$ \frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3} $ is equal to :

The value of $\sum\limits_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right)$ is

The common difference of the A.P.: $a_1, a_2, \ldots, a_{\mathrm{m}}$ is 13 more than the common difference of the A.P.: $b_1, b_2, \ldots, b_n$. If $b_{31}=-277, b_{43}=-385$ and $a_{78}=327$, then $a_1$ is equal to

Let $a_1, a_2, a_3, a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1 a_2 a_3 a_4+l^4=361$, then the largest term of the A.P. is equal to