In an AP, a = 7, and the sum from the 21st term to the 40th term is 3680. Find d.
Answer and explanation
Correct answer: 6
The sum from the 21st to the 40th term contains 20 terms. The first term of this block is a + 20d = 7 + 20d, and the last is a + 39d = 7 + 39d. Using the AP sum formula for these 20 terms, 3680 = 20/2[(7 + 20d) + (7 + 39d)] = 10(14 + 59d). Therefore 368 = 14 + 59d, so 59d = 354 and d = 6. Equivalently, S₄₀ − S₂₀ = 3680 gives the same equation. Hence option A is correct. The common mistake is to use 20d as the last-term multiplier instead of 39d, forgetting that the original sequence starts at the first term.
Frequently asked questions
What is the correct answer to this question?
6
Why is this the correct answer?
The sum from the 21st to the 40th term contains 20 terms. The first term of this block is a + 20d = 7 + 20d, and the last is a + 39d = 7 + 39d. Using the AP sum formula for these 20 terms, 3680 = 20/2[(7 + 20d) + (7 + 39d)] = 10(14 + 59d). Therefore 368 = 14 + 59d, so 59d = 354 and d = 6. Equivalently, S₄₀ − S₂₀ = 3680 gives the same equation. Hence option A is correct. The common mistake is to use 20d as the last-term multiplier instead of 39d, forgetting that the original sequence starts at the first term.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
