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Geometric Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions1 Mark
MCQ
Given that $x > 0,$ the sum $\sum\limits^\infty_{\text{n}=1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{\text{n}-1}$ equals :
  • A
    $\text{x}$
  • ✓
    $\text{x}+1$
  • C
    $\frac{\text{x}}{2\text{x}+1}$
  • D
    $\frac{\text{x}+1}{2\text{x}+1}$
✓

Answer

Correct option: B.
$\text{x}+1$
$\sum\limits^\infty_{​​\text{x}+1}\Big(\frac{\text{x}}{\text{x}+1}\Big)^{(\text{n}-1)}$
$=1+\Big(\frac{\text{x}}{\text{x}+1}\Big)+\Big(\frac{\text{x}}{\text{x}+1}\Big)^2+\Big(\frac{\text{x}}{\text{x}+1}\Big)^3+\Big(\frac{\text{x}}{\text{x}+1}\Big)^4+\dots\infty$
$=\frac{1}{1-\big(\frac{\text{x}}{\text{x}+1}\big)}$
$\Big[\because$ it is a $G.P$. with $a = 1$ and $\text{r}=\Big(\frac{\text{x}}{\text{x}+1}\Big)\Big]$
$=\frac{(\text{x}+1)}{(\text{x}+1-\text{x})}$
$=\frac{(\text{x}+1)}{1}=(\text{x}+1)$

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