<p>Let three consecutive terms of a G.P. be \(\dfrac{a}{r}\), \(a\) and \(ar\). Given \(\dfrac{a}{r} \times a \times ar = a^3 = 512\). If 4 is added to the first and second terms, the resulting terms together with the third term form an A.P. Find the sum of the three consecutive terms of G.P.</p>
Step-by-Step Solution
Key Concept: Use the product condition to find a, then apply the A.P. condition on the modified terms to find r, recognizing that the product of three G.P. terms always equals the middle term cubed.
<p><strong>Step 1:</strong> Find the value of a.</p><p>Given: Product of three terms = a³ = 512</p><p>Therefore: a³ = 512 ⟹ a = 8</p><p>The three G.P. terms are: 8/r, 8, 8r</p><p><strong>Step 2:</strong> Apply the A.P. condition.</p><p>After adding 4 to first and second terms, the three terms are:</p><p>(8/r + 4), (8 + 4), 8r</p><p>Which simplifies to: (8/r + 4), 12, 8r</p><p><strong>Step 3:</strong> Use the A.P. property 2(middle) = first + third.</p><p>2(12) = (8/r + 4) + 8r</p><p>24 = 8/r + 4 + 8r</p><p>20 = 8/r + 8r</p><p>Multiply by r: 20r = 8 + 8r²</p><p>8r² - 20r + 8 = 0</p><p>2r² - 5r + 2 = 0</p><p>(2r - 1)(r - 2) = 0</p><p>Therefore: r = 1/2 or r = 2</p><p><strong>Step 4:</strong> Calculate the sum for each case.</p><p>Case 1 (r = 2): Terms are 4, 8, 16 → Sum = 28</p><p>Case 2 (r = 1/2): Terms are 16, 8, 4 → Sum = 28</p><p><strong>∴ Answer: 28</strong></p>
Correct Answer: 28