Find the sum to n terms in the geometric progression x3, x5, x7, … (if x ≠ ±1 )

Asked by Abhisek | 1 year ago |  100

##### Solution :-

Given G.P. is x3, x5, x7, …

Here, we have a = x3 and r = $$\dfrac{x^5}{x^3}=x^2$$

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

$$\dfrac{[(x^31-(x^2)^n]}{1-x^2}$$

$$\dfrac{x^3(1-x^{2n})}{1-x^2}$$

Answered by Pragya Singh | 1 year ago

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