0.mathop {423}limits^{,,,,, bullet , bullet , | vedclass

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

Question

(a) We have $0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = 0.4232323.......$

$ = 0.4 + 0.023 + 0.00023 + 0.0000023 + ........\in

Vedclass · Accepted Answer

$\frac{{419}}{{990}}$

Vedclass · Answer

(a) We have $0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = 0.4232323.......$

$ = 0.4 + 0.023 + 0.00023 + 0.0000023 + ........\infty $

$ = \frac{4}{{10}} + \frac{{23}}{{{{10}^3}}} + \frac{{23}}{{{{10}^5}}} + \frac{{23}}{{{{10}^7}}}........\infty $

$ = \frac{4}{{10}} + \frac{{23}}{{{{10}^3}}}\left[ {1 + \frac{1}{{{{10}^2}}} + \frac{1}{{{{10}^4}}} + ..........\infty } \right]$

$ = \frac{4}{{10}} + \frac{{23}}{{1000}}\left( {\frac{1}{{1 - \frac{1}{{{{10}^2}}}}}} \right) = \frac{4}{{10}} + \frac{{23}}{{990}} = \frac{{419}}{{990}}$.

$0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = $

Similar Questions

In a geometric progression, if the ratio of the sum of first $5$ terms to the sum of their reciprocals is $49$, and the sum of the first and the third term is $35$ . Then the first term of this geometric progression is

The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.

The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of its squares of first three terms is $33033$, then the sum of these three terms is equal to

Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.