Answer: (a) first term \(\displaystyle \frac52\), common ratio \(\displaystyle \frac12\); (b) \(13\).

Answer: (a) first term \(\displaystyle \frac52\), common ratio \(\displaystyle \frac12\); (b) \(13\).

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

The fifteenth term is

\(ar^{14}.\)

The twelfth term is

\(ar^{11}.\)

It is given that

\(ar^{14}=\frac18ar^{11}.\)

Assuming \(a\neq0\), this gives

\(r^3=\frac18.\)

Therefore

\(r=\frac12.\)

The sum to infinity is

\(\frac{a}{1-r}=5.\)

Substitute \(r=\frac12\):

\(\frac{a}{1-\frac12}=5.\)

So

\(a=\frac52.\)

(b) The sum of the first \(n\) terms is

\(S_n=\frac{a(1-r^n)}{1-r}.\)

Using \(a=\frac52\) and \(r=\frac12\),

\(S_n=5\left(1-\left(\frac12\right)^n\right).\)

We need

\(5\left(1-\left(\frac12\right)^n\right)\gt 4.999.\)

Divide by \(5\):

\(1-\left(\frac12\right)^n\gt 0.9998.\)

So

\(\left(\frac12\right)^n\lt 0.0002.\)

Checking powers of \(\frac12\),

\(\left(\frac12\right)^{12}=\frac1{4096}\gt 0.0002\)

but

\(\left(\frac12\right)^{13}=\frac1{8192}\lt 0.0002.\)

Therefore the least number of terms is

\(13.\)