2.overline{357} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $2.\overline{357} = 2.357357357...$This can be written as $2 + 0.357357357...$$= 2 + \frac{357}{10^3} + \frac{357}{10^6} + \frac{357}{1

Vedclass · Accepted Answer

$\frac{2355}{999}$

Vedclass · Answer

(C) Given that $2.\overline{357} = 2.357357357...$This can be written as $2 + 0.357357357...$$= 2 + \frac{357}{10^3} + \frac{357}{10^6} + \frac{357}{10^9} + ...$This is an infinite geometric series with the first term $a = \frac{357}{1000}$ and common ratio $r = \frac{1}{1000}$.The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.$S = \frac{\frac{357}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{357}{1000}}{\frac{999}{1000}} = \frac{357}{999}$.Therefore,$2.\overline{357} = 2 + \frac{357}{999} = \frac{2 	imes 999 + 357}{999} = \frac{1998 + 357}{999} = \frac{2355}{999}$.

$2.\overline{357} = $

Similar Questions

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

The first term of a $G.P.$ is $7$,the last term is $448$,and the sum of all terms is $889$. Then the common ratio is:

Find the $n^{th}$ term of the series:
$\frac{1}{n} + \frac{n+1}{n} + \frac{2n+1}{n} + \ldots$

If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x = 1}^n f(x) = 120$,then the value of $n$ is

The sum of infinite terms of a $G.P.$ is $x$ and on squaring each term of it,the sum becomes $y$. Then the common ratio of this series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module