$\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots$ upto infinite terms, is equal to

Let $729,81,9,1, \ldots$ be a sequence and $\mathrm{P}_n$ denote the product of the first $n$ terms of this sequence.

If $2 \sum\limits_{n=1}^{40}\left(\mathrm{P}_n\right)^{\frac{1}{n}}=\frac{3^\alpha-1}{3^\beta}$ and $\operatorname{gcd}(\alpha, \beta)=1$, then

$\alpha+\beta$ is equal to

Consider an A.P.: $a_1, a_2, \ldots, a_{\mathrm{n}} ; a_1>0$. If $a_2-a_1=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_1$, and $\sum\limits_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum\limits_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to

Let $\sum\limits_{k=1}^n a_k=\alpha n^2+\beta n$. If $a_{10}=59$ and $a_6=7 a_1$, then $\alpha+\beta$ is equal to :