If the sum of the first n terms of an arithmetic progression is S_n = 8n² − 3n, find the sum of the 51st term to the 70th term.
Answer and explanation
Correct answer: 19,140
The governing concept is that the sum of consecutive terms from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 51st through the 70th term, so calculate S_70 − S_50. We have S_70 = 8(70)² − 3(70) = 8(4900) − 210 = 39,200 − 210 = 38,990. Similarly, S_50 = 8(50)² − 3(50) = 20,000 − 150 = 19,850. Therefore, the required sum is 38,990 − 19,850 = 19,140. Hence option D is correct. Option A, B, and C result from incorrect substitution or subtraction; using S_51 would also include the wrong range boundary.
Frequently asked questions
What is the correct answer to this question?
19,140
Why is this the correct answer?
The governing concept is that the sum of consecutive terms from the rth term through the sth term is S_s − S_{r−1}. Here the required range is from the 51st through the 70th term, so calculate S_70 − S_50. We have S_70 = 8(70)² − 3(70) = 8(4900) − 210 = 39,200 − 210 = 38,990. Similarly, S_50 = 8(50)² − 3(50) = 20,000 − 150 = 19,850. Therefore, the required sum is 38,990 − 19,850 = 19,140. Hence option D is correct. Option A, B, and C result from incorrect substitution or subtraction; using S_51 would also include the wrong range boundary.
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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