If ∑ n = 210,

Solution:

(a) 2870

Given:

∑n = 210

$\Rightarrow n\left(\frac{n+1}{2}\right)=210$

$\Rightarrow n^{2}+n-420=0$

$\Rightarrow(n-20)(n+21)=0$

$\Rightarrow n=20$$\quad(\because n>0)$

Now,

$\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$

$\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$

$\Rightarrow \frac{n(n+1)}{2} \times \frac{(2 n+1)}{3}$

$\Rightarrow(210) \times\left(\frac{41}{3}\right)$

$\Rightarrow(70) \times(41)$

$\Rightarrow 2870$