Answer
$a_6 = \frac{19}{5}$
$a_7 = 5$
$a_8 = \frac{19}{5}$
Work Step by Step
With arithmetic sequences, the middle term of three consecutive terms can be found by taking the arithmetic mean of the two terms that are given.
For this exercise, we can use the third term as the middle term. If we take the arithmetic mean of the fifth and ninth terms, then we will get the seventh term.
The arithmetic mean is given by the formula:
arithmetic mean = $\frac{x + y}{2}$
Substitute the fifth and ninth terms into the formula:
arithmetic mean = $a_7 = \frac{\frac{13}{5} + \frac{37}{5}}{2}$
Add to simplify:
arithmetic mean = $a_7 = \frac{\frac{50}{5}}{2}$
To simplify this fraction, multiply by the reciprocal of the denominator:
$a_7 = \frac{50}{5} \bullet \frac{1}{2}$
Simplify:
$a_7 = \frac{50}{10}$
Simplify the fraction:
$a_7 = 5$
To find the sixth term, $a_6$, apply the arithmetic mean formula again, using the fifth and seventh terms:
$a_6 = \frac{\frac{13}{5} + 5}{2}$
Convert $5$ to an equivalent fraction:
$a_6 = \frac{\frac{13}{5} + \frac{25}{5}}{2}$
Add to simplify:
$a_6 = \frac{\frac{38}{5}}{2}$
To simplify this fraction, multiply by the reciprocal of the denominator:
$a_6 = \frac{38}{5} \bullet \frac{1}{2}$
Simplify:
$a_6 = \frac{38}{10}$
Divide both numerator and denominator by their greatest common factor, $2$:
$a_6 = \frac{19}{5}$
To find the eighth term, $a_8$, apply the arithmetic mean formula again, using the seventh and ninth terms this time:
$a_8 = \frac{5 + \frac{37}{5}}{2}$
Convert $10$ to an equivalent fraction:
$a_8 = \frac{\frac{25}{5} + \frac{37}{5}}{2}$
Add to simplify:
$a_8 = \frac{\frac{62}{5}}{2}$
To simplify this fraction, multiply by the reciprocal of the denominator:
$a_8 = \frac{62}{5} \bullet \frac{1}{2}$
Simplify:
$a_8 = \frac{62}{10}$
Divide both numerator and denominator by their greatest common factor, $2$:
$a_8 = \frac{31}{5}$
