Finding common ratio from two sums

Question

I'm struggling with this very basic question on the binomial theorem:

The sum of the first and second terms of a geometric progression is 12, and the sum of the third and fourth term is 48. Find the two possible values of the common ratio and the corresponding values of the first term.

So I tried approaching this in two ways. The first:

$$ u_1 + u_2 = a + ar = 12 \\ u_3 + u_4 = ar^2 + ar^3 = 48 \\ 4a(1 + r) = ar^2 + ar^3 \\ 4 + 4r = r^2 + r^3 \\ r^3 + r^2 - 4r - 4 =0 $$

Okay, no solution there. So then I tried:

$$ S_2 = \frac{a(r^2 - 1)}{r-1} = 12 \\ S_4 = \frac{a(r^4 - 1)}{r-1} = 48 \\ \frac{4a(r^2 - 1)}{r-1} = \frac{a(r^4 - 1)}{r-1} \\ 4r^2 - 4 = r^4 - 1 \\ r^4 - 4r^2 + 3 =0 $$

Am I missing something obvious here?

Answer 1

From your first two equations we get $$\frac{48}{12}=\frac{ar^2+ar^3}{a+ar}=r^2.$$

Answer 2

Continuing with the notation you've used, given $a + ar = 12$, $ar^2 +ar^3 = 48$, we see that

$ar^2 + ar^3 = r^2(a + ar)$

$\implies 48 = r^2*12$

$\implies r = \pm 2$

Back-substituting this gives us $r = 2$ and $a = 4$ or $r=-2$ and $a = -12$.