Search any question & find its solution
Question:
Answered & Verified by Expert
What is the sum of first eight terms of the series
$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots . . ?$
Options:
$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots . . ?$
Solution:
1015 Upvotes
Verified Answer
The correct answer is:
$\frac{85}{128}$
$a=1$ $\mathrm{r}=-\frac{1}{2}( < 1)$
$\therefore \quad$ Sum of Ist 8 terms is :-
$\mathrm{S}_{8}=\frac{1\left[1-\left(\frac{-1}{2}\right)^{8}\right]}{1+\frac{1}{2}} \quad\left[\right.$ For $\left.\mathrm{G} . \mathrm{P} . \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}\right]$
$=\frac{\frac{1}{1}-\frac{1}{256}}{\frac{3}{2}}=\frac{85}{128}$
$\therefore \quad$ Sum of Ist 8 terms is :-
$\mathrm{S}_{8}=\frac{1\left[1-\left(\frac{-1}{2}\right)^{8}\right]}{1+\frac{1}{2}} \quad\left[\right.$ For $\left.\mathrm{G} . \mathrm{P} . \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}\right]$
$=\frac{\frac{1}{1}-\frac{1}{256}}{\frac{3}{2}}=\frac{85}{128}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.