Question

Find a12 of the sequence 1/4, 7/12, 11/12, 5/4

Answer 1

1/4 = 3/12, and 5/4 = 15/12, so it looks like there's a common difference between terms of 4/12 = 1/3. The the $n$-th term in the sequence is given recursively by

$\begin{cases}a_1=\frac14\\a_n=a_{n-1}+\frac13&\text{for }n>1\end{cases}$

By substitution, we get

$a_n=a_{n-1}+\dfrac13\implies a_n=\left(a_{n-2}+\dfrac13\right)+\dfrac13$

$a_n=a_{n-2}+\dfrac23$

and doing this again and again until we stop with an expression containing $a_1$, we find that

$a_n=a_1+\dfrac{n-1}3$

$a_n=\dfrac{4n-1}{12}$

Then the 12th term in the sequence is

$a_{12}=\dfrac{4\cdot12-1}{12}=\boxed{\dfrac{47}{12}}$