Answer: (a) \(a=8\), \(d=-3\); (b) \(r=\dfrac12\) or \(r=-\dfrac32\), and the convergent sum to infinity is \(8\).

We start with the main method. Identify whether the sequence is arithmetic or geometric, then apply the appropriate formula for the required sum or term.

For the arithmetic progression, use

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

The sum of the first \(30\) terms is \(-1065\), so

\(\frac{30}{2}(2a+29d)=-1065.\)

Hence

\(2a+29d=-71.\)

The sum of the next \(20\) terms is \(-2210\), so the sum of the first \(50\) terms is

\(-1065-2210=-3275.\)

Therefore

\(\frac{50}{2}(2a+49d)=-3275.\)

So

\(2a+49d=-131.\)

Subtract the first equation from the second:

\(20d=-60\), so \(d=-3\).

Then

\(2a+29(-3)=-71\), so \(2a=16\), and \(a=8\).

For part (b), the first three terms of the geometric progression are

\(4,\quad4r,\quad4r^2.\)

Their sum is \(7\), so

\(4+4r+4r^2=7.\)

This gives

\(4r^2+4r-3=0.\)

Factorise:

\((2r-1)(2r+3)=0.\)

Thus

\(r=\frac12\quad\text{or}\quad r=-\frac32.\)

A geometric progression is convergent only when \(|r|\lt 1\), so the convergent value is \(r=\frac12\).

The sum to infinity is

\(S_\infty=\frac{4}{1-\frac12}=8.\)

This completes the solution and gives the required result.