If a_1 (0), a_2, a_3, a_4, a_5 are in a G.P., a_2+a_4 = 2a_3+1, and 3a_2+a_3 = 2a_4, then…

Question

If $a_1 (>0), a_2, a_3, a_4, a_5$ are in a G.P., $a_2+a_4 = 2a_3+1$, and $3a_2+a_3 = 2a_4$, then $a_2+a_4+2a_5$ is equal to \_\_\_\_\_\_\_\_.

Answer: 40

Accepted Answer

With ratio $r$: $ar(r-1)^2 = 1$ (from the first condition) and $3+r=2r^2$ (from the second, after dividing by $ar\ne0$), giving $r=\tfrac32$ or $r=-1$.

Since $a_1>0$ requires $a=\dfrac{1}{r(r-1)^2}>0$, testing $r=\tfrac32$: $a=\dfrac{8}{3}$.

$$a_2=4,\quad a_3=6,\quad a_4=9,\quad a_5=\frac{27}{2}.$$

$a_2+a_4+2a_5 = 4+9+27 = 40.$

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