Question

There are 20 consecutive even numbers. how much bigger is the sum of the 10 larger ones than the sum of the ten smaller ones?

Answer 1

Let the numbers be:
$2n,2(n+1),...,2(n+19)\\\text{Sum of the 10 biggest ones:}\\2(n+10)+...+2(n+19)=2((n+10)+...+(n+19))\\=2\times10\frac{n+10+n+19}{2}\\=10(2n+29)\text{the sum of an arithmetic sequence}$
The sum of the ten smaller one in the same manner:
$2(n)+...+2(n+9)=2(n+...+(n+9))\\=2\times10\frac{n+n+9}{2}$
Compute the difference of  the above two sums:
$\\10(2n+29)-10\(2n+9)=290-90=200$

The bigger sum is 200 more  bigger than the sum o the ten smaller sum.