Answer: (a) \((a,r)=(4,\frac32)\) or \((25,-\frac35)\); convergent sum \(\frac{125}{8}\). (b) \(119382\).

(a) Let the first term be \(a\) and the common ratio be \(r\). Then

\(a+ar=10\)

and

\(ar^2=9.\)

From \(a(1+r)=10\),

\(a=\frac{10}{1+r}.\)

Substitute this into \(ar^2=9\):

\(\frac{10r^2}{1+r}=9.\)

So

\(10r^2=9+9r,\)

and hence

\(10r^2-9r-9=0.\)

Factorising or using the quadratic formula gives

\(r=\frac32\quad\text{or}\quad r=-\frac35.\)

If \(r=\frac32\), then

\(a=\frac{10}{1+\frac32}=4.\)

If \(r=-\frac35\), then

\(a=\frac{10}{1-\frac35}=25.\)

The convergent progression is the one with \(r=-\frac35\), since \(|r|\lt 1\). Its sum to infinity is

\(S_\infty=\frac{a}{1-r}=\frac{25}{1+\frac35}=\frac{125}{8}.\)

(b) For the arithmetic progression,

\(u_4=u_1+3d.\)

So

\(14=-10+3d,\)

giving

\(d=8.\)

Now

\(u_{100}=-10+99(8)=782\)

and

\(u_{200}=-10+199(8)=1582.\)

There are \(200-100+1=101\) terms from \(u_{100}\) to \(u_{200}\).

Therefore

\(u_{100}+u_{101}+\cdots+u_{200}=\frac{101}{2}(782+1582).\)

Hence

\(\boxed{119382}.\)