elynnea4xl

2023-02-18

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

Talon Mcfarland

Beginner2023-02-19Added 9 answers

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, $f\left(x\right)=a1\times {r}^{\left(n-1\right)}$

1. The first of the sequence is $-6.25.$

2. The second term of the sequence is,

$a2=(-6.25)\times {(1.25)}^{(2-1)}a2=(-6.25)\times (1.25)a2=-7.81$

The second term of the sequence is$-7.81.$

3. The third term of the sequence is,

$a3=(-6.25)\times {(1.25)}^{(3-1)}a3=(-6.25)\times {(1.25)}^{2}a3=(-6.25)\times 1.56a3=-9.81$

The third term of the sequence is$-9.81.$

4. The fourth term of the sequence is,

$a4=(-6.25)\times {(1.25)}^{(4-1)}a4=(-6.25)\times {(1.25)}^{3}a4=(-6.25)\times 1.95a4=-12.20$

The fourth term of the sequence is$-12.20.$

5. The fifth term of the sequence is,

$a5=(-6.25)\times {(1.25)}^{(5-1)}a5=(-6.25)\times {(1.25)}^{4}a5=(-6.25)\times 2.44a5=-15.25$

The fifth term of the sequence is $-15.25.$

6. The sixth term of the sequence is,

$a6=(-6.25)\times {(1.25)}^{(6-1)}a6=(-6.25)\times {(1.25)}^{5}a6=(-6.25)\times 3.05a6=-19.07$

The sixth term of the sequence is $-19.07.$

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, $f\left(x\right)=a1\times {r}^{\left(n-1\right)}$

1. The first of the sequence is $-6.25.$

2. The second term of the sequence is,

$a2=(-6.25)\times {(1.25)}^{(2-1)}a2=(-6.25)\times (1.25)a2=-7.81$

The second term of the sequence is$-7.81.$

3. The third term of the sequence is,

$a3=(-6.25)\times {(1.25)}^{(3-1)}a3=(-6.25)\times {(1.25)}^{2}a3=(-6.25)\times 1.56a3=-9.81$

The third term of the sequence is$-9.81.$

4. The fourth term of the sequence is,

$a4=(-6.25)\times {(1.25)}^{(4-1)}a4=(-6.25)\times {(1.25)}^{3}a4=(-6.25)\times 1.95a4=-12.20$

The fourth term of the sequence is$-12.20.$

5. The fifth term of the sequence is,

$a5=(-6.25)\times {(1.25)}^{(5-1)}a5=(-6.25)\times {(1.25)}^{4}a5=(-6.25)\times 2.44a5=-15.25$

The fifth term of the sequence is $-15.25.$

6. The sixth term of the sequence is,

$a6=(-6.25)\times {(1.25)}^{(6-1)}a6=(-6.25)\times {(1.25)}^{5}a6=(-6.25)\times 3.05a6=-19.07$

The sixth term of the sequence is $-19.07.$