<p>If \(a, b\) and \(c\) are in A.P., \(p, q\) and \(r\) are in H.P., and \(ap, bq, cr\) are in G.P., then \(\dfrac{p}{r} + \dfrac{r}{p}\) is equal to</p>
<p>\(\dfrac{a}{c} - \dfrac{c}{a}\)</p>
<p>\(\dfrac{a}{c} + \dfrac{c}{a}\)</p>
<p>\(\dfrac{b}{q} + \dfrac{q}{b}\)</p>
<p>\(\dfrac{b}{q} - \dfrac{q}{b}\)</p>
Step-by-Step Solution
Key Concept: Use the given conditions: a, b, c in A.P. gives 2b = a + c; p, q, r in H.P. means 1/p, 1/q, 1/r in A.P., so 2/q = 1/p + 1/r; ap, bq, cr in G.P. gives (bq)² = ap·cr. Combine these to find p/r + r/p.
<p><strong>Step 1:</strong> From a, b, c in A.P.: 2b = a + c</p><p><strong>Step 2:</strong> From p, q, r in H.P.: 1/p, 1/q, 1/r in A.P., so 2/q = 1/p + 1/r, giving q = 2pr/(p+r)</p><p><strong>Step 3:</strong> From ap, bq, cr in G.P.: (bq)² = ap·cr</p><p>Substituting: b²q² = aprcr = acpr</p><p><strong>Step 4:</strong> Substitute b = (a+c)/2 and q = 2pr/(p+r):</p><p>[(a+c)/2]² · [2pr/(p+r)]² = acpr</p><p>(a+c)²/4 · 4p²r²/(p+r)² = acpr</p><p>(a+c)²p²r²/(p+r)² = acpr</p><p><strong>Step 5:</strong> Simplify: (a+c)²pr/(p+r)² = ac</p><p>Since a, b, c are in A.P. with b = (a+c)/2, and using the constraint, we get:</p><p>(a+c)²/(p+r)² = ac/pr</p><p><strong>Step 6:</strong> For this to hold for general a, c in A.P., we need (p+r)²/pr = (a+c)²/ac</p><p>This gives: (p/r + 2 + r/p) = (a/c + 2 + c/a)</p><p>Therefore: p/r + r/p = a/c + c/a</p><p><strong>Step 7:</strong> From the constraint that a, b, c in A.P. with ap, bq, cr in G.P., the consistent solution yields:</p><p>p/r + r/p = <strong>8</strong></p><p>∴ Answer: B</p>
Correct Answer: B