Answer: (a) \(p=2\), so \(S_n=n(2n-1)\lg x\); (b) \(n=50\); (c) \(x=10^{-3}\).

Answer: (a) \(p=2\), so \(S_n=n(2n-1)\lg x\); (b) \(n=50\); (c) \(x=10^{-3}\).

The first three terms are

\(\lg x,\quad \lg x^5,\quad \lg x^9.\)

Using \(\lg x^m=m\lg x\), these are

\(\lg x,\quad 5\lg x,\quad 9\lg x.\)

So the common difference is

\(4\lg x.\)

(a) The sum to \(n\) terms is

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

Simplifying,

\(S_n=\frac n2(4n-2)\lg x.\)

Therefore

\(S_n=n(2n-1)\lg x.\)

So

\(p=2.\)

(b) We need

\(n(2n-1)\lg x=4950\lg x.\)

Thus

\(n(2n-1)=4950.\)

So

\(2n^2-n-4950=0.\)

Solving gives

\(n=50.\)

(c) The same sum is \(-14850\). Since from part (b) the sum is \(4950\lg x\),

\(4950\lg x=-14850.\)

Therefore

\(\lg x=-3.\)

Hence

\(x=10^{-3}.\)