The sum of $100$ terms of the series $.9 + .09 + .009.........$ will be
$1 - {\left( {\frac{1}{{10}}} \right)^{100}}$
$1 + {\left( {\frac{1}{{10}}} \right)^{100}}$
$1 - {\left( {\frac{1}{{10}}} \right)^{106}}$
$1 + {\left( {\frac{1}{{10}}} \right)^{106}}$
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals
Let the positive numbers $a _1, a _2, a _3, a _4$ and $a _5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{ m }{ n }$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$, then $m + n$ is equal to $.........$.
The ${20^{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + .......$ will be
If $x, {G_1},{G_2},\;y$ be the consecutive terms of a $G.P.$, then the value of ${G_1}\,{G_2}$ will be
The solution of the equation $1 + a + {a^2} + {a^3} + ....... + {a^x}$ $ = (1 + a)(1 + {a^2})(1 + {a^4})$ is given by $x$ is equal to