Question

If $a, x$ are real numbers and $|a|<1,|x|<1$, then $1+(1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty$ is equal to
(A) $\frac{1}{(1-a)(1-a x)}$
(B) $\frac{1}{(1-a)(1-x)}$
(C) $\frac{1}{(1-x)(1-a x)}$
(D) $\frac{1}{(1+a x)(1-a)}$

Answer 1

Ans: (C)
 $: \frac{1}{1-x}+\frac{a x}{1-x}+\frac{a^{2} x^{2}}{1-x}+\ldots .=\frac{1}{1-x} \times\left(1+a x+a^{2} x^{2}+\ldots\right)=\frac{1}{1-x} \cdot \frac{1}{1-a x}$

Answer 2

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Answer 3

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