Question

let S be the sum P be the product and r the sum of reciprokens of n terms of a G.P. P square R to the power n equal to S to the power n

Answer 1

Step-by-step explanation:

Let ' $\large{\sf{\red{a}}}$' be the first term and ' $\large{\sf{\green{r}}}$' be the common ratio of G.P

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$\sf \therefore S\:=\: \dfrac{a(1-r^n)}{1-r}$

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$\sf P\:=\: (a)(ar)(ar^2)\dots \dots (ar^{n-1})$

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$\sf \quad=\:a^{n}r^{1+2+3+\dots \dots +(n-1)}$

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$\sf \quad=\:a^{n}r^{\frac{n(n-1)}{2}}\qquad$$\sf \left[\because\: \sum n\:=\: \dfrac{n(n+1)}{2}\right]$

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$\sf R\:=\: \dfrac{1}{a}+ \dfrac{1}{ar}+ \dfrac{1}{ar^2}+ \dots \dots \dfrac{1}{ar^{n-1}}$

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$\sf \quad=\: \dfrac{1}{a} \left[\dfrac{1- \left(\dfrac{1}{r}\right)^n}{1-\dfrac{1}{r}}\right]$$\sf \:=\: \dfrac{1}{a} \left[\dfrac{\dfrac{r^n-1}{r^n}}{\dfrac{(r-1)}{r}}\right]$

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$\sf \quad=\: \dfrac{1}{a}\left[\dfrac{r^{n-1}}{r^n}\times \dfrac{r}{(r-1)}\right]$ $\sf \:=\: \dfrac{(1-r^n)}{ar^{n-1} (1-r)}$

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$\sf R\:=\: \dfrac{-(1-r^n)}{-(1-r)ar^{n-1}}\:=\: \dfrac{(1-r^n)}{ar^{n-1}(1-r)}$

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$\sf \therefore\dfrac{S}{R}\:=\: \dfrac{\dfrac{a(1-r^n)}{(1-r)}}{\dfrac{(1-r^n)}{ar^{n-1}(1-r)}}\:=\:a^2r^{n-1}$

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$\sf \therefore\left(\dfrac{S}{R}\right)^2\:=\:(a^2r^{n-1})^n</p><p> ‎ ‎ ‎ ‎ ‎ ‎ ‎</p><p></p><p>[tex]\sf \quad=\:a^{2n}r^{n(n-1)}$

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$\sf \quad=\: \left[a^nr^{\dfrac{n(n-1)}{2}}\right]^2$

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$\sf \therefore\dfrac{S}{R}\:=\:p^2$

Answer 2

Answer:

Ans is

${p}^{2}$