elynnea4xl

2023-02-18

Find the six terms of a finite sequence where ${a}_{1}=-6.25$ and $r=1.25$

Talon Mcfarland

The first of the sequence is $-6.25.$
The second term of the sequence is $-7.81.$
The third term of the sequence is $-9.81.$
The fourth term of the sequence is $-12.20.$
The fifth term of the sequence is $-15.25.$
The sixth term of the sequence is $-19.07.$
We have to determine, the six terms of a finite sequence and graph these terms.
According to the question, The first term of the sequence $-6.25.$ and the common ratio is $=1.25$
The six terms of finite sequence would be as per the following equation is determined by the formula, $f\left(x\right)=a1×{r}^{\left(n-1\right)}$
1. The first of the sequence is $-6.25.$
2. The second term of the sequence is,
$a2=\left(-6.25\right)×{\left(1.25\right)}^{\left(2-1\right)}a2=\left(-6.25\right)×\left(1.25\right)a2=-7.81$
The second term of the sequence is$-7.81.$
3. The third term of the sequence is,
$a3=\left(-6.25\right)×{\left(1.25\right)}^{\left(3-1\right)}a3=\left(-6.25\right)×{\left(1.25\right)}^{2}a3=\left(-6.25\right)×1.56a3=-9.81$
The third term of the sequence is$-9.81.$
4. The fourth term of the sequence is,
$a4=\left(-6.25\right)×{\left(1.25\right)}^{\left(4-1\right)}a4=\left(-6.25\right)×{\left(1.25\right)}^{3}a4=\left(-6.25\right)×1.95a4=-12.20$
The fourth term of the sequence is$-12.20.$
5. The fifth term of the sequence is,
$a5=\left(-6.25\right)×{\left(1.25\right)}^{\left(5-1\right)}a5=\left(-6.25\right)×{\left(1.25\right)}^{4}a5=\left(-6.25\right)×2.44a5=-15.25$
The fifth term of the sequence is $-15.25.$
6. The sixth term of the sequence is,
$a6=\left(-6.25\right)×{\left(1.25\right)}^{\left(6-1\right)}a6=\left(-6.25\right)×{\left(1.25\right)}^{5}a6=\left(-6.25\right)×3.05a6=-19.07$
The sixth term of the sequence is $-19.07.$

Do you have a similar question?