The mean of 10 numbers 7 × 8, 10 × 10, 13 × 12, 16 × 14, is

Given that

$7, 10, 13 - - \text{ and } 8, 10, 12, - - - \in A . P$

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

Here $a = 7, d = 3 \text{ and } a = 8, d = 2$

${t}_{r} = 3r + 4 \text{ and } {t}_{r} = 2r + 6$

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

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

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