Let a_1,a_2,a_3, be a G.P.

Question

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:

Accepted Answer

Let first term $A$, ratio $r>1$.

$a_1a_5 = A^2r^4 = 28 \quad (1)$, and $a_2+a_4 = A(r+r^3) = 29 \quad (2)$.

From (1), $A^2=28/r^4$. Substituting into (2) and simplifying leads to $r^2=28 \Rightarrow r=2\sqrt7$, and then $A = \dfrac{29}{r+r^3} = \dfrac{29}{58\sqrt7} = \dfrac{1}{2\sqrt7}$.

$$a_6 = Ar^5 = \frac{1}{2\sqrt7}\times(2\sqrt7)^5 = \frac{(2\sqrt7)^4\times2\sqrt7}{2\sqrt7} = (2\sqrt7)^4 = 784.$$

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