Answer: (a)(ii) \(r=\frac14\); (b) the sum of the first 20 terms of \(A\) is \(600\).

Use the appropriate formula for an arithmetic or geometric sequence, and check the common difference or common ratio before substituting.

The second, fourteenth and seventeenth terms of \(A\) are

These are consecutive terms of the geometric progression \(G\).

(a)(i) Therefore the common ratio can be written as

\(r=\frac{a+13d}{a+d}\)

and also as

\(r=\frac{a+16d}{a+13d}.\)

Equating these gives

\(\frac{a+13d}{a+d}=\frac{a+16d}{a+13d}.\)

Hence

\((a+13d)^2=(a+d)(a+16d).\)

Expanding,

\(a^2+26ad+169d^2=a^2+17ad+16d^2.\)

So

\(9ad+153d^2=0.\)

Factorising,

\(9d(a+17d)=0.\)

Given that \(d\ne0\),

\(a+17d=0.\)

Therefore

\(a=-17d.\)

(a)(ii) Substitute \(a=-17d\) into

\(r=\frac{a+13d}{a+d}.\)

This gives

\(r=\frac{-17d+13d}{-17d+d} =\frac{-4d}{-16d} =\frac14.\)

(b) The first term of \(G\) is the second term of \(A\), so

\(q=a+d.\)

The sum to infinity of \(G\) is

\(\frac{q}{1-r}=\frac{256}{3}.\)

Since \(r=\frac14\),

\(\frac{q}{1-\frac14}=\frac{256}{3}.\)

So

\(\frac{q}{\frac34}=\frac{256}{3},\)

which gives

\(q=64.\)

Hence

\(a+d=64.\)

Using \(a=-17d\),

\(-17d+d=64.\)

So

\(-16d=64,\)

and therefore

\(d=-4.\)

Then

\(a=68.\)

The sum of the first 20 terms of \(A\) is

\(S_{20}=\frac{20}{2}\bigl(2a+19d\bigr).\)

Thus

\(S_{20}=10\bigl(2(68)+19(-4)\bigr).\)

So

\(S_{20}=10(136-76)=600.\)