Interpret the given list of 10 numbers.

The numbers are written as: $7\times 8,\ 10\times 10,\ 13\times 12,\ 16\times 14,\ \dots$

This means each term is a product of two arithmetic progressions:

• First factors: $7,10,13,16,\dots$
• Second factors: $8,10,12,14,\dots$

For the $n$-th term: $a_n=(7+3(n-1))(8+2(n-1))$

Simplify the general term.

$a_n=(3n+4)(2n+6)$

Expanding, $a_n=6n^2+26n+24$

Mean of 10 numbers.

The mean is $\bar a=\frac{1}{10}\sum_{n=1}^{10} a_n$

So, $\bar a=\frac{1}{10}\sum_{n=1}^{10}(6n^2+26n+24)$

$\bar a=\frac{1}{10}\left(6\sum_{n=1}^{10}n^2+26\sum_{n=1}^{10}n+24\sum_{n=1}^{10}1\right)$

Use standard sums.

$\sum_{n=1}^{10} n=\frac{10\cdot 11}{2}=55$ $\sum_{n=1}^{10} n^2=\frac{10\cdot 11\cdot 21}{6}=385$ $\sum_{n=1}^{10}1=10$

Substitute: $\bar a=\frac{1}{10}\left(6\cdot 385+26\cdot 55+24\cdot 10\right)$

$\bar a=\frac{1}{10}(2310+1430+240)$

$\bar a=\frac{3980}{10}=398$

Therefore, the mean of the 10 numbers is: $\boxed{398}$

Comparison with stored answer.

Stored correct answer = $398$.

Our derived answer matches the stored answer.