Answer: (a)(i) \(r=-\dfrac23\); (a)(ii) \(S_7=6.35\). (b)(i) \(\log_x3\); (b)(ii) \(S_n=\dfrac{n(n+1)}2\log_x3\); (b)(iii) \(n=78\); (b)(iv) \(x=27\).

(a)(i) For a geometric progression with first term \(10\) and common ratio \(r\),

\(S_\infty=\frac{10}{1-r}.\)

Since \(S_\infty=6\),

\(6=\frac{10}{1-r}.\)

Thus

\(6(1-r)=10,\qquad 6-6r=10,\)

so

\(r=-\frac23.\)

(a)(ii) The sum of the first 7 terms is

\(S_7=\frac{10(1-r^7)}{1-r}.\)

Substituting \(r=-\frac23\),

This gives

\(S_7=6.35\)

to 2 decimal places.

(b)(i) The first three terms are

\(\log_x3,\quad \log_x(3^2),\quad \log_x(3^3).\)

Using \(\log_x(3^k)=k\log_x3\), these are

\(\log_x3,\quad 2\log_x3,\quad 3\log_x3.\)

So the common difference is

\(\log_x3.\)

(b)(ii) The sum to \(n\) terms of an arithmetic progression is

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

Here \(a=\log_x3\) and \(d=\log_x3\), so

\(S_n=\frac n2\left(2\log_x3+(n-1)\log_x3\right).\)

Therefore

\(S_n=\frac{n(n+1)}2\log_x3.\)

(b)(iii) Given that

\(S_n=3081\log_x3,\)

we have

\(\frac{n(n+1)}2\log_x3=3081\log_x3.\)

Since \(\log_x3\neq0\),

\(\frac{n(n+1)}2=3081.\)

So

\(n^2+n-6162=0.\)

Solving gives the positive integer value

\(n=78.\)

(b)(iv) The same sum is also \(1027\), so

\(1027=3081\log_x3.\)

Hence

\(\log_x3=\frac13.\)

This means

\(x^{1/3}=3,\)

and therefore

\(x=27.\)