What is the sum of the first 31 terms of the AP 29, 43, 57, …?
Answer and explanation
Correct answer: 7409
Option A is correct. The sequence is an arithmetic progression because the difference between consecutive terms is constant: d = 43 − 29 = 14 and 57 − 43 = 14. Here the first term is a = 29 and the number of terms is n = 31. Use S_n = n/2 [2a + (n − 1)d]. Substitution gives S_31 = 31/2 [2(29) + (31 − 1)(14)] = 31/2 [58 + 420] = 31/2 × 478 = 31 × 239 = 7409. As a check, the 31st term is 29 + 30 × 14 = 449, and the average of the first and last terms is (29 + 449)/2 = 239; multiplying this average by 31 again gives 7409. Therefore A is unambiguous. The other values result from arithmetic or formula errors.
Frequently asked questions
What is the correct answer to this question?
7409
Why is this the correct answer?
Option A is correct. The sequence is an arithmetic progression because the difference between consecutive terms is constant: d = 43 − 29 = 14 and 57 − 43 = 14. Here the first term is a = 29 and the number of terms is n = 31. Use S_n = n/2 [2a + (n − 1)d]. Substitution gives S_31 = 31/2 [2(29) + (31 − 1)(14)] = 31/2 [58 + 420] = 31/2 × 478 = 31 × 239 = 7409. As a check, the 31st term is 29 + 30 × 14 = 449, and the average of the first and last terms is (29 + 449)/2 = 239; multiplying this average by 31 again gives 7409. Therefore A is unambiguous. The other values result from arithmetic or formula errors.
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.
