Let $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive terms. If $a_1a_5=28$ and $a_2+a_4=29$, then $a_6$ is equal to:

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $a_3^2=a_1a_5=28$, so $a_3=\sqrt{28}$. $a_2+a_4=a_3\left(r+\frac1r\right)=29$ gives $r=\sqrt{28}$ (increasing G.P.). $a_6=a_3r^3=\sqrt{28}\cdot28\sqrt{28}=784$. Answer (4).

Q. Let $a_1,a_2,a_3,\ldots$ be a G.P. of increasing positive terms. If $a_1a_5=28$ and $a_2+a_4=29$, then $a_6$ is equal to:

Explanation

Reviewed by Dr. Abubacker

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