Question

In AP find the sum of n term of 0.3+0.33+0.333+......

Answer 1

0.3+0.33+0.333+……n terms.

First we take 3 as common from the series,

So, 3×{0.1+0.11+0.111+…. n terms}

Now we multiply it by 9 and also divide it by 9 to neutralize the net value

3/9×{0.9+0.99+0.999+…..n terms}

3/9×{(1–0.1)+(1–0.01)+(1–0.001)+……n terms}

1 added upto n times to become ’n' , so we have

3/9×{n -(0.1+0.01+0.001+…… n terms) }

The term we get above (0.1+0.01+0.001+…..n terms) is in GP.

With a=0.1 & R= 0.1 so sum of the series is

Sn= a(1-r^n)/(1-r)

Sn=0.1(1–0.1^n)/(1–0.1)

Sn=(1–0.1^n)/9

So the some of the series is

3/9×{n-1/9(1–0.1^n) }

1/27×{9n-1+0.1^n}