Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2R= Sn

Asked by Pragya Singh | 1 year ago |  78

##### Solution :-

Let the terms in G.P. be a, ar, ar2, ar3, … arn – 1

Form the question, we have

$$s_n=\dfrac{a(1-r^n)}{1-r}$$

$$P=a^n\times r^{1+2+..+n-1}$$

Since the sum of first n natural numbers is

$$\dfrac{n(n+1)}{2}$$

$$P=a^n\times r^{\dfrac{n(n-1)}{2}}$$

$$R=\dfrac{1}{a}+\dfrac{1}{ar}+..+\dfrac{1}{ar^{n-1}}$$

$$\dfrac{r^{n-1}+r^{n-2}+...r+1}{ar^{n-1}}$$

Since 1,r,.... $$r^{n-1}$$ forms a G.P.,

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

$$\dfrac{(r^n-1)}{ar^{n-1}(r-1)}$$

Then,

$$P^2R^2=a^{2n}r^{n(n-1)}$$
$$\dfrac{(r^n-1)^n}{a^nr^{n-1}(r-1)}$$

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

Therefore, $$P^2R^2=S^n$$

Answered by Pragya Singh | 1 year ago

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