<p>Let three positive numbers \(a, b, c\) (in order) be in HP such that \(a + c = 8\). If \(\langle t_n \rangle\) is a geometric progression with common ratio 3 where \(t_1 = a - \dfrac{b}{2}\), \(t_2 = \dfrac{b}{2}\) and \(t_3 = c - \dfrac{b}{2}\), then find the value of \(t_7\left(\dfrac{2}{3}\right)^6\).</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in HP, their reciprocals are in AP. Combined with the GP property of t₁, t₂, t₃ (where the ratio condition t₂/t₁ = t₃/t₂ holds), we can set up equations to find a, b, c, then determine t₇.
<p><strong>Step 1: Use HP condition</strong></p><p>Since a, b, c are in HP, we have 1/a, 1/b, 1/c in AP.</p><p>Therefore: 1/b = (1/a + 1/c)/2</p><p>This gives: b = 2ac/(a+c)</p><p>Given a + c = 8, we get: <strong>b = 2ac/8 = ac/4</strong></p><p><strong>Step 2: Use GP condition</strong></p><p>Since {tₙ} is a GP with common ratio 3:</p><p>t₂/t₁ = 3 and t₃/t₂ = 3</p><p>From t₂/t₁ = 3:</p><p>(b/2)/(a - b/2) = 3</p><p>b/2 = 3(a - b/2)</p><p>b/2 = 3a - 3b/2</p><p>2b = 3a</p><p><strong>b = 3a/2</strong></p><p><strong>Step 3: Verify with t₃/t₂ = 3</strong></p><p>From t₃/t₂ = 3:</p><p>(c - b/2)/(b/2) = 3</p><p>c - b/2 = 3b/2</p><p>c = 2b</p><p><strong>c = 2b</strong></p><p><strong>Step 4: Solve the system</strong></p><p>From b = 3a/2 and c = 2b:</p><p>c = 2(3a/2) = 3a</p><p>Using a + c = 8:</p><p>a + 3a = 8</p><p>4a = 8</p><p><strong>a = 2</strong></p><p>Therefore: b = 3(2)/2 = 3 and c = 6</p><p>Verify: a + c = 2 + 6 = 8 ✓ and b = ac/4 = 12/4 = 3 ✓</p><p><strong>Step 5: Find t₁</strong></p><p>t₁ = a - b/2 = 2 - 3/2 = 1/2</p><p><strong>Step 6: Find t₇</strong></p><p>Since tₙ = t₁ · 3^(n-1):</p><p>t₇ = (1/2) · 3⁶ = 729/2</p><p><strong>Step 7: Calculate t₇(2/3)⁶</strong></p><p>t₇(2/3)⁶ = (729/2) · (2/3)⁶</p><p>= (729/2) · (2⁶/3⁶)</p><p>= (729/2) · (64/729)</p><p>= (729 · 64)/(2 · 729)</p><p>= 64/2</p><p>= 32</p><p><strong>∴ Answer: 32</strong></p>
Correct Answer: 32