Consider the following addition problem : 3P + 4P + PP + PP = RQ2; where P, Q and R are different digits. What is the arithmetic mean of all such possible sums ?

• The equation is $3P + 4P + 10P + P + 10P + P = RQ2$, which simplifies to $35P = RQ2$.

• Since the result ends in 2, $5P$ must end in 2, which is impossible for any integer digit $P$ (products of 5 end in 0 or 5).

• Re-evaluating the standard interpretation of such cryptarithmetic problems where 'PP' represents the two-digit number $11P$, the sum is $3P + 4P + 11P + 11P = 29P$.

• For $29P$ to end in 2, $P$ must be 4 (since $29 \times 4 = 116$, ends in 6 - wait, $29 \times 4 = 116$). Let's check $P=4$: $12+16+44+44=116$ (ends in 6). Let's check $P=8$: $24+32+88+88=232$ (ends in 2). So $P=8$.

• If $P=8$, the sum is $232$. Here $R=2, Q=3$. $P, Q, R$ are 8, 3, 2 (all different). This is the only valid solution.

• The question asks for the arithmetic mean of all such possible sums. Since there is only one valid sum (232), the mean is 232. However, looking at the options, let's re-read carefully. Is it possible '3P' means $3 \times P$? Yes. Is it possible the question implies multiple solutions? Let's check other P. $29P$ ends in 2. $9P$ ends in 2. $P=8$ is the only digit. Sum = 232. Option (c) is 202. Let's re-calculate $3P+4P+PP+PP$. If $P=4$, sum=116. If $P=8$, sum=232. Neither is 202. Let's look at the options again. 102, 120, 202, 220. Perhaps I misinterpreted the terms. $3P$ is a two digit number? No, usually single digit. What if $3P$ means the number with tens 3 and units P? i.e., $30+P$? Then $(30+P) + (40+P) + (10P+P) + (10P+P) = 70 + 2P + 22P = 70 + 24P$. Ends in 2. $24P$ ends in 2. $4P$ ends in 2. $P=3$ or $P=8$. If $P=3$, Sum = $70 + 72 = 142$. $R=1, Q=4, P=3$. Distinct. If $P=8$, Sum = $70 + 192 = 262$. $R=2, Q=6, P=8$. Distinct. Sums are 142 and 262. Mean = $(142+262)/2 = 404/2 = 202$. This matches option (c).