Question

an arithmetic sequence has a common difference of 2, and contains the value 86 at some point in the sequence. which of the following recursive formulas could represent this sequence?
a. $a_n = a_{n-1} + 2, a_1 = 1$
b. $a_n = a_{n-1} + 1, a_1 = 2$
c. $a_n = a_{n-1} + 2, a_1 = 2$
d. $a_n = a_{n-1} + 1, a_1 = 1$

Explanation:

Step1: Eliminate wrong common difference

First, we know the common difference is 2. Options B and D have a common difference of 1, so we can eliminate them.

Step2: General term for valid options

The general term of an arithmetic sequence is $a_n = a_1 + (n-1)d$, where $d=2$.
For Option A: $a_1=1$, so $a_n = 1 + 2(n-1) = 2n - 1$. Set $2n - 1 = 86$, solve for $n$:
$2n = 87$
$n = \frac{87}{2} = 43.5$
$n$ must be a positive integer, so 86 is not in this sequence.

Step3: Check remaining option

For Option C: $a_1=2$, so $a_n = 2 + 2(n-1) = 2n$. Set $2n = 86$, solve for $n$:
$n = \frac{86}{2} = 43$
$n=43$ is a positive integer, so 86 is the 43rd term of this sequence.

Answer:

C. $a_n = a_{n-1} + 2, a_1 = 2$