [ 1 - 2^4 ] / [ 1 - 2^(1/10) ] =
-10 [ 1 - 2^4 ] =
-10 (-15) =
150
