Prove that 3S=2Q+P for a geometric progression

Question

I did Arithmetic Progression and Geometric Progression two years ago but this is challenging to me. If 27 is the Pth term,8 is the Qth term and 12 is the Sth term of a G.P,then show that 3S=2Q+P

Answer 1

As usual, let $a$ be the first term of the geometric progression, and let $r$ be the common ratio.

So we know that

$$ar^{P-1}=27, \qquad ar^{Q-1}=8, \qquad ar^{S-1}=12.$$

Now note that $12^3=(8^2)(27)$.

It follows that $$(ar^{S-1})^3 =(ar^{Q-1})^2(ar^{P-1}).\tag{1}$$ Expand, and the result falls out. On the left, we have $a^3r^{3S-3}$ and on the right we have $a^3 r^{2Q+P-3}$. Cancel the $a^3$. We get that $r^{3S-3}=r^{2Q+P-3}$ and therefore $3S-3=2Q+P-3$.

Remark: We cheated a little. The equality $12^3=(8^2)(27)$ looks as if it was pulled out of a hat, but it really wasn't. By looking at (1), one can see that the desired result is equivalent to the assertion that the cube of $Ar^{S-1}$ is equal to the square of $ar^{Q-1}$ times $ar^{P-1}$. But the condition $(12)^3=(8^2)(27)$ is easy to verify by multiplying, or, better, factoring.