**(i)** 5 + 55 + 555 + …

Let S_{n} = 5 + 55 + 555 + ….. up to n terms

= \( \dfrac{5}{9}[9+99+999+....to\;n \;terms]\)

= \( \dfrac{5}{9}[(10-1)+(10^2-1)+(10^3-1)\)

\( +....to\;n \;terms]\)

= \( \dfrac{5}{9}[(10+10^2+10^3+....to\;n\;terms)\)

\( -(1+1+to\;n\;terms)]\)

= \( \dfrac{5}{9}[\dfrac{10(10^n-1)}{10-1}-n]\)

= \( \dfrac{5}{9}[\dfrac{10(10^n-1)}{9}-n]\)

= \( \dfrac{50}{81}(10^n-1)-\dfrac{5n}{9}\)

Therefore, the sum of n terms of the given series is

\( \dfrac{50}{81}(10^n-1)-\dfrac{5n}{9}\)

**(ii)** Given, .6 + .66 + . 666 + …

Let S_{n} = 06. + 0.66 + 0.666 + … up to n term

\( \dfrac{6}{9}[(1-\dfrac{1}{10})+(1-\dfrac{1}{10^2})+(1-\dfrac{1}{10^3})+....to\;n\;terms]\)

= \( \dfrac{2}{3}[(1+1+...to\;n\;terms)-\)

\( \dfrac{1}{10}(1+\dfrac{1}{10}+\dfrac{1}{10^2}to\;n\;terms)]\)

\( =\dfrac{2}{3}[n-10(1-\dfrac{(\dfrac{1}{10})^n}{1-(\dfrac{1}{10})})]\)

= \( \dfrac{2}{3}n-\dfrac{2}{30}\times \dfrac{10}{9}(1-10^{-n})\)

= \( \dfrac{2}{3}n-\dfrac{2}{27}(1-10^{-n})\)

Therefore, the sum of n terms of the given series is

\( \dfrac{2}{3}n-\dfrac{2}{27}(1-10^{-n})\)

Answered by Abhisek | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).