Tricky AP- GP

Ah! It was easy yet nice!

$Geometric mean of 2 numbers(b,c(here)) in GP(a,b,c) = \sqrt{ac}$

$b = \sqrt{ac}$

Now, $a, a + d, a + 2d + 3, a + 3d + 8$ are in G.P.

Therefore, $a+d = \sqrt{a(a+2d+3)}$

${a+d}^{2} = {a}^{2} + 2ad + 3a$

${a}^{2} + {d}^{2} + 2ad = {a}^{2} + 2ad + 3a$

$a = \frac{{d}^{2}}{3}$

Also, $a + 2d + 3 = \sqrt{(a+d)(a+3d+8)}$

${a}^{2} + 4{d}^{2} + 9 + 4ad + 12d + 6a = {a}^{2} + 3ad + 8a + ad + 3{d}^{2} + 8d$

${d}^{2} + 4d - 2a = -9$

Putting $a = \frac{{d}^{2}}{3}$ ,

$3{d}^{2} + 12d -2{d}^{2} = -27$

${d}^{2} + 12d + 27 = 0$

Now using quadratic formula, the roots of the equation are $-9, -3$

Now, d can be -9, - 3

Now, a is respectively 27, 3.

But for a = 3, d= -3,

the A.P. is 3, 0, -3, -6 and therefore the first 2 terms don't form a G.P.

For a = 27, d = -9,

the A.P. is 27, 18, 9, 0

Now, the G.P. will be 27, 18, 21, 8 with r = $\frac{2}{3}$

Hence, sum of A.P. is 27 + 18 + 9 = $\boxed{54}$