[ [Geometric progression]

It is known that the sum of the first n terms of a geometric progression consisting of positive numbers is $\mathrm{S}$, and the sum of the reciprocals of the first n terms of this progression is R. Find the product of the first n terms of this progression.

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Use the formula for the sum of the first n terms of a geometric progression; the reciprocals of the terms of a geometric progression also form a geometric progression.

## Solution

Let a be the first term of the progression, and q be its common ratio. Then, by the formula for the sum of the first n terms of a geometric progression, we get that $S=a+aq+aq^2+\ldots+aq^{n-1}=a(q^n-1)/(q-1)$. The reciprocals $1/a$, $1/(aq)$, ..., $1/(aq^{n-1})$ of the terms of the geometric progression also form a geometric progression with the first term $1/a$ and common ratio $1/q$. Hence, $R=1/a+1/(aq)+1/(aq^2)+\ldots+1/(aq^{n-1})=1/a(1/q^n-1)/(1/q-1)=(1/(aq^{n-1}))(q^n-1)/(q-1)$. From the expressions obtained for S and R, it follows that $S/R=a^2q^{(n-1)}$. The product of the first n terms of the progression is $P=a(aq)(aq^2)\ldots(aq^{n-1})=a^nq^{1+2+\ldots+(n-1)}=a^nq^{n(n-1)/2}$. Raising the equality $S/R=a^2q^{(n-1)}$ to the power of $(n/2)$, we get that $(S/R)^{n/2}=a^nq^{n(n-1)/2}=P$.

## Answer

$(S/R)^{n/2}$