The arithmetic progression is given by the formula аn = 2n -14 whether the number 412

| May 4, 2021 | education

The arithmetic progression is given by the formula аn = 2n -14 whether the number 412 is a member of this progressions.

First, let’s check whether the sequence {аn}, n = 1, 2, …, given by the formula аn = 2 * n – 14, is an arithmetic progression. As you know, the characteristic property of an arithmetic progression requires the equality аn + аn + 2 = 2 * аn + 1 for any n = 1, 2,…. Suppose that n is any natural number, then we have: 2 * n – 14 + 2 * (n + 2) – 14 = 2 * n + 2 * n + 2 – 28 = 2 * (2 * (n + 1) – 14) = 2 * аn + 1. Hence, the sequence {аn}, n = 1, 2, …, given by the formula аn = 2 * n – 14, is really an arithmetic progression.
Let’s say that the number 412 is a member of this arithmetic progression. Then, for some natural number n, the following equality must hold: 2 * n – 14 = 412 or 2 * n = 412 + 14 = 426, whence n = 426: 2 = 213. Since the number 213 is a natural number, the number 412 is indeed ( 213rd) member of the given arithmetic progression.
Answer: Yes, the number 412 is the (213rd) member of this arithmetic progression.



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