To find the first three terms of the sequence defined by the recursive formula [tex]\(a_n = n^2 - a_{n-1}\)[/tex] with the initial condition [tex]\(a_5 = 14 \frac{1}{4}\)[/tex], we need to work backward. The aim is to find [tex]\(a_4\)[/tex], [tex]\(a_3\)[/tex], [tex]\(a_2\)[/tex], and [tex]\(a_1\)[/tex] sequentially.
### Step-by-step solution:
1. Initial Condition:
We know that [tex]\(a_5 = 14 \frac{1}{4}\)[/tex]. Converting [tex]\(14 \frac{1}{4}\)[/tex] to an improper fraction, we get:
[tex]\[
a_5 = 14 \frac{1}{4} = \frac{57}{4}
\][/tex]
2. Finding [tex]\(a_4\)[/tex]:
According to the recursive formula [tex]\(a_n = n^2 - a_{n-1}\)[/tex], substituting [tex]\(a_n\)[/tex] with [tex]\(a_5\)[/tex] and [tex]\(n = 5\)[/tex], we get:
[tex]\[
a_5 = 25 - a_4 \Rightarrow \frac{57}{4} = 25 - a_4
\][/tex]
Solving this equation for [tex]\(a_4\)[/tex]:
[tex]\[
a_4 = 25 - \frac{57}{4} = 25 - 14.25 = 10.75
\][/tex]
3. Finding [tex]\(a_3\)[/tex]:
Using the same formula for [tex]\(n = 4\)[/tex], we substitute [tex]\(a_4\)[/tex]:
[tex]\[
a_4 = 16 - a_3 \Rightarrow 10.75 = 16 - a_3
\][/tex]
Solving this equation for [tex]\(a_3\)[/tex]:
[tex]\[
a_3 = 16 - 10.75 = 5.25
\][/tex]
4. Finding [tex]\(a_2\)[/tex]:
Next, for [tex]\(n = 3\)[/tex], we substitute [tex]\(a_3\)[/tex]:
[tex]\[
a_3 = 9 - a_2 \Rightarrow 5.25 = 9 - a_2
\][/tex]
Solving this equation for [tex]\(a_2\)[/tex]:
[tex]\[
a_2 = 9 - 5.25 = 3.75
\][/tex]
5. Finding [tex]\(a_1\)[/tex]:
Finally, for [tex]\(n = 2\)[/tex], we substitute [tex]\(a_2\)[/tex]:
[tex]\[
a_2 = 4 - a_1 \Rightarrow 3.75 = 4 - a_1
\][/tex]
Solving this equation for [tex]\(a_1\)[/tex]:
[tex]\[
a_1 = 4 - 3.75 = 0.25
\][/tex]
The first three terms of the sequence are therefore:
[tex]\[
a_1 = 0.25, \quad a_2 = 3.75, \quad a_3 = 5.25
\][/tex]
The correct answer from the given choices is:
[tex]\[
\boxed{0.25, 3.75, 5.25}
\][/tex]
