Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be a G.P. of positive real numbers such that the A.M. of $a_{2}$ and $a_{4}$ is $117$ and the G.M. of $a_{2}$ and $a_{4}$ is $108$ . Then the A.M. of $a_{1}$ and $a_{5}$ is:

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Solution

$a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are in G.P.
A.M. of $a_{2} & a_{4} = 117$
$a_{2} + a_{4} = 234$
G.M. of $a_{2} & a_{4} = 108 = a_{3} a_{2} a_{4} = {(108)}^{2} (a_{2} {- a}_{4})^{2} = (a_{2} {+ a}_{4})^{2} - 4 a_{2} a_{4} = {(234)}^{2} - {(216)}^{2} = (450) (18) a_{2} {- a}_{4} = 30 \times 3 = 90 a_{2} = 162, a_{4} = 72 r = \frac{a_{3}}{a_{2}} = \frac{108}{162} = \frac{2}{3} a_{1} = 243, a_{5} = \frac{2}{3} (72) = 48$
A.M $= \frac{a_{1} + a_{5}}{2} = 145.5$

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