$\frac{4}{3} p+2, \frac{p}{2}-1$, and $\frac{4}{3} p-2$ are consecutive terms of an arithmetic progression.

Find the value of $p$.
\[
p=
\]

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Answer

Final Answer: $\boxed{-1.2}$

Steps

Step 1 ：The difference between the second and first term is $\frac{p}{2}-1 - \frac{4}{3}p - 2$. The difference between the third and second term is $\frac{4}{3}p - 2 - \frac{p}{2} + 1$.

Step 2 ：Setting these two expressions equal to each other gives us an equation in terms of p: $-0.833333333333333*p - 3 = 0.833333333333333*p - 1$

Step 3 ：Solving this equation gives us the value of p: $p = -1.2$

Step 4 ：Final Answer: $\boxed{-1.2}$

$\frac{4}{3} p+2, \frac{p}{2}-1$, and $\frac{4}{3} p-2$ are consecutive terms of an arithmetic progression. Find the value of $p$.\[p=\] Show Calculator Stuck? Use a hint.Report a problem

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$\frac{4}{3} p+2, \frac{p}{2}-1$, and $\frac{4}{3} p-2$ are consecutive terms of an arithmetic progression.

Find the value of $p$.
\[
p=
\]

Show Calculator

Stuck? Use a hint.
Report a problem