Answer
$a_2 = \pm 324$
$a_3 = 108$
$a_4 = \pm 36$
Work Step by Step
The middle term of three consecutive terms of a geometric sequence can be found using the geometric mean formula.
For this exercise, we can use the third term as the middle term. If we take the geometric mean of the first and fifth terms, then we will get the third term.
The geometric mean is given by the formula:
geometric mean = $\sqrt {a_1 \bullet a_5}$
Substitute $a_1$ and $a_5$ into the equation:
$a_3 = \sqrt {972 \bullet 12}$
Multiply terms within the radicand first:
$a_3 = \sqrt {11,664}$
Take the square root. Remember that the third term must be of the same sign as the first and fifth terms because every other term must have the same sign; therefore, the third term must be positive as well:
$a_3 = 108$
To get the value of $a_2$, take the geometric mean of $a_1$ and $a_3$ into the equation:
$a_2 = \sqrt {972 \bullet 108}$
Multiply terms within the radicand first:
$a_2 = \sqrt {104,976}$
Take the square root:
$a_2 = \pm 324$
To get the value of $a_4$, take the geometric mean of $a_3$ and $a_5$ into the equation:
$a_4 = \sqrt {108 \bullet 12}$
Multiply terms within the radicand first:
$a_4 = \sqrt {1296}$
Take the square root:
$a_4 = \pm 36$
