If twice the $11^{th}$ term of an Arithmetic Progression is equal to seven times its $21^{st}$ term,then its $25^{th}$ term is .......

Let $a_n, n \geq 1$,be an arithmetic progression with first term $2$ and common difference $4$. Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum_{n=1}^{10} M_n$ is

The ${n^{th}}$ term of the series $3 \cdot 8 + 6 \cdot 11 + 9 \cdot 14 + 12 \cdot 17 + \dots$ will be

If the sides of a right-angled triangle are in $A.P.$,then the sides are proportional to

Let $a_1, a_2, a_3, \ldots$ be in an $A.P.$ such that $\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1$,where $a_1 \neq 0$. If $\sum_{k=1}^{n} a_k = 0$,then $n$ is:

If $\frac{3 + 5 + 7 + \dots + (2n + 1)}{5 + 8 + 11 + \dots + (3n + 2)} = 7$,find the value of $n$. (Note: The original problem implies $n$ terms in the numerator and $10$ terms in the denominator as per the provided solution structure).