Question

WILL GIVE BRAILIEST IF ANSWERED FAST

Answer 1

Answer:

$a_n = \dfrac{1}{28}(-2)^{n-1}$

$a_7 = \dfrac{16}{7}$

Step-by-step explanation:

We are given a geometric series, where each term has a common ratio between it and the previous consecutive term:

$\dfrac{1}{28},\ \dfrac{-1}{14},\ \dfrac{1}{7},\ ...$

We can solve for the common ratio by dividing any term by the previous consecutive term:

$r = \dfrac{-1}{14} \div\dfrac{1}{28}$

$r = \dfrac{-1}{14} \times\dfrac{28}{1}$

$r = \dfrac{-28}{14}$

$r=-2$

We can identify the first term as:

$\dfrac{1}{28}$

So, the explicit formula for the geometric sequence is:

$a_n = a_1 \cdot r^{n-1}$

$\boxed{a_n = \dfrac{1}{28}(-2)^{n-1}}$

Using this to solve for the 7th term, $a_7$:

$a_7 = \dfrac{1}{28}(-2)^{7-1}$

$a_7 = \dfrac{1}{28}(-2)^{6}$

$a_7 = \dfrac{1}{28}(64)$

$a_7 = \dfrac{64}{28}$

$\boxed{a_7 = \dfrac{16}{7}}$