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Let the series be a, ar, ar² …..
Given: Sum_∞ = a/(1 − r) = 80 …(i)
Also, a + ar = a(1 + r) = 35 …(ii)
80(1 – r) (1 + r) = 35
1 – r² = 35/80 = 7/16
r² = 9/16
r = ± 3/4
Substituting r = 3/4 in eq.(i) we get a = 20
These values of a and r, won’t lead the sum of series to 100 as sum of infinite terms is 80 only.
Substituting r = − 3/4 in eq.(i) we get a = 140.
a₁ = 140;
a₂ = 140 × − 3/4 = −105;
a₃ = − 105 × − 3/4 = 78.75;
a₄ = − 78.75 × − 3/4 = − 59.0625;
a₅ = − 59.0625 × − 3/4 = 44.296875 & so on.
Adding first 5 terms of the series, we get
a₁ + a₂ + a₃ + a₄ + a₅ = 98.984375 ≈ 100
So, the sum of first 5 terms of the series when r = − 3/4 and a = 140 gives the sum close to 100.