Just tense and a half

Algebra Level 2

1 a = 1 2 ( 1 10 + 1 100 + 1 1000 + ⋯ ) \dfrac 1a = \frac 12 \left(\dfrac 1{10} + \dfrac 1{100}+\dfrac 1{1000} + \cdots \right)

Find a a .


The answer is 18.

View wiki
Aakash Nln by Aakash Nln , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Jun 30, 2016

1 a = 1 2 ( 1 10 + 1 100 + 1 1000 + ⋯ ) = 1 2 ⋅ 1 10 ( 1 1 − 1 10 ) = 1 20 × 10 9 = 1 18 ⟹ a = 18 \begin{aligned} \frac 1a & = \frac 12 \left( \frac 1{10} + \frac 1{100} + \frac 1{1000} + \cdots \right) \\ & = \frac 12 \cdot \frac 1{10} \left( \frac 1{1-\frac 1{10}}\right) \\ & = \frac 1{20} \times \frac {10}9 \\ & = \frac 1{18} \\ \implies a & = \boxed{18} \end{aligned}

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Sudoku Subbu
Jul 2, 2016

Let us consider the given equation 1 a = 1 2 ( 1 10 + 1 100 + 1 1000 … ) \frac{1}{a} = \frac{1}{2} \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \dots \right) The Value in the bracket follows infinite G.P T h e r e f o r e , ( 1 10 + 1 100 + 1 1000 … ) = 1 10 1 − 1 10 = 1 9 Therefore, \space \left( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \dots \right) = \frac{\frac{1}{10}}{1-\frac{1}{10}} = \frac{1}{9} = > 1 a = 1 2 ( 1 9 ) =>\frac{1}{a}=\frac{1}{2} \left(\frac{1}{9} \right) = > a = 18 => \space a=18

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...