Answer: (a)(i) \(-\frac{1}{131072}\). (a)(ii) \(\frac83\). (b)(i) \(7\). (b)(ii) \(1000\).

(a) The sequence is geometric with first term \(4\) and common ratio

\(r=-\frac12.\)

(i) The 20th term is

\(u_{20}=4\left(-\frac12\right)^{19}.\)

Therefore

\(u_{20}=-\frac{4}{2^{19}}=-\frac{1}{131072}.\)

So

\(\boxed{u_{20}=-\frac{1}{131072}}.\)

(ii) The sum to infinity exists because

\(|r|=\frac12\lt 1.\)

The sum to infinity is

\(S_\infty=\frac{a}{1-r}=\frac{4}{1-(-\frac12)}=\frac{4}{\frac32}=\frac83.\)

Thus

\(\boxed{S_\infty=\frac83}.\)

(b) Let the first term be \(a\) and the common difference be \(d\).

The tenth term is \(a+9d\), and the second term is \(a+d\). Given that the tenth term is 15 times the second term,

\(a+9d=15(a+d).\)

Hence

\(7a+3d=0.\)

The sum of the first 6 terms is

\(S_6=\frac62(2a+5d)=87.\)

So

\(2a+5d=29.\)

From \(7a+3d=0\),

\(a=-\frac{3d}{7}.\)

Substitute into \(2a+5d=29\):

\(2\left(-\frac{3d}{7}\right)+5d=29.\)

Thus

\(\frac{29d}{7}=29,\)

so

\(\boxed{d=7}.\)

Then \(a=-3\), so the \(n\)th term is

\(u_n=a+(n-1)d=-3+7(n-1)=7n-10.\)

Given that \(u_n=6990\),

\(7n-10=6990.\)

Therefore

\(7n=7000,\)

and hence

\(\boxed{n=1000}.\)