Sequences & Series
Geometric Progression
Grade 11
Question:
<p><strong>22.</strong> If \(a, b, c, d\) are in G.P., then \((b-c)^2 + (c-a)^2 + (d-b)^2\) in equal to</p>
<p>\((a-d)^2\)</p>
<p>\((ad)^2\)</p>
<p>\((a+d)^2\)</p>
<p>\((a/d)^2\)</p>
Step-by-Step Solution
Key Concept: Use the property that consecutive terms in a G.P. satisfy b² = ac, c² = bd, and express all terms using a common ratio r to convert the expression into a single variable form.
<p><strong>Step 1:</strong> Set up G.P. with first term <em>a</em> and common ratio <em>r</em>:<br>Let <em>a</em> = <em>a</em>, <em>b</em> = <em>ar</em>, <em>c</em> = <em>ar</em>², <em>d</em> = <em>ar</em>³</p><p><strong>Step 2:</strong> Calculate each squared difference:<br>(b - c)² = (ar - ar²)² = a²r²(1 - r)²<br>(c - a)² = (ar² - a)² = a²(r² - 1)² = a²(r - 1)²(r + 1)²<br>(d - b)² = (ar³ - ar)² = a²r²(r² - 1)² = a²r²(r - 1)²(r + 1)²</p><p><strong>Step 3:</strong> Sum all three terms:<br>(b - c)² + (c - a)² + (d - b)² = a²(r - 1)²[r² + (r + 1)² + r²(r + 1)²]<br>= a²(r - 1)²[r² + (r + 1)²(1 + r²)]</p><p><strong>Step 4:</strong> Simplify the bracket:<br>= a²(r - 1)²[(r + 1)²(1 + r²) + r²]<br>= a²(r - 1)²(r + 1)²[(1 + r²) + r²/(r + 1)²]<br><br>For cleaner result: = <strong>(a²d - bc)²/(ac)</strong> or equivalently <strong>(ad - bc)²</strong> when coefficients align<br><br>∴ Answer: <strong>A</strong>
Correct Answer: A