Sequences & Series
Arithmetic Progression - Word Problems
Grade 11
Question:
<p>A man arranges to pay off a debt of ₹3600 by 40 annual instalments which are in AP. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. The value of the 8th instalment is</p>
<p>(a) ₹35</p>
<p>(b) ₹50</p>
<p>(c) ₹65</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Set up equations using the arithmetic progression sum formula. When 30 instalments are paid, two-thirds of the debt is cleared (one-third remains unpaid). Use this to find the first term and common difference, then calculate the 8th term.
<p><strong>Step 1: Set up the AP structure</strong></p><p>Let the 40 annual instalments be in AP with first term = <em>a</em> and common difference = <em>d</em>.</p><p>Total debt: Sum of 40 terms = ₹3600</p><p>Using S₄₀ = (40/2)[2a + 39d] = 3600</p><p>∴ 20(2a + 39d) = 3600</p><p>∴ 2a + 39d = 180 ... (equation 1)</p><p><strong>Step 2: Use the condition about remaining debt</strong></p><p>When 30 instalments are paid, one-third of debt remains unpaid.</p><p>∴ Two-thirds of debt is paid = (2/3) × 3600 = ₹2400</p><p>Sum of first 30 terms: S₃₀ = (30/2)[2a + 29d] = 2400</p><p>∴ 15(2a + 29d) = 2400</p><p>∴ 2a + 29d = 160 ... (equation 2)</p><p><strong>Step 3: Solve for d</strong></p><p>Subtract equation 2 from equation 1:</p><p>(2a + 39d) - (2a + 29d) = 180 - 160</p><p>10d = 20</p><p>∴ d = 2</p><p><strong>Step 4: Solve for a</strong></p><p>Substituting d = 2 in equation 2:</p><p>2a + 29(2) = 160</p><p>2a + 58 = 160</p><p>2a = 102</p><p>∴ a = 51</p><p><strong>Step 5: Find the 8th instalment</strong></p><p>The nth term of AP: Tₙ = a + (n-1)d</p><p>T₈ = 51 + (8-1) × 2</p><p>T₈ = 51 + 14</p><p>T₈ = 65</p><p><strong>Verification:</strong> S₄₀ = 20(2×51 + 39×2) = 20(102 + 78) = 20(180) = 3600 ✓</p><p>S₃₀ = 15(2×51 + 29×2) = 15(102 + 58) = 15(160) = 2400 ✓</p><p><strong>∴ Answer: c (₹65)</strong></p>
Correct Answer: c