Answer: (a) \(22\). (b) \(x=\mathrm{e}^4\).

The first three terms are

\(2\ln(x^3)=6\ln x,\)

\(5\ln(x^2)=10\ln x,\)

\(2\ln(x^7)=14\ln x.\)

So the arithmetic progression has first term \(6\ln x\) and common difference \(4\ln x\).

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

\(S_n=\frac n2\left(2(6\ln x)+(n-1)(4\ln x)\right).\)

Thus

\(S_n=\frac n2(4n+8)\ln x=(2n^2+4n)\ln x.\)

Also

\(43\ln(x^{24})=1032\ln x.\)

Since \(x\gt 1\), \(\ln x\gt 0\), so

\(2n^2+4n\gt 1032.\)

That is

\(n^2+2n-516\gt 0.\)

The positive root of \(n^2+2n-516=0\) is

\(-1+\sqrt{517}=21.7\ldots.\)

So the least integer value of \(n\) is

\(22.\)

(b) The 25th term is

\(6\ln x+24(4\ln x)=102\ln x.\)

Given that this equals \(408\),

\(102\ln x=408.\)

So \(\ln x=4\), and hence

\(x=\mathrm{e}^4.\)