Question

The terms of a particular sequence are determined according to the following rule: If the value of a given term $t$ is an odd positive integer, then the value of the following term is $3t -9$; if the value of a given term $t$ is an even positive integer, then the value of the following term is $2t -7$. Suppose that the terms of the sequence alternate between two positive integers $(a, b, a, b, \dots )$. What is the sum of the two positive integers

Answer 1

More plainly, the sequence is defined recursively by

`a_(n+1) = \begin{cases} 3a_n - 9 & \text{if } a_n \text{ is odd} \\ 2a_n - 7 & \text{if } a_n \text{ is even} \end{cases}`

and some starting value
`a_1`.

We're given that the sequence alternates between two constants,
`a` and
`b`, so that
`a_1 = a`.

• If
`a` is even, then the second term
`b` must be odd, since

`a_2 = 2a_1 - 7`

by the given rule, and 2×(even) - (odd) = (odd). So

`a_2 = 2a-7 = b`

In turn, the third term is even, since we jump back to
`a`. From the given rule,

`a_3 = 3a_2 - 9`

and so

`3b-9 = 3(2a-7)-9 = a \implies 6a-30=a \implies 5a=30 \implies a=6`

`3b-9 = 6 \implies 3b = 15 \implies b = 5`

Then the sum of the two integers is
`a+b=\boxed{11}`

• You end up with the same answer in the case of odd
`a`, so I'll omit this part of the solution. (It's almost identical as the even case.)