Respuesta :
Let the numbers be:
[tex]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}[/tex]
The sum of the ten smaller one in the same manner:
[tex]2(n)+...+2(n+9)=2(n+...+(n+9))\\=2\times10\frac{n+n+9}{2}[/tex]
Compute the difference of the above two sums:
[tex] \\10(2n+29)-10\(2n+9)=290-90=200[/tex]
The bigger sum is 200 more bigger than the sum o the ten smaller sum.
[tex]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}[/tex]
The sum of the ten smaller one in the same manner:
[tex]2(n)+...+2(n+9)=2(n+...+(n+9))\\=2\times10\frac{n+n+9}{2}[/tex]
Compute the difference of the above two sums:
[tex] \\10(2n+29)-10\(2n+9)=290-90=200[/tex]
The bigger sum is 200 more bigger than the sum o the ten smaller sum.
