Given sequence: 8, 88, 888, 8888…
This sequence is not a G.P.
But, it can be changed to G.P. by writing the terms as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
= \( \dfrac{8}{9}[9+99+999+9999\)
\( +....to\;n\;terms]\)
\( \dfrac{8}{9}[(10-1)+(10^2-1)+(10^3-1)\)
\( +(10^4-1)+.......to\;n\;terms]\)
= \( \dfrac{8}{9}[10+10^2+....to\;n\;terms\)
\( -(1+1+1+..n\;tems)]\)
\( \dfrac{8}{9}[\dfrac{10(10^n-1)}{10-1}-n]\)
\( \dfrac{8}{9}[\dfrac{10(10^n-1)}{9}-n]\)
\( \dfrac{80}{81}(10^n-1)-\dfrac{8}{9}n\)
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 \( \dfrac{81}{2}\).