Consider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Find the $15^{th}$ term $(T_{15})$ of this sequence.

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$. If $B - 2A = 100\lambda$,then $\lambda$ is equal to:

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

Let $a_{1}, a_{2}, \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S_{n} = a_{1} + a_{2} + \ldots + a_{n}$. If $a_{1} = 1$,$a_{n} = 300$ and $15 \leq n \leq 50$,then the ordered pair $(S_{n-4}, a_{n-4})$ is equal to

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

Find the $10^{th}$ term of the geometric series $5+25+125+\ldots$