Sum Inside Sum?

Let the first term be $\large \displaystyle a=1$ , common ratio $\large \displaystyle r$ , and infinite sum $\large \displaystyle S$ .

Let any term of the IGP be $\large \displaystyle ar^n$ .

Then the sum of the terms following this term $\large \displaystyle = \frac{ar^{n+1}}{1-r}$ .

According to the question,

$\large \displaystyle ar^n = \frac{ar^{n+1}}{1-r}$

$\large \displaystyle \frac{r^{n+1}}{r^n}=1-r$

$\large \displaystyle \frac{r^n \times r}{r^n} = 1- r$

$\large \displaystyle r=1-r$

$\large \displaystyle \boxed{r=\frac{1}{2}}$

Hence, the sum of this IGP = $\large \displaystyle \frac{a}{1-r} = \frac{1}{1-\frac{1}{2}} = \boxed{2}$

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Question Source: R. D. Sharma - Mathematics For Class XI - Geometric Progressions - Ex 20.4 - Level 2 - Q. 10 (Pg. 20.39)

It is a verbal question as 1=sum of terms following it and sum of GP=1+ sum of terms following it=2.