Lots of people sent in the solution that the sixth term of the Fibonacci sequence starting with $2$ and $38$ is $196$ and you found other sequences with $196$ as one of the terms. Exactly how many other Fibonacci sequences contain the term $196$? A lot of solutions as Jimmy rightly pointed out!

We are only looking for positive whole numbers. The terms increase quickly so $196$ has to be one of the first few terms.

The simplest Fibonacci sequence is:

$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 253, \ldots$

and we denote the $n$th term of this sequence by $F(n)$.

Starting with the terms $a, b$ (for $a$ and $b$ positive whole numbers and $a < b$) we get the general Fib sequence:

$a$, $b$, $a+b$, $a+2b$, $2a+3b$, $3a+5b$, $5a+8b$, $8a+13b$, $13a+21b$, $21a+34b$, $34a+55b$, $55a+89b$, $89a+144b$, $\ldots$

The $n$th term of the general Fib sequence $f(n) = aF(n-2) + bF(n-1)$ and note that, if the term $196$ occurs in the sequence, it can't be beyond the twelfth term as after that the terms are too large.

Here are some sequences containing $196$.

Sequences with $196$ as the first term

$196, b, 196+b, 196+2b, \ldots$

for $b > 196$

Sequences with $196$ as the second term

$a, 196, \ldots$

for $1 < a < 195$

Sequences with 196 as the third term

$1, 195, 196, \ldots$
$2, 194, 196, \ldots$
$...$
$97, 99, 196, \ldots$
$98, 98, 196, \ldots$ etc

So far we see that there are infinitely many sequences with $196$ as the first term; exactly $195$ with $196$ as the second term; exactly $98$ with $196$ as the third term.

To find all the remaining sequences containing $196$ we have to find whole numbers $a$ and $b$ where:

$a + 2b = 196$
or
$2a + 3b = 196$
or
$3a + 5b = 196$
etc.

In general we have to find whole number values of $a$ and $b$ satisfying

$aF(n-2) + bF(n-1) = 196$.

and so we need to find whole number solutions to these equations for $n = 4, 5, 6, \ldots12$.

We shall consider one remaining case and leave the rest to the reader.

For $n = 6$ we seek values of $a$ and $b$ such that $3a + 5b = 196$

There are no solutions for $a = 1$ because then b would not be a whole number. We have already seen that $a = 2$ and $b = 38$ gives $196$ as the sixth term. For larger values of $a$ we have to take smaller values of $b$. For $a = 3$ or $4$ or $5$ or $6$ there are again no solutions because b has to be a whole number.

For $a = 7$ we have:

$21 + 5b$	$=$	$196$
$5b$	$=$	$175$
$b$	$=$	$35$

giving the sequence $7, 35, 42, 77, 119, 196, \ldots$

To find the remaining solutions for $n = 6$ we increase $a$ by steps of $5$ and decrease $b$ by steps of $3$.
There are five solutions for $n = 6$, which are:

• $2, 38, \ldots$
• $7, 35, \ldots$
• $12, 32, \ldots$
• $17, 29,\ldots$
• $22, 26,\ldots$

You can use the same method to find the solutions for $n = 4, 5, 7, \ldots 12$