Question

The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate the common difference for each value of p.​

Answer 1

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$

Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

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;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for d for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for p. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for d. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, p is 5/2 and d is -2.

Likewise, for the second case:

$\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}$

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.