In an AP, a = 3 and the sum from the 11th term to the 20th term is 465. Find d.
Answer and explanation
Correct answer: 3
The governing concept is finding a range sum through partial sums, together with the AP sum formula. Since the first term is a = 3, S₂₀ = 20/2 [2(3) + 19d] = 10(6 + 19d), and S₁₀ = 10/2 [2(3) + 9d] = 5(6 + 9d). The sum from the 11th through the 20th term is S₂₀ - S₁₀, so 10(6 + 19d) - 5(6 + 9d) = 465. This simplifies to 60 + 190d - 30 - 45d = 465, hence 30 + 145d = 465, 145d = 435 and d = 3. Therefore option C is correct. The other choices result from using the wrong number of terms or subtracting the wrong partial sum.
Frequently asked questions
What is the correct answer to this question?
3
Why is this the correct answer?
The governing concept is finding a range sum through partial sums, together with the AP sum formula. Since the first term is a = 3, S₂₀ = 20/2 [2(3) + 19d] = 10(6 + 19d), and S₁₀ = 10/2 [2(3) + 9d] = 5(6 + 9d). The sum from the 11th through the 20th term is S₂₀ - S₁₀, so 10(6 + 19d) - 5(6 + 9d) = 465. This simplifies to 60 + 190d - 30 - 45d = 465, hence 30 + 145d = 465, 145d = 435 and d = 3. Therefore option C is correct. The other choices result from using the wrong number of terms or subtracting the wrong partial sum.
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.
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