Answer: (a) \(a=40\), \(d=-26\); (b) first term \(=\mathrm e^{2x}\), common ratio \(=\dfrac{\mathrm e^x}{2}\).

We start with the main method. Identify whether the sequence is arithmetic or geometric, then apply the appropriate formula for the required sum or term.

For the arithmetic progression, the first three terms are

\(a,\quad a+d,\quad a+2d.\)

Their sum is \(42\), so

\(a+(a+d)+(a+2d)=42.\)

This gives

\(3a+3d=42\), so

\(a+d=14.\)

The product of the first three terms is \(-6720\), so

\(a(a+d)(a+2d)=-6720.\)

Using \(a+d=14\),

\(14a(a+2d)=-6720.\)

Hence

\(a(a+2d)=-480.\)

Now \(a+d=14\), so \(a=14-d\) and \(a+2d=14+d\).

Substitute into \(a(a+2d)=-480\):

\((14-d)(14+d)=-480.\)

Thus

\(196-d^2=-480.\)

So

\(d^2=676\), giving \(d=\pm26\).

Since \(a=14-d\) and \(a\) is positive, \(d=-26\).

Therefore

\(a=40,\quad d=-26.\)

For the geometric progression, let the first term be \(A\) and the common ratio be \(r\).

The third term is \(Ar^2\), so

\(Ar^2=\frac{\mathrm e^{4x}}{4}.\)

The tenth term is \(Ar^9\), so

\(Ar^9=\frac{\mathrm e^{11x}}{512}.\)

Divide the second equation by the first:

\(r^7=\frac{\mathrm e^{11x}/512}{\mathrm e^{4x}/4}.\)

Hence

\(r^7=\frac{\mathrm e^{7x}}{128}=\left(\frac{\mathrm e^x}{2}\right)^7.\)

Therefore

\(r=\frac{\mathrm e^x}{2}.\)

Now use \(Ar^2=\frac{\mathrm e^{4x}}{4}\):

\(A\left(\frac{\mathrm e^x}{2}\right)^2=\frac{\mathrm e^{4x}}{4}.\)

So

\(A\frac{\mathrm e^{2x}}{4}=\frac{\mathrm e^{4x}}{4}.\)

Thus

\(A=\mathrm e^{2x}.\)

This completes the solution and gives the required result.