Answer:
First Case:
[tex]\displaystyle p=\frac{5}{2}\text{ and } d=-2[/tex]
Second Case:
[tex]\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}[/tex]
Step-by-step explanation:
We know that the first three terms of an arithmetic series are:
[tex]6p+2, 4p^2-10, \text{ and } 4p+3[/tex]
Since this is an arithmetic sequence, each subsequent term is d more than the previous term, where d is our common difference.
Therefore, we can write the second term as;
[tex]4p^2-10=(6p+2)+d[/tex]
And, likewise, for the third term:
[tex]4p+3=(6p+2)+2d[/tex]
Let's solve for d for each of the equations.
Subtracting in the first equation yields:
[tex]d=4p^2-6p-12[/tex]
And for the second equation:
[tex]2d=-2p+1[/tex]
To avoid fractions, let's multiply the first equation by 2. Hence:
[tex]2d=8p^2-12p-24[/tex]
Therefore:
[tex]8p^2-12p-24=-2p+1[/tex]
Simplifying yields:
[tex]8p^2-10p-25=0[/tex]
Solve for p. We can factor:
[tex]8p^2+10p-20p-25=0[/tex]
Factor:
[tex]2p(4p+5)-5(4p+5)=0[/tex]
Grouping:
[tex](2p-5)(4p+5)=0[/tex]
Zero Product Property:
[tex]\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}[/tex]
Then, we can use the second equation to solve for d. So:
[tex]2d_1=-2p_1+1[/tex]
Substituting:
[tex]\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}[/tex]
So, for the first case, p is 5/2 and d is -2.
Likewise, for the second case:
[tex]\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}[/tex]
So, for the second case, p is -5/4, and d is 7/4.
By using the values, we can determine our series.
For Case 1, we will have:
17, 15, 13.
For Case 2, we will have:
-11/2, -15/4, -2.
