Question

Find the oth term of the geometric sequence 5,--25, 125,

Answer 1

Given the geometric progression below

$5,-25,125,\ldots$

The nth term of a geometric progression is given below

$T_n=ar^{n-1},\begin{cases}a=\text{first term} \\ r=\text{common ratio}\end{cases}$

From the geometric progression, we can deduce the following

$\begin{gathered} T_1=a=5 \\ T_2=ar=-25 \\ T_3=ar^2=125 \end{gathered}$

To find the value of r, we will take ratios of two consecutive terms

$\begin{gathered} \frac{T_2}{T_1}=\frac{ar}{a}=\frac{-25}{5} \\ \Rightarrow r=-5 \end{gathered}$

To find the 9th term of the geometric, we will have that;

$\begin{gathered} T_9=ar^8=5\times(-5)^8=5\times390625 \\ =1953125 \end{gathered}$

Hence, the 9th term of the geometric progression is 1953125