Answer: \(a=5,\ d=4\); \(35\) terms.

(a) The third term is

\(a+2d=13.\)

The tenth term is

\(a+9d=41.\)

Subtract the first equation from the second:

\(7d=28.\)

So

\(d=4.\)

Substitute into \(a+2d=13\):

\(a+8=13,\)

so

\(a=5.\)

(b) The sum of the first \(n\) terms of an arithmetic progression is

\(S_n=\frac n2\{2a+(n-1)d\}.\)

Substitute \(a=5\), \(d=4\) and \(S_n=2555\):

\(2555=\frac n2\{10+4(n-1)\}.\)

So

\(2555=\frac n2(4n+6).\)

Hence

\(2555=n(2n+3).\)

Therefore

\(2n^2+3n-2555=0.\)

Factorising,

\((n-35)(2n+73)=0.\)

Since the number of terms must be positive,

\(n=35.\)

(c) Using

\(S_n=\frac n2(4n+6)=2n^2+3n,\)

we have

\(S_{2k}=2(2k)^2+3(2k)=8k^2+6k.\)

Also,

\(S_k=2k^2+3k.\)

Therefore

\(S_{2k}-S_k=(8k^2+6k)-(2k^2+3k).\)

Simplifying,

\(S_{2k}-S_k=6k^2+3k.\)

Factorising,

\(S_{2k}-S_k=3k(2k+1)=3k(1+2k).\)