In the AP 5, 9, 13, ..., how many first terms have sum 434?
Answer and explanation
Correct answer: 14
The original numerical target 425 is inconsistent with this AP because S_n = n/2[2(5) + (n − 1)4] = n(2n + 3), and no listed integer n gives 425. Correcting the target to 434 makes the question valid. For 434, solve n(2n + 3) = 434, or 2n^2 + 3n − 434 = 0. Factoring gives (n − 14)(2n + 31) = 0, so the positive value is n = 14. Verification: the 14th term is 5 + 13(4) = 57, and S_14 = 14/2(5 + 57) = 7 × 62 = 434. Thus option C is correct.
Frequently asked questions
What is the correct answer to this question?
14
Why is this the correct answer?
The original numerical target 425 is inconsistent with this AP because S_n = n/2[2(5) + (n − 1)4] = n(2n + 3), and no listed integer n gives 425. Correcting the target to 434 makes the question valid. For 434, solve n(2n + 3) = 434, or 2n^2 + 3n − 434 = 0. Factoring gives (n − 14)(2n + 31) = 0, so the positive value is n = 14. Verification: the 14th term is 5 + 13(4) = 57, and S_14 = 14/2(5 + 57) = 7 × 62 = 434. Thus option C is correct.
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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