If $a_1, a_2, \ldots, a_n$ is a sequence of positive numbers which are in A.P. with common difference $d$ and $a_1 + a_4 + a_7 + \ldots + a_{16} = 147$ then $a_1 + a_{16} = M$ and $a_1 + a_6 + a_{11} + a_{16} = N$.
Maximum value of $a_1 a_2 \ldots a_{16} = \left(\dfrac{S}{W}\right)^{16}$ (where $S$ and $W$ are coprime), then:
Step-by-Step Solution
Key Concept: The core idea is to first use the properties of arithmetic progressions (AP), specifically the sum formula $S_k = \frac{k}{2}(a_1 + a_k)$, to find relationships between terms. Then, apply the AM-GM inequality, recognizing that for positive terms in an AP with a fixed sum (or fixed first and last term), the product is maximized when all terms are equal, implying a common difference of zero.
Step 1: Find $M = a_1 + a_{16}$.
The sum $a_1 + a_4 + a_7 + \ldots + a_{16}$ is an AP with first term $a_1$, common difference $3d$, and 6 terms.
$$a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = \frac{6}{2}(a_1 + a_{16}) = 3(a_1 + a_{16}) = 147$$
$$\Rightarrow a_1 + a_{16} = 49 = M$$
Step 2: Find $N = a_1 + a_6 + a_{11} + a_{16}$.
$$a_1 + a_6 + a_{11} + a_{16} = \frac{4}{2}(a_1 + a_{16}) = 2 \times 49 = 98 = N$$
Step 3: Find the maximum value of $a_1 a_2 \ldots a_{16}$.
By AM-GM inequality, the product $a_1 a_2 \ldots a_{16}$ is maximized when all terms are equal, i.e., $a_1 = a_2 = \ldots = a_{16}$.
Since $a_1 + a_{16} = 49$ and all terms are equal, each term $= \dfrac{49}{2}$.
$$\text{Maximum of } a_1 a_2 \ldots a_{16} = \left(\frac{49}{2}\right)^{16}$$
So $S = 49$ and $W = 2$, which are coprime.
Step 4: Verify all options.
$M = 49$ ✓, $N = 98$ ✓, $S = 49$ ✓, $W = 2$ ✓.
Correct Answer: 1, 2, 3, 4