To find three consecutive odd integers that have a sum of 63, let's denote the integers using the expression [tex]\( 2n + 1 \)[/tex], where [tex]\( n \)[/tex] is an integer.
1. The first odd integer can be expressed as [tex]\( 2n + 1 \)[/tex].
2. The second consecutive odd integer can be expressed as [tex]\( 2n + 3 \)[/tex] (since it is two more than the first).
3. The third consecutive odd integer can be expressed as [tex]\( 2n + 5 \)[/tex] (since it is two more than the second).
The sum of these three consecutive odd integers is given as 63:
[tex]\[
(2n + 1) + (2n + 3) + (2n + 5) = 63
\][/tex]
Next, we simplify this equation:
[tex]\[
2n + 1 + 2n + 3 + 2n + 5 = 63
\][/tex]
Combining like terms:
[tex]\[
6n + 9 = 63
\][/tex]
To isolate [tex]\( n \)[/tex], we first subtract 9 from both sides of the equation:
[tex]\[
6n = 54
\][/tex]
Then, we divide by 6:
[tex]\[
n = 9
\][/tex]
Now that we have [tex]\( n = 9 \)[/tex], we can find the three consecutive odd integers:
1. First integer: [tex]\( 2n + 1 \)[/tex]
[tex]\[
2(9) + 1 = 19
\][/tex]
2. Second integer: [tex]\( 2n + 3 \)[/tex]
[tex]\[
2(9) + 3 = 21
\][/tex]
3. Third integer: [tex]\( 2n + 5 \)[/tex]
[tex]\[
2(9) + 5 = 23
\][/tex]
Therefore, the three consecutive odd integers are [tex]\( 19, 21, \)[/tex] and [tex]\( 23 \)[/tex].
