Answer: First term \(=10-2d\), second term \(=10-d\), and \(S_{200}=80000\).

If the third term is \(10\), then working backwards by the common difference gives

\(\text{second term}=10-d,\)

and

\(\text{first term}=10-2d.\)

The sum of the squares of the first three terms is \(140\), so

\((10-2d)^2+(10-d)^2+10^2=140.\)

Expanding,

\(100-40d+4d^2+100-20d+d^2+100=140.\)

So

\(5d^2-60d+160=0.\)

Divide by \(5\):

\(d^2-12d+32=0.\)

Factorising,

\((d-4)(d-8)=0.\)

Thus \(d=4\) or \(d=8\). The smaller value is \(d=4\).

Then the first term is

\(10-2(4)=2.\)

The sum of the first \(200\) terms is

\(S_{200}=\frac{200}{2}\{2(2)+199(4)\}.\)

Therefore

\(S_{200}=100(4+796)=80000.\)