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.)