Given the geometric progression below
[tex]5,-25,125,\ldots[/tex]
The nth term of a geometric progression is given below
[tex]T_n=ar^{n-1},\begin{cases}a=\text{first term} \\ r=\text{common ratio}\end{cases}[/tex]
From the geometric progression, we can deduce the following
[tex]\begin{gathered} T_1=a=5 \\ T_2=ar=-25 \\ T_3=ar^2=125 \end{gathered}[/tex]
To find the value of r, we will take ratios of two consecutive terms
[tex]\begin{gathered} \frac{T_2}{T_1}=\frac{ar}{a}=\frac{-25}{5} \\ \Rightarrow r=-5 \end{gathered}[/tex]
To find the 9th term of the geometric, we will have that;
[tex]\begin{gathered} T_9=ar^8=5\times(-5)^8=5\times390625 \\ =1953125 \end{gathered}[/tex]
Hence, the 9th term of the geometric progression is 1953125
