1st,since this GP is convergent that means the common ratio r <1
2nd, sum of a GP = a₁(1-rⁿ)/(1-r), where a₁ = 1st term and n=number of terms
3rd, for any convergent GP, r<1 and the sum of all terms =a₁/(1-r): Why?
[since r<1 → lim rⁿ when n→∞, =0 in the formula of the 2nd)]
Now let's solve :
a) Sum = a₁(1-r³)/(1-r) = 19 (sum of the first 3 terms)
b) Σ(Sum) = a₁/(1-r) = 27 (sum of all terms of this CONVERGENT GP)
Divide a) by b):
[a₁(1-r³)/(1-r)] / [a₁/(1-r)] = 19 /27 ↔ [a₁(1-r³)/(1-r)] x [(1-r)/a₁]=19/27.
Simplify:
(1-r³) = 19/27
-r³ = 19/27 - 1
r³ = 8/27
r = ∛(8/27)
r = 2/3 and a₁ = 9 (Plug r in the Σ sum)
Hence first term a₁ = 9
and common ration r =2/3
