If p is a constant and an1 + an = pn (n 1) for all positive integer

We know \(a_n = pn (n − 1) - a_{n−1} \), so

\(a_{31} = p(31)(30) - a_{30}\)

and

\(a_{30} = p(30)(29) - a_{29}\).

If we now substitute this expression for a_30 into the expression for a_31, we find

\(\\
\begin{align*}\\
a_{31} &= p(31)(30) - (p(30)(29) - a_{29}) \\\\
a_{31} - a_{29} &= 30p(31 - 29) \\\\
a_{31} - a_{29} &= 60p \\
\end{align*}\\
\)

so Statement 1 tells us 60p = 120, and p = 2.

Statement 2 won't be sufficient, because if you just choose any value for p, the equation in the stem will let you work out the other terms in the sequence, and p can be anything. If it wasn't clear that was true, we can also come up with two example values for p: p might be 0, in which case the equation \(a_{n−1} + a_n = pn (n − 1)\) just tells us that the sum of any two consecutive terms of the sequence is zero, and if a_2 = 6, then the sequence is just 6, -6, 6, -6, 6, -6, etc (the sequence must start at a_0 if the rule works for all positive integers n including n =1, so the first term would be 6). We also know, because DS statements are consistent, that p = 2 must work.

So the answer is A.