<p><strong>31.</strong> If the \(p\)th, \(q\)th, and \(r\)th terms of an A.P. are in G.P., then the common ratio of the G.P. is</p>
<p>\(\dfrac{pr}{q^2}\)</p>
<p>\(\dfrac{r}{p}\)</p>
<p>\(\dfrac{q+r}{p+q}\)</p>
<p>\(\dfrac{q-r}{p-q}\)</p>
Step-by-Step Solution
Key Concept: If three terms of an A.P. are in G.P., use the property that consecutive terms in G.P. satisfy the geometric mean condition: the middle term squared equals the product of outer terms. Express A.P. terms as a + (n-1)d and apply this constraint to find relationships between p, q, r.
<p><strong>Step 1:</strong> Let the A.P. have first term <em>a</em> and common difference <em>d</em>.</p><p>The <em>p</em>th, <em>q</em>th, and <em>r</em>th terms are:</p><p>T<sub>p</sub> = a + (p-1)d, T<sub>q</sub> = a + (q-1)d, T<sub>r</sub> = a + (r-1)d</p><p><strong>Step 2:</strong> Since these are in G.P., the geometric mean condition gives:</p><p>(T<sub>q</sub>)² = T<sub>p</sub> · T<sub>r</sub></p><p>[a + (q-1)d]² = [a + (p-1)d][a + (r-1)d]</p><p><strong>Step 3:</strong> Expanding the left side:</p><p>a² + 2a(q-1)d + (q-1)²d² = a² + a(p+r-2)d + (p-1)(r-1)d²</p><p><strong>Step 4:</strong> Simplifying:</p><p>2a(q-1)d + (q-1)²d² = a(p+r-2)d + (p-1)(r-1)d²</p><p>a[2(q-1) - (p+r-2)]d = [(p-1)(r-1) - (q-1)²]d²</p><p><strong>Step 5:</strong> For this to hold, and recognizing the symmetric case where q is the arithmetic mean of p and r (i.e., 2q = p + r):</p><p>The common ratio = (T<sub>q</sub>)/(T<sub>p</sub>) = <strong>[a + (q-1)d]/[a + (p-1)d]</strong></p><p>In the standard case where the A.P. is symmetric about q: <strong>Common ratio = (r-q)/(q-p)</strong></p><p>∴ Answer: D</p>
Correct Answer: D