Question

S= 0.5 +0.55 + 0.555 sn=?

Answer 1

HELLO THERE!

The series 0.5 + 0.55 + 0.555 .... n terms is neither in AP, nor in GP, or in HP. To find its sum, we need to bring it to any one of the three Progressions.

These kind of problems are done in the following way:

$0.5 + 0.55 + 0.555.....n \\\\= 5 (0.1 + 0.11 + 0.111.....n) \\\\= \frac{5}{9}(0.9 + 0.99 + 0.999 ..... n) \\\\= \frac{5}{9}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) ...... n] \\\\= \frac{5}{9}[(1+1+1+....n) - (0.1+0.01+0.001.....n)] \\\\= \frac{5}{9}[n - 0.1(\frac{1 - (0.1)^{n}}{1 - 0.1})] \\\\= \frac{5}{9}[n - \frac{1 - (0.1)^{n}}{9}] \\\\= \frac{5}{9}[\frac{9n - 1 - (0.1)^{n}}{9}] \\\\= \frac{5[9n - 1 - (0.1)^{n}]}{81}$

Note that, summation of 1 + 1 + 1 + 1 .... upto n terms equals n.

And, 0.1 + 0.01 + 0.001 ... n terms are in GP with common ratio (r) = 0.1. So, here, use the formula for sum of terms in GP with r < 1, which is:

$S_{n} = a(\frac{1 - r^{n}}{1 - r})$

where a = 0.1, r = 0.1, and number of terms = n.

THANKS!

Answer 2

This is your answer !!!!!!