Question:

<div class="solution"> <p><strong>Step 1:</strong> To find the common ratio of the given Geometric Progression (G.P.), we first need to understand the properties of a G.P. In a G.P., each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.</p> <p><strong>Step 2:</strong> Given that the first term of the G.P. is 5 and one of the terms is \(5 \cdot 2^8 \cdot 3^{16}\), we can use the formula for the nth term of a G.P.: \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. Since \(5 \cdot 2^8 \cdot 3^{16}\) is a term in the sequence, we can set \(a_n = 5 \cdot 2^8 \cdot 3^{16}\) and \(a_1 = 5\), and solve for \(r\). We notice that \(5 \cdot 2^8 \cdot 3^{16} = 5 \cdot (2^4 \cdot 3^8)^2 = 5 \cdot (2^4 \cdot 3^8)^2 = 5 \cdot (16 \cdot 3^8)^2 = 5 \cdot (16 \cdot 6561)^2 = 5 \cdot (104976)^2\), but to simplify the calculation, let's consider the relationship of the given term with the first term directly in terms of powers of primes.</p> <p><strong>Step 3:</strong> The given term is \(5 \cdot 2^8 \cdot 3^{16}\). For this to be a term in the G.P. with the first term being 5, the common ratio must include the factors \(2^8\) and \(3^{16}\) in such a way that when multiplied by 5 (the first term) and raised to some power (the term number minus one), it results in \(5 \cdot 2^8 \cdot 3^{16}\). This implies that the common ratio, \(r\), could be of the form \(2^a \cdot 3^b\), where \(a\) and \(b\) are such that \(r^{n-1} = 2^8 \cdot 3^{16}\) for some \(n\). Given that all terms are natural numbers, \(r\) itself must be a natural number, suggesting \(r = 2^a \cdot 3^b\) where \(a\) and \(b\) are positive integers.</p> <p><strong>Step 4:</strong> To find a possible common ratio, let's consider how \(5 \cdot 2^8 \cdot 3^{16}\) can be expressed as \(5 \cdot r^{n-1}\). If \(r = 2^a \cdot 3^b\), then \(r^{n-1} = (2^a \cdot 3^b)^{n-1} = 2^{a(n-1)} \cdot 3^{b(n-1)}\). We want this to equal \(2^8 \cdot 3^{16}\), so \(a(n-1) = 8\) and \(b(n-1) = 16\). One simple solution is \(a = 2\), \(b = 4\), and \(n-1 = 4\), which gives \(r = 2^2 \cdot 3^4 = 4 \cdot 81 = 324\). Thus, a possible common ratio is \(r = 324\), but we need to check if this aligns with any of the given options, considering the form of the options provided.</p> <p><strong>Answer:</strong> The possible common ratio of the G.P. that matches one of the given options, considering our derivation and the specific form of the options provided, is \((324)^2\), which corresponds to option (d).</p> <div class="key-concept"><strong>Key Concept:</strong> Understanding the properties of a Geometric Progression, specifically how each term is generated by multiplying the previous term by a common ratio, and applying this understanding to derive the common ratio from given terms.</div> </div>