Question:

Step 1: Find the sum $a_1 + a_4 + a_7 + \ldots + a_{16}$. The terms are $a_1, a_4, a_7, a_{10}, a_{13}, a_{16}$ — that is 6 terms with common difference $3d$. Their sum $= \dfrac{6}{2}(a_1 + a_{16}) = 3(a_1 + a_{16}) = 147$. So $a_1 + a_{16} = 49$, i.e., $M = 49$. ✓ Step 2: Find $N = a_1 + a_6 + a_{11} + a_{16}$. These 4 terms: $a_1 + a_{16} = 49$ and $a_6 + a_{11} = a_1 + 5d + a_1 + 10d = 2a_1 + 15d = a_1 + a_{16} = 49$. So $N = 49 + 49 = 98$. ✓ Step 3: Find the maximum of $a_1 a_2 \cdots a_{16}$. By AM-GM inequality, $a_1 a_2 \cdots a_{16} \leq \left(\dfrac{a_1 + a_2 + \cdots + a_{16}}{16}\right)^{16}$. The sum $a_1 + a_{16} = 49$, so $a_1 + a_2 + \cdots + a_{16} = \dfrac{16}{2}(a_1 + a_{16}) = 8 \times 49 = 392$. Thus $\dfrac{a_1 + \cdots + a_{16}}{16} = \dfrac{392}{16} = \dfrac{49}{2}$. Step 4: Maximum value $= \left(\dfrac{49}{2}\right)^{16}$. So $S = 49$ and $W = 2$. Since $\gcd(49, 2) = 1$, they are coprime. ✓ Step 5: Check option (b): $N = 98$. ✓ But the answer key shows options (a), (c), (d) as correct, meaning $N = 98$ is listed as option (b) which should also be correct. All four options appear correct based on calculation.