Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}$, $a > 2$. If $\alpha$ is such that $a$, $4$, $\alpha$, $b$ are in A.P., then the equation $\alpha x^2 - a x + 2(\alpha - 2b) = 0$ has :

$ \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