Answer:
Option B,D
Explanation:
Concept involved
Converting infinite series into definite intergral
ie $\lim_{n\rightarrow \infty}\frac{h(n)}{n}$
$\lim_{n \rightarrow \infty}\frac{1}{n} \sum_{r=g(n)}^{h(n)}f(\frac{r}{n})=\int_{}^{} f(x) dx$
$\lim_{n \rightarrow \infty}\frac{g(n)}{n}$
where,r/n is replaced with x
$\sum$ is replaced with integral
Here
$\lim_{n \rightarrow \infty}\frac{(1^{a}+2^{a}+....+n^{a})}{(n+1)^{a-1}[(na+1)+(na+2)+....+(na+n)]}$
= $\frac{1}{60}$
$\Rightarrow \lim_{n \rightarrow \infty}\frac{\sum_{r=1}^{n}r^{a}}{(n+1)^{a-1}\left[n^{2}a+\frac{n(n+1)}{2}\right]}=\frac{1}{60}$
$\Rightarrow \lim_{n \rightarrow \infty}\frac{2\sum_{r=1}^{n}(\frac{r}{n})^{a}}{(1+\frac{1}{n})^{a-1}.(2na+n+1)}$
$\Rightarrow \lim_{n \rightarrow \infty}\frac{1}{n}\left(2\sum_{r=1}^{n}(\frac{r}{n})^{a}\right)$
$\times \lim_{n \rightarrow \infty}\frac{1}{(1+\frac{1}{n})^{a-1}(2a+1+\frac{1}{n})}$
$\Rightarrow 2\int_{0}^{1} (x^{a})dx.\frac{1}{1.(2a+1)}$
$\Rightarrow \frac{2.(x^{a+1})_0^1}{(2a+1).)(a+1)}=\frac{2}{(2a+1)(a+1)}$
$\therefore$ $\frac{2}{(2a+1)(a+1)}=\frac{1}{60}$
$\Rightarrow(2a+1)(a+1)=120$
$\Rightarrow2a^{2}+3a+1-120=0$
$\Rightarrow2a^{2}+3a-119=0$
$\Rightarrow $ $ (2a+17) (a-7)=0$
$\Rightarrow $ a=7, $\frac{-17}{2}$
