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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Sequences and Series

NDA 2018 · Mathematics · Sequences and Series

If x=1-y+y^2-y^3+ . up to infinite terms, where |y| < 1 then which...

If x=1−y+y2−y3+…x=1-y+y^2-y^3+\ldots. up to infinite terms, where ∣y∣<1|y| < 1 then which one of the following is correct?

Options

  1. Ax=11+yx=\frac{1}{1+y}
  2. Bx=y1−yx=\frac{y}{1-y}
  3. Cx=y1+yx=\frac{y}{1+y}
  4. Dx=y1−yx=\frac{y}{1-y}

Correct answer

A. x=11+yx=\frac{1}{1+y}

Step-by-step solution

x=1−y+y2−y3+…x=1-y+y^2-y^3+\ldots. up to infinite terms. We can see that the given series is geometric progression, with a=1andr=−y\mathrm{a}=1 and \mathrm{r}=-\mathrm{y}∴S∞=a1−r=11−(−y)=11+y∴x=11+y.\begin{aligned} & \therefore \mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-(-\mathrm{y})}=\frac{1}{1+\mathrm{y}} \\ & \therefore \mathrm{x}=\frac{1}{1+\mathrm{y}}. \end{aligned}$$
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