Answer: (a) first term \(4\), common difference \(3\); (b) \(1216\).

Answer: (a) first term \(4\), common difference \(3\); (b) \(1216\).

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

The third term is \(a+2d\), so

\(a+2d=10.\)

The sum of the first \(8\) terms is

\(S_8=\frac82(2a+7d).\)

Since \(S_8=116\),

\(4(2a+7d)=116,\)

so

\(2a+7d=29.\)

From \(a+2d=10\), \(2a+4d=20\). Subtract this from \(2a+7d=29\):

\(3d=9,\qquad d=3.\)

Then

\(a+2(3)=10,\qquad a=4.\)

For the sum of \(19\) terms starting with the twelfth term, use the terms \(u_{12}\) to \(u_{30}\).

Now

\(u_{12}=4+11(3)=37,\)

and

\(u_{30}=4+29(3)=91.\)

Therefore the required sum is

\(\frac{19}{2}(37+91)=\frac{19}{2}(128)=1216.\)