Answer: first term \(176\), common difference \(-3\); least \(n=119\).

Let the first term be \(a\) and the common difference be \(d\).

The seventh term is

\(a+6d=158.\)

The tenth term is

\(a+9d=149.\)

Subtracting the first equation from the second gives

\(3d=-9,\)

so

\(d=-3.\)

Then

\(a+6(-3)=158,\)

so

\(a=176.\)

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

\(S_n=\frac{n}{2}\{2a+(n-1)d\}.\)

Substituting \(a=176\) and \(d=-3\),

\(S_n=\frac{n}{2}\{352-3(n-1)\} =\frac{n}{2}(355-3n).\)

For the sum to be negative, since \(n\gt 0\),

\(355-3n\lt 0.\)

Therefore

\(n\gt \frac{355}{3}=118\frac13.\)

The least integer value is

\(n=119.\)