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.

Answer 1

Not sure just what you're after. If the sums in the ratio 3:1 are just the first five terms, then

a^2(1-r^10)/(1-r^2)
------------------------- = 3
a(1-r^5)/(1-r)

ar = 4/3
a = 4/(3r), so plug that in, and

(4/(3r))^2(1-r^10)/(1-r^2)
------------------------- = 3
4/(3r)(1-r^5)/(1-r)

Simplify that and you get
4r^4 - 4r^3 + 4r^2 - 13r + 4 = 0
Solve that by your favorite method and you get r =0.33445

Check: r^2 = .11186

ar = 4/3
a = 3.98667
S(r^2)/S(r) = (1-r)/(1-r^2) ≈ 3.0

Ok so far.
S5 = 3.98667(1-.33445^5)/(1-.33445) = 5.96497

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If the sums in the ratio 3:1 are the complete sums,

a/(1-r^2)
------------- = 1/(1+r) = 3
a/(1-r)

3+3r = 1
r = -2/3
ar = 4/3, so a = -2
S5 = -2(1-(-2/3)^5)/(1-(-2/3)) = -110/81