Answer: (a) \(d=6\). (b) \(r=3\).

(a) For an arithmetic progression with first term \(3\) and common difference \(d\),

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

Thus

\(S_{10}=\dfrac{10}{2}(6+9d)=5(6+9d),\)

and

\(S_5=\dfrac52(6+4d).\)

It is given that \(S_{10}=4S_5\), so

\(5(6+9d)=4\cdot\dfrac52(6+4d).\)

Expanding,

\(30+45d=60+40d.\)

Therefore

\(5d=30,\)

so

\(d=6.\)

(b) Let the first term of the arithmetic progression be \(a\), and let its common difference be \(d\). Then its first, second and fifth terms are

\(a,\quad a+d,\quad a+4d.\)

These are the first three terms of a geometric progression. If the common ratio is \(r\), then

\(a+d=ar,\qquad a+4d=ar^2.\)

From the first equation,

\(d=a(r-1).\)

Substitute this into the second equation:

\(a+4a(r-1)=ar^2.\)

Since the first term is non-zero, \(a\ne0\), so divide by \(a\):

\(1+4(r-1)=r^2.\)

Hence

\(r^2-4r+3=0.\)

Factorising,

\((r-1)(r-3)=0.\)

The common ratio is not \(1\), so

\(r=3.\)