Let a,b,c be in G.P.

Question

Let $a,b,c$ be in G.P. with common ratio $r$, where $a
e0$ and $0<r\le\dfrac12$. If $3a, 7b, 15c$ are the first three terms of an A.P., then the 4th term of this A.P. is:

Accepted Answer

$14(ar) = 3a+15ar^2 \Rightarrow 15r^2-14r+3=0 \Rightarrow r=\dfrac13$ or $\dfrac35$; since $r\le\dfrac12$, $r=\dfrac13$.

Fourth term $=15ar^2+7ar-3a = a(15r^2+7r-3) = a\left(\dfrac{15}{9}+\dfrac73-3\right) = a.$

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