Answer: AP: first term \(5\), common difference \(3\). GP: sum of first 10 terms \(=-1023\).

Let the first term of the arithmetic progression be \(a\), and let its common difference be \(d\).

The first four terms are

\(a,\quad a+d,\quad a+2d,\quad a+3d.\)

Their sum is

\(4a+6d=38.\)

Divide by \(2\):

\(2a+3d=19.\)

The next four terms are

\(a+4d,\quad a+5d,\quad a+6d,\quad a+7d.\)

Their sum is

\(4a+22d=86.\)

Divide by \(2\):

\(2a+11d=43.\)

Subtract the first equation from the second:

\(8d=24,\)

so \(d=3\). Then

\(2a+3(3)=19,\)

so \(2a=10\), and \(a=5\).

For the geometric progression, let the first term be \(A\) and the common ratio be \(r\).

The third term is

\(Ar^2=12,\)

and the sixth term is

\(Ar^5=-96.\)

Divide the sixth term by the third term:

\(\frac{Ar^5}{Ar^2}=r^3=\frac{-96}{12}=-8.\)

Thus

\(r=-2.\)

Using \(Ar^2=12\),

\(A(4)=12,\)

so \(A=3\).

The sum of the first \(10\) terms is

\(S_{10}=\frac{A(1-r^{10})}{1-r}.\)

Therefore

\(S_{10}=\frac{3(1-(-2)^{10})}{1-(-2)} =\frac{3(1-1024)}{3} =-1023.\)