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Kryger [21]
3 years ago
15

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

Mathematics
1 answer:
UNO [17]3 years ago
7 0

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}}

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