Question

Let S be the sum, P the product and R the sum of the reciprocals of n terms in a G.P. Prove that $straight P squared straight R to the power of straight n space equals space straight S to the power of straight n.$

Solution

Let a be the first term and r be the common ratio of a G.P.
S = the sum of n terms of G.P. = a + ar + .........ar^n-1

∴ $straight S equals fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction$ ...(i)
P = the product of n terms of G.P. = a. ar. ar² ....... ar^n-1

∴ P = $straight a to the power of straight n straight r to the power of 1 plus 2 plus..... plus left parenthesis straight n minus 1 right parenthesis end exponent space equals space straight a to the power of straight n straight r to the power of fraction numerator straight n minus 1 over denominator 2 end fraction left parenthesis 1 plus straight n minus 1 right parenthesis end exponent$
$rightwards double arrow$ $straight P space equals space straight a to the power of straight n straight r to the power of fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction end exponent$ ...(ii)
 R = the sum of the reciprocals of n terms of G.P. = $space space 1 over straight a plus 1 over ar plus 1 over ar squared plus... space plus space 1 over ar to the power of straight n minus 1 end exponent$
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Now, $straight S over straight R space equals space fraction numerator begin display style fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction end style over denominator begin display style fraction numerator straight r over denominator straight a left parenthesis straight r minus 1 right parenthesis space end fraction end style open parentheses begin display style fraction numerator straight r to the power of straight n minus 1 over denominator straight r to the power of straight n end fraction end style close parentheses end fraction space equals space fraction numerator straight a squared straight r to the power of straight n over denominator straight r end fraction space equals space straight a squared straight r to the power of straight n minus 1 end exponent$
L.H.S. = $open parentheses straight S over straight R close parentheses to the power of straight n space equals space left parenthesis straight a squared straight r to the power of straight n minus 1 end exponent right parenthesis to the power of straight n space equals space straight a to the power of 2 straight n end exponent space straight r to the power of straight n left parenthesis straight n minus 1 right parenthesis end exponent$
R.H.S. = $straight P squared space equals space open square brackets straight a to the power of straight n space space straight r to the power of fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction end exponent close square brackets squared space equals space straight a to the power of 2 straight n space space space end exponent straight r to the power of straight n left parenthesis straight n minus 1 right parenthesis end exponent$