InterviewSolution
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InterviewSolution
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If `S`be the sum, `P`the product and `R`the sum of the reciprocals of `n`terms of a G.P. prove that `(S/R)^n=P^2dot` |
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Answer» Let the given GP be `a, ar, ar^(2),..., ar^((n-1))`. Then, `S=(a(1-r^(n)))/((1-r))` ..(i) and `P=[axxarxxar^(2) xx...xxar^((n-1))]` `rArr P=a^(n)r^([1+2+3+...+(n-1)])=a^(n) r^(1/2 (n-1) n)` `rArr P=a^(n)r^(1/2 (n-1) n)` `rArr P^(2)=a^(2n)r^((n-1)n)` ...(ii) And, `R={1/a+1/(ar)+...+1/(ar^(n-1))}=(1/a). ({1/r^(n)-1})/({1/r-1})` `rArr R=(r/a). ((1-r^(n)))/((1-r).r^(n))` `rArr R=((1-r^(n)))/(a(1-r).r^((n-1)))`. ...(iii) On dividing (i) by (iii), we get `S/R=(a(1-r^(n)))/((1-r)). (a(1-r).r^((n-1)))/((1-r^(n)))=a^(2) r^((n-1))` `:. (S/R)^(n)={a^(2)r^((n-1))}^(n)=a^(2n).r^((n-1)n)=P^(2)` [using (ii)]. Hence, `P^(2)=(S/R)^(n)`. |
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