Question 15 of 44intermediate🔧 ApplyShort Answer3 marks

Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Correct Answer

65th term

Exercise: EXERCISE 5.2 | Q: 11 | (Page Page 14)
For More Understanding

Explanation

This question requires finding a specific term in an AP that exceeds another term by a given value. Students need to first identify the AP parameters (first term and common difference), then calculate the 54th term, add 132 to it, and finally determine which term equals that value. The solution demonstrates application of the nth term formula.

Solution Steps

  1. Step 1: Identify the AP parameters. First term a = 3. Common difference d = 15 - 3 = 12.

  2. Step 2: Find the 54th term using the formula an=a+(n−1)da_n = a + (n-1)d. So, a54=3+(54−1)×12=3+53×12=3+636=639a_{54} = 3 + (54-1) \times 12 = 3 + 53 \times 12 = 3 + 636 = 639.

  3. Step 3: Calculate the required term value. The required term = 639+132=771639 + 132 = 771.

  4. Step 4: Find which term equals 771. Using an=a+(n−1)da_n = a + (n-1)d: 771=3+(n−1)×12771 = 3 + (n-1) \times 12. So, (n−1)×12=768(n-1) \times 12 = 768, giving n−1=64n-1 = 64. Therefore, n = 65.

  5. Step 5: Verification: a65=3+64×12=3+768=771a_{65} = 3 + 64 \times 12 = 3 + 768 = 771. And 771−639=132771 - 639 = 132. ✓

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