Question

Find the 7th term of the geometric sequence with a3= -45 and a5= -405

Answer 1

Answer:

Step-by-step explanation:

let a be the first term and r be common ratio.

$a_{3}=ar^2=-45\\a_{5}=ar^4=-405\\divide\\\frac{ar^4}{ar^2} =\frac{-405}{-45} =9\\r^2=9\\r=\pm3\\9a=-45\\a=-5\\a_{7}=ar^{7-1}=ar^6=ar^4 \times r^2=-405 \times 9=-3645$