What will $Rs. 500$ amount to in $10$ years after its deposit in a bank which pays an annual interest rate of $10\%$ compounded annually?

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,find the values of $a$ and $r$.

For the functions $f(\theta) = \alpha \tan^2 \theta + \beta \cot^2 \theta$ and $g(\theta) = \alpha \sin^2 \theta + \beta \cos^2 \theta$, where $\alpha > \beta > 0$, let $\min_{0 < \theta < \pi/2} f(\theta) = \max_{0 < \theta < \pi} g(\theta)$. If the first term of a $G$.$P$. is $(\frac{\alpha}{2\beta})$, its common ratio is $(\frac{2\beta}{\alpha})$ and the sum of its first $10$ terms is $\frac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is equal to . . . . . . .

If the $10^{th}$ term of a geometric progression is $9$ and the $4^{th}$ term is $4$,then its $7^{th}$ term is: