The sum of first $20$ terms of the sequence $0.7,0.77,0.777, . . . $ is
$\frac{7}{{18}}\left( {179 - {{10}^{ - 20}}} \right)$
$\;\frac{7}{9}\left( {99 - {{10}^{ - 20}}} \right)$
$\;\frac{7}{{81}}\left( {179 + {{10}^{ - 20}}} \right)$
$\;\frac{7}{9}\left( {99 + {{10}^{ - 20}}} \right)$
What will $Rs.$ $500$ amounts to in $10$ years after its deposit in a bank which pays annual interest rate of $10 \%$ compounded annually?
Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and
$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$
If $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots . .+\frac{10240}{3}=2^{ n } \cdot m$, where $m$ is odd, then $m . n$ is equal to
Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to
If the ${p^{th}}$,${q^{th}}$ and ${r^{th}}$ term of a $G.P.$ are $a,\;b,\;c$ respectively, then ${a^{q - r}}{b^{r - p}}{c^{p - q}}$ is equal to