# Arithmetic and Geometric Progressions

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Sequence and Series

## Sum of Series

Find the sum of the series .4 + .44 + .444……. to n terms
1. 5.69
2. 14.44
3. $\frac{4}{81}[9n-1+\frac{1}{10^n}]$
4. $\frac{4}{81}[n + 1]$

Choice C. $\frac{4}{81}[9n-1+\frac{1}{10^n}]$

## Detailed Solution

.4 + .44 + .444+……. to n terms

= 4x[0.1 +0.11 +0.111+……..to n terms]

= $\frac{4}{9}$ * [0.9 +0.99 + 0.999+…….to n terms] (Multiplying and dividing by 9)

= $\frac{4}{9}$ * [(1-0.1)+(1-0.01)+(1-0.001)+…….to n terms]

= $\frac{4}{9}$ * [(1+1+1…..to n terms)-(0.1+0.01+0.001…to n terms)]

= $\frac{4}{9} * [n - \frac{\frac{1}{10} * (1 - \frac{1}{10^n})}{1 - \frac{1}{10}}]$

= $\frac{4}{9} * [n - \frac{10^n - 1}{9 * 10^n}]$

= $\frac{4}{9} * [n - \frac{1}{9} * (1 - \frac{1}{10^n})]$

= $\frac{4}{81} * [9n - (1 - \frac{1}{10^n})]$

= $\frac{4}{81} * [9n - 1 + \frac{1}{10^n}]$

Correct Answer: $\frac{4}{81}[9n-1+\frac{1}{10^n}]$

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## More questions from Progressions

With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. Reinforce these ideas with these questions.