Answer: (a)(i) \(\frac18\); (a)(ii) \(\frac{1023}{8}\); (a)(iii) \(128\); (b) \(27\).

Answer: (a)(i) \(\frac18\); (a)(ii) \(\frac{1023}{8}\); (a)(iii) \(128\); (b) \(27\).

(a)(i) The 10th term of the geometric progression is

\(u_{10}=64(0.5)^9.\)

Since \(0.5=\frac12\),

(a)(ii) The sum of the first 10 terms is

\(S_{10}=\frac{64(1-(0.5)^{10})}{1-0.5}.\)

So

(a)(iii) Since \(|r|\lt 1\), the sum to infinity is

\(S_\infty=\frac{64}{1-0.5}=128.\)

(b) For the arithmetic progression,

\(S_n=\frac n2\left(2a+(n-1)d\right).\)

So

\(S_{20}=10(2a+19d),\qquad S_{10}=5(2a+9d).\)

Using \(S_{20}-400=2S_{10}\),

\(10(2a+19d)-400=2\cdot5(2a+9d).\)

Expanding gives

\(20a+190d-400=20a+90d.\)

Hence

\(100d=400,\)

so

\(d=4.\)

Also

\(u_1:u_6=a:(a+5d)=1:5.\)

Therefore

\(5a=a+5d.\)

So

\(4a=5d=20,\)

and hence \(a=5\).

The first three terms are

\(5,\quad 9,\quad 13.\)

Their sum is

\(5+9+13=27.\)