The recurring decimal 0-125125125- is expressed as a rational number as mk- Find k-1m- Given mk is in its simplest form

Let x=0.125125125⋯ 1000x=125.125125125⋯ Subtract both the equations, we get 999x=125.0000000=125 x= 125 999 m=125 and k=999 ...

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Standard XII

Mathematics

Binary Operation

The recurring...

Question

The recurring decimal $0.125125125 \dots$ is expressed as a rational number as $\frac{m}{k}$ . Find $\frac{k + 1}{m}$ . (Given $\frac{m}{k}$ is in its simplest form)

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Solution

Let $x = 0.125125125 \dots$
$1000 x = 125.125125125 \dots$
Subtract both the equations, we get
$999 x = 125.0000000 = 125$
$x = \frac{125}{999}$
$m = 125$ and $k = 999$
$\frac{k + 1}{m} = \frac{1000}{125} = 8$

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The recurring decimal 0.125125125⋯ is expressed as a rational number as mk. Find k+1m. (Given mk is in its simplest form)

Let x=0.125125125⋯1000x=125.125125125⋯Subtract both the equations, we get999x=125.0000000=125x=125999m=125 and k=999k+1m=1000125=8

The recurring decimal $0.125125125 \dots$ is expressed as a rational number as $\frac{m}{k}$ . Find $\frac{k + 1}{m}$ . (Given $\frac{m}{k}$ is in its simplest form)

Let $x = 0.125125125 \dots$
$1000 x = 125.125125125 \dots$
Subtract both the equations, we get
$999 x = 125.0000000 = 125$
$x = \frac{125}{999}$
$m = 125$ and $k = 999$
$\frac{k + 1}{m} = \frac{1000}{125} = 8$