(i) 5 + 55 + 555 + …
Let Sn = 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 Sn = 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) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).