Answer: (a) first term \(2\), common ratio \(1.5\); (b) \(99253\).

Answer: (a) first term \(2\), common ratio \(1.5\); (b) \(99253\).

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

\(ar^2=4.5.\)

The sixth term is

\(ar^5=15.1875.\)

Divide the sixth-term equation by the third-term equation:

\(r^3=\frac{15.1875}{4.5}.\)

Since

\(\frac{15.1875}{4.5}=3.375=\frac{27}{8},\)

we get

\(r=\frac32=1.5.\)

Now use \(ar^2=4.5\):

\(a\left(\frac32\right)^2=4.5.\)

So

\(a\cdot\frac94=4.5, \qquad a=2.\)

(b) The sixteenth term is

\(u_{16}=2(1.5)^{15}.\)

The sum of ten terms starting with the sixteenth term is

\(2(1.5)^{15}\frac{1-(1.5)^{10}}{1-1.5}.\)

Evaluating this gives

\(99253.097\ldots\)

Therefore the required sum to the nearest integer is

\(99253.\)