Find the sum of the first five terms of an infinite geometric progression with a common ratio |r|<1 if the second term is (4/3) and the ratio of the sum of the squares of the terms of the progression to the sum of the terms of the progression is 3:1.
Call the first term "a" so....the second term is
ar =4/3 → a = 4/[3r]
And we know that
[ a^2 + (ar)^2 + (ar^2)^2 + ...... + (ar^n)^2 ] / [a + ar + ar^2 +......+ ar^n] = 3
The sum of the series in the denominator can be written as : a^2 / [1 - r^2]
And the sum of the series in the numerator can be written as : a / [ 1 - r]
And we have that
( a^2 / [1 - r^2] ) / ( a / [ 1 - r] ) = 3 simplify
a(1 -r) / (1 - r^2) = 3
a(1 - r) / [ (1 - r) (1 + r)] = 3
a/(1 + r) = 3
a = 3(1 + r) sub for "a"
4/[3r] = 3(1 + r)
4 = 9r(1 + r)
4 = 9r + 9r^2 rearrange
9r^2 + 9r - 4 = 0 factor
(3r + 4) (3r - 1) = 0 set each factor to 0 and r = 1/3 or r = -4/3
But .... l -4/3 l > 1 so reject this solution
So r = 1/3 and a = 4/ [ 3 *1/3] = 4
And the sum of the series = 4 / [1 - 1/3] = 4 / (2/3) = 12 / 2 = 6
And the sum of the squares of the series =
4^2 / [1 - (1/3)^2] = 16 / [ 8/9] = 16 * 9 / 8 = (16/8) * 9 = 2 * 9 = 18
And 18 / 6 = 3
And the sum of the first 5 terms =
4 [1 - (1/3)^5] / (1 - 1/3) = 4 [ 1 - 1/243] / [2/3] = 6 [ 242/243] = 2*242 / 81 = 484 / 81
