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} (T_2)/(T_1)=(ar)/(a)=(-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*(-5)^8=5*390625 \\ =1953125 \end{gathered}`

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