Mathematics - Statistics Question with Solution | TestHub

Given that 

$7,10,13-- \& 8,10,12,--- \in A.P$

$r^{th }$term of an $A.P \Rightarrow t_r = a +\left(r-1\right)d$

Here $a = 7, d = 3 \& a = 8, d =2$

$t_r =3r +4 \& t_r =2r +6$

$\sum _{r=1}^{10}(3r+4)(2r+6)$
$=\sum _{r=1}^{10}\left(6r^2+26r+24\right)=6\sum _{r=1}^{10}{r}^2+26\sum _{r=1}^{10}r+\sum _{r=1}^{10}24$

$\sum _{r=1}^{n}1 = n, \sum _{r=1}^{n}r = \frac{n\left(n+1\right)}{2}, \sum _{r=1}^n{r}^2 = \frac{n\left(n+1\right)\left(2n+1\right)}{6}, \sum _{r=1}^n{r}^3 ={\left[ \frac{n\left(n+1\right)}{2}\right]}^2$

$=6 \left(\frac{10\times 11\times 21}{6}\right)+26\left(\frac{10\times 11}{2}\right)+24\left(10\right)$
$=2310+1430+240=3980$
mean $=398$