In this specific case, the initial term (a) is 5 and the common ratio (r) is -2
Henceforth, after determining what a and r are, we use the formula for the nth term. Which is:
[tex]a_n = a \times r^{n-1} [/tex]
Therefore, the 14th term is :
[tex]\implies 5 \times { (- 2)}^{14 - 1} \\\implies 5 \times {( - 2)}^{13} \\ =\boxed { -40,960} [/tex]
Hope it helps!
