Geometric sequence problem

A1 + A1R = 12

A1R^2 + A1R^3 = 48

R^2(A1 + A1R) = 48

R^2(12) = 48

R^2 = 4

R = 2

R =-2 (Eliminate -2: problem states that all terms are positive)

A1(1+R) = 12

A1(1+2) = 12

A1 = 4

S = [A1(1-R^n)]/(1-R)

S = [4(1-2^8)]/(1-2)

S = 4(-255) / -1

S = 1020

$4(x_1 + x_2) = x_3 + x_4$ This pattern will continue for all subsequent pairs of terms, as this is a geometric sequence. Here's the idea: $x_3 + x_4 = r^2x_1 + r^2x_2 = r^2(x_1 + x_2) \\ x_5 + x_6 = r^2(x_3 + x_4) \\ x_7 + x_8 = r^2(x_5 + x_6)$ In this case, $r^2 = 4,$ so the sum of the first $8$ terms is $S_8 = 12 + 4 (12) + 16 (12) + 64 (12) \\ S_8 = \boxed{1020}$

Side note: in the problem, we are told that all terms are strictly positive. In fact, this information is irrelevant! Let's say the series could have negative terms; then from the solution above, $r = -2.$ If you solve the series this way, the sum still becomes $1020.$ (The first two terms are $-12,\ 24.$ ) So why is it irrelevant? From the solution above, $r$ doesn't actually make any difference; $r^2$ does. So $r = \pm\sqrt{r^2}.$ Either way, the sum is only affected by $r^2,$ so it will come out to be the same!

Write the expression in terms of a and r (a is the first term and r is the ratio). The product of the first 2 terms is 12 (write in terms of a and r). The product of the 3rd and 4th term is 48 (write in terms of a and r). Divide and r=2, substitute into either equations and a=4. Then use the sum formula for partial geometric sequence.

a1 + a2 = 12 i.e ar + ar² = 12 and a3 + a4 = 48 i.e. ar² + ar³ = 48

divide r² = 48/12 = 4 i.e r = 2 so a = 4

Sn = a[r^n – 1]/(r – 1) = S8 = 4]2^8 – 1]/1 = 1020