Answer:
First Case:
![\displaystyle p=\frac{5}{2}\text{ and } d=-2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20p%3D%5Cfrac%7B5%7D%7B2%7D%5Ctext%7B%20and%20%7D%20d%3D-2)
Second Case:
![\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20p%3D-%5Cfrac%7B5%7D%7B4%7D%5Ctext%7B%20and%20%7D%20d%3D%5Cfrac%7B7%7D%7B4%7D)
Step-by-step explanation:
We know that the first three terms of an arithmetic series are:
![6p+2, 4p^2-10, \text{ and } 4p+3](https://tex.z-dn.net/?f=6p%2B2%2C%204p%5E2-10%2C%20%5Ctext%7B%20and%20%7D%204p%2B3)
Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.
Therefore, we can write the second term as;
![4p^2-10=(6p+2)+d](https://tex.z-dn.net/?f=4p%5E2-10%3D%286p%2B2%29%2Bd)
And, likewise, for the third term:
![4p+3=(6p+2)+2d](https://tex.z-dn.net/?f=4p%2B3%3D%286p%2B2%29%2B2d)
Let's solve for <em>d</em> for each of the equations.
Subtracting in the first equation yields:
![d=4p^2-6p-12](https://tex.z-dn.net/?f=d%3D4p%5E2-6p-12)
And for the second equation:
![2d=-2p+1](https://tex.z-dn.net/?f=2d%3D-2p%2B1)
To avoid fractions, let's multiply the first equation by 2. Hence:
![2d=8p^2-12p-24](https://tex.z-dn.net/?f=2d%3D8p%5E2-12p-24)
Therefore:
![8p^2-12p-24=-2p+1](https://tex.z-dn.net/?f=8p%5E2-12p-24%3D-2p%2B1)
Simplifying yields:
![8p^2-10p-25=0](https://tex.z-dn.net/?f=8p%5E2-10p-25%3D0)
Solve for <em>p</em>. We can factor:
![8p^2+10p-20p-25=0](https://tex.z-dn.net/?f=8p%5E2%2B10p-20p-25%3D0)
Factor:
![2p(4p+5)-5(4p+5)=0](https://tex.z-dn.net/?f=2p%284p%2B5%29-5%284p%2B5%29%3D0)
Grouping:
![(2p-5)(4p+5)=0](https://tex.z-dn.net/?f=%282p-5%29%284p%2B5%29%3D0)
Zero Product Property:
![\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20p_1%3D%5Cfrac%7B5%7D%7B2%7D%20%5Ctext%7B%20or%20%7D%20p_2%3D-%5Cfrac%7B5%7D%7B4%7D)
Then, we can use the second equation to solve for <em>d</em>. So:
![2d_1=-2p_1+1](https://tex.z-dn.net/?f=2d_1%3D-2p_1%2B1)
Substituting:
![\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%202d_1%26%3D-2%28%5Cfrac%7B5%7D%7B2%7D%29%2B1%20%5C%5C%202d_1%26%3D-5%2B1%20%5C%5C%202d_1%26%3D-4%20%5C%5C%20d_1%26%3D-2%5Cend%7Baligned%7D)
So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.
Likewise, for the second case:
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%202d_2%26%3D-2%28-%5Cfrac%7B5%7D%7B4%7D%29%2B1%20%5C%5C%202d_2%26%3D%5Cfrac%7B5%7D%7B2%7D%2B1%20%5C%5C%202d_2%26%3D%5Cfrac%7B7%7D%7B2%7D%20%5C%5C%20d_2%26%3D%5Cfrac%7B7%7D%7B4%7D%5Cend%7Baligned%7D)
So, for the second case, <em>p </em>is -5/4, and <em>d</em> 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.