The Sufficient Sums

Let r = common ratio and a = first term

Sum of first 3 terms = a + a r + a r^2 = 8

a * (1 + r + r^2) = 8 => 1 + r + r^2 = 8/a

Sum of first 6 terms = a + a r + a r^2 + a r^3 + a r^4 + a*r^5 = 12

a * (1 + r + r^2) + a * (r^3 + r^4 + r^5) = 12

Since first term is = 8, then a * (r^3 + r^4 + r^5) = 4

a * r^3 * ( 1 + r + r^2) = 4

a * r^3 * (8/a) = 4

r^3 * 8 = 4

r^3 = 0.5

r = 0.79

The sum of the first $n$ -terms of a geometric progression is $S_n = \dfrac{a(r^n-1)}{r-1} \space \text{or} \space S_n = \dfrac{a(1-r^n)}{1-r}$ , where $a$ is the first term of the GP, $r$ be the common ratio, $S_n$ be the sum of the first $n$ -terms of a GP.

Given that $S_3 = \dfrac{a(r^3-1)}{r-1} = 8 \space ...(1)$

$S_6 = \dfrac{a(r^6-1)}{r-1} = 12 \space ...(2)$

$(1):(2)$ gives $\dfrac{r^3-1}{r^6-1} = \dfrac{2}{3}$

By factoring and dividing, we have

$\dfrac{1}{r^3 +1 } = \dfrac{2}{3}$

$r^3 +1 = \dfrac{3}{2}$

$r^3 = \dfrac{1}{2}$

$r = \sqrt[3]{ \dfrac{1}{2}} \approx 0.79$

$\text {Case 1}$ :-
Let the numbers be $\dfrac {a}{r}$ , $a$ and $ar$

$\dfrac {a}{r} + a + ar = 8$
$\dfrac {a + ar + ar^2}{r} = 8$
$a(1+r+r^2) = 8r$
Let $(1 + r + r^2)$ be $x$ $\therefore ax = 8r$
$a = \dfrac {8r}{x} \rightarrow \boxed {1}$

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$\text {Case 2}$ :-
Let the numbers be $\dfrac {a}{r^2}$ , $\dfrac {a}{r}$ , $a$ , $ar$ , $ar^2$ and $ar^3$

$\dfrac {a}{r^2}$ + $\dfrac {a}{r}$ + $a$ + $ar$ + $ar^2$ + $ar^3$ = 12)
$\dfrac {a + ar + ar^2 + ar^3 + ar^4 + ar^5}{r^2} = 12$
$a(1 + r + r^2 + r^3 + r^4 + r^5) = 12r^2$ Substituting the value of $x$ we get:-
$a(x + xr^3) = 12r^2$
$a = \dfrac{12r^2}{x(1+r^3)} \rightarrow \boxed {2}$

From $\boxed {1}$ and $\boxed {2}$ we get:-
$\dfrac {8r}{x} = \dfrac {12r^2}{x(1+r^3)}$
$2 + 2r^3 = 3$
$2r^3 = 1$
$r^3 = \dfrac {1}{2}$
$r = \sqrt[3]{\dfrac {1}{2}}$
$r \approx \boxed{0.79}$ .

a(r^3-1)/(r-1)=8 ---------(1) a(r^6-1)/(r-1)=12------(2) Simplifying L.H.S of eq(2) we get a{(r^3-1)(r^3+1)/(r-1)}=8(r^3+1)=12 r^3= (12/8)-1=0.5 Now r = cubroot of 0.5 =0.79

Cheers!