The mean of 10 numbers
$$7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14, \ldots$$ is _________.

We are given ten numbers that follow the pattern $$7 \times 8,\; 10 \times 10,\; 13 \times 12,\; 16 \times 14,\; \ldots$$ and we have to find their mean.

First we observe the pattern in each factor:

• The first factor starts at $$7$$ and increases by $$3$$ each time: $$7,\; 10,\; 13,\; 16,\; \ldots$$

• The second factor starts at $$8$$ and increases by $$2$$ each time: $$8,\; 10,\; 12,\; 14,\; \ldots$$

So for the $$n^{\text{th}}$$ term (counting the first term as $$n=1$$) we can write

$$a_n=\bigl(7+3(n-1)\bigr)\bigl(8+2(n-1)\bigr).$$

Because we need ten terms, it will be convenient to let $$n$$ run from $$0$$ to $$9$$ instead of $$1$$ to $$10$$; that re-indexes nothing but simplifies the algebra. Thus we define

$$A_n=\bigl(7+3n\bigr)\bigl(8+2n\bigr),\qquad n=0,1,2,\ldots,9.$$

We now expand each product algebraically:

$$\begin{aligned} A_n&=\bigl(7+3n\bigr)\bigl(8+2n\bigr)\\ &=7\cdot8+7\cdot2n+3n\cdot8+3n\cdot2n\\ &=56+14n+24n+6n^{2}\\ &=56+38n+6n^{2}. \end{aligned}$$

The sum of all ten terms is therefore

$$S=\sum_{n=0}^{9}A_n=\sum_{n=0}^{9}\bigl(56+38n+6n^{2}\bigr).$$

We separate the sum into three simpler sums:

$$S=56\sum_{n=0}^{9}1+38\sum_{n=0}^{9}n+6\sum_{n=0}^{9}n^{2}.$$

We now evaluate each of these standard sums one by one.

1. The sum of ten ones:

$$\sum_{n=0}^{9}1=10.$$

2. The sum of the first nine positive integers (starting from zero):
The formula for the sum of the first $$m$$ integers is $$\dfrac{m(m+1)}{2}.$$ Here $$m=9$$, so

$$\sum_{n=0}^{9}n=\frac{9\cdot10}{2}=45.$$

3. The sum of the squares of the first nine positive integers (starting from zero):
The formula for the sum of squares is $$\dfrac{m(m+1)(2m+1)}{6}.$$ Again $$m=9$$, hence

$$\sum_{n=0}^{9}n^{2}=\frac{9\cdot10\cdot19}{6}=285.$$

Substituting these values back into $$S$$ we get

$$\begin{aligned} S&=56(10)+38(45)+6(285)\\ &=560+1710+1710\\ &=3980. \end{aligned}$$

The mean (average) of the ten numbers is the total sum divided by 10:

$$\text{Mean}=\frac{S}{10}=\frac{3980}{10}=398.$$

So, the answer is $$398$$.