Question

Find the sum of the series 0.6 + 0.66 + 0.666 + ... .

Or

The product of the three numbers in GP is 216. If 2, 8 and 6 be added to them, then the results are in Al, find the numbers.

Answer 1

We have, 0.6 + 0.66 + 0.666 + . . . upto n terms

$=6\times 0.1+6\times 0.111+6\times 0.111+\cdots$ upto n terms

= 6[0.1 + 0.11 + 0.111 + . . . upto n terms]

$=\frac{6}{9}$ [0.9 + 0.99 + 0.999 + upto n terms]

[multiplying numerator and denominator by 9]

$=\frac{2}{3}[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\cdots \text{upto n terms}]$

$=\frac{2}{3}[(1-\frac{1}{10})+(1-\frac{1}{100})+(1-\frac{1}{100})+\cdots \text{upto n terms}]$

$=\frac{2}{3}\left[\right(1+1+1+\cdots \text{n terms})-(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+\cdots \text{n terms})]$

$=\frac{2}{3}[n-\frac{1}{10}\left\{\frac{1-{\left(\frac{1}{10}\right)}^n}{1-\frac{1}{10}}\right\}]$

$[\because \text{sum of GP}=a\frac{(1-r^n)}{(1-r)},|r|<1]$

$=\frac{2}{3}[n-\frac{1}{10}\left\{\frac{1-{\left(\frac{1}{10}\right)}^n}{\frac{9}{10}}\right\}]$

$=\frac{2}{3}[n-\frac{1}{9}\{1-{\left(\frac{1}{10}\right)}^n\}]=\frac{2}{3}n-\frac{2}{27}(1-{10}^{-n})$

Let the numbers in GP be a, ar and $a{r}^2$.

It is given that, the product if these numbers is 216.

$\therefore a.ar.a{r}^2=216$

$\Rightarrow a^3{r}^3=216\Rightarrow ar=6$ [taking cube root]

$\Rightarrow a=\frac{6}{r}$ . . .(i)

It is also given that a + 2, ar + 8 and $a{r}^2$ + 6 are in AP.

$\therefore 2(ar+8)=a+2+a{r}^2+6$

$\Rightarrow 2ar+16=a+a{r}^2+8$

$\Rightarrow a{r}^2-2ar+a-8=0$ . . . (ii)

Putting $a=\frac{6}{r}$ in Eq. (ii), we get

$\frac{6}{r}\left(r^2\right)-2\times \frac{6}{r}\times r+\frac{6}{r}-8=0$

$\Rightarrow 6r-12+\frac{6}{r}-8=0$

$\Rightarrow 6r^2-12r+6-8r=0$

$\Rightarrow 6r^2-20r+6=0$

$\Rightarrow 3r^2-10r+3=0$ [dividing by 2]

$\Rightarrow (3r-1)(r-3)=0\Rightarrow r=\frac{1}{3},3$

When r = 3, $a=\frac{6}{3}=2$

and when $r=\frac{1}{3},a=\frac{6}{\frac{1}{3}}=18$

Hence, the numbers are 2, 6, 18 or 18, 6, 2.