Answer: (a) \(64\). (b) \(x=4\) or \(x=39\), giving common ratios \(-\frac12\) and \(\frac23\), so the sum to infinity exists in both cases.

Answer: (a) \(64\). (b) \(x=4\) or \(x=39\), giving common ratios \(-\frac12\) and \(\frac23\), so the sum to infinity exists in both cases.

(a) The sum of the first \(n\) terms of an arithmetic progression is

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

Using \(S_{20}=1100\):

\(\displaystyle \frac{20}{2}(2a+19d)=1100\),

so

\(2a+19d=110\).

Using \(S_{70}=14350\):

\(\displaystyle \frac{70}{2}(2a+69d)=14350\),

so

\(2a+69d=410\).

Subtracting the two equations gives

\(50d=300\),

so \(d=6\).

Then

\(2a+19(6)=110\),

so \(2a=-4\) and \(a=-2\).

The 12th term is

\(a+11d=-2+66=64\).

(b) Since the three terms form a geometric progression, the ratio between consecutive terms is the same:

Cross-multiply:

\((x-9)^2=\frac12(x+1)(x+6)\).

Multiplying by 2 gives

\(2(x-9)^2=(x+1)(x+6)\).

Expand:

\(2(x^2-18x+81)=x^2+7x+6\).

So

\(2x^2-36x+162=x^2+7x+6\),

and hence

\(x^2-43x+156=0\).

Factorise:

\((x-4)(x-39)=0\).

Thus \(x=4\) or \(x=39\).

If \(x=4\), the terms are \(10,-5,\frac52\), so the common ratio is \(-\frac12\).

If \(x=39\), the terms are \(45,30,20\), so the common ratio is \(\frac23\).

In both cases \(|r|\lt 1\), so the sum to infinity exists.