If aₙ = kn + 17 and a₂₀ − a₇ = 104, what is a₃₁?
Answer and explanation
Correct answer: 265
Substitute the given formula into the difference condition. We have a₂₀ = 20k + 17 and a₇ = 7k + 17. Therefore a₂₀ − a₇ = (20k + 17) − (7k + 17) = 13k. The constant terms cancel, so 13k = 104 and k = 8. Now evaluate the required term: a₃₁ = 31k + 17 = 31(8) + 17 = 248 + 17 = 265. Hence option B is correct. The other options result from using an incorrect value of k, miscalculating 31 × 8, or adding the constant incorrectly. The structure is linear in n, which is consistent with an arithmetic progression.
Frequently asked questions
What is the correct answer to this question?
265
Why is this the correct answer?
Substitute the given formula into the difference condition. We have a₂₀ = 20k + 17 and a₇ = 7k + 17. Therefore a₂₀ − a₇ = (20k + 17) − (7k + 17) = 13k. The constant terms cancel, so 13k = 104 and k = 8. Now evaluate the required term: a₃₁ = 31k + 17 = 31(8) + 17 = 248 + 17 = 265. Hence option B is correct. The other options result from using an incorrect value of k, miscalculating 31 × 8, or adding the constant incorrectly. The structure is linear in n, which is consistent with an arithmetic progression.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
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