Question
p+1, 2p-10,1-4p² are consecutive terms of an ap find the possible value of p
Answers
GPT-4o mini
To find the value of \(p\) such that \(p + 1\), \(2p - 10\), and \(1 - 4p^2\) are consecutive terms of an arithmetic progression (AP), we need to apply the property of AP that states the middle term is the average of the other two terms. Therefore, we set up the equation:
\[
2p - 10 = \frac{(p + 1) + (1 - 4p^2)}{2}
\]
Now, we'll simplify and solve for \(p\).
1. Start by simplifying the right side:
\[
\frac{(p + 1) + (1 - 4p^2)}{2} = \frac{p + 1 + 1 - 4p^2}{2} = \frac{p - 4p^2 + 2}{2}
\]
2. Therefore, the equation becomes:
\[
2p - 10 = \frac{p - 4p^2 + 2}{2}
\]
3. To eliminate the fraction, multiply through by 2:
\[
4p - 20 = p - 4p^2 + 2
\]
4. Rearranging gives us:
\[
4p - p + 4p^2 - 20 - 2 = 0
\]
\[
4p^2 + 3p - 22 = 0
\]
5. Now, we solve this quadratic equation using the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 4\), \(b = 3\), and \(c = -22\):
\[
p = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-22)}}{2 \cdot 4}
\]
\[
p = \frac{-3 \pm \sqrt{9 + 352}}{8}
\]
\[
p = \frac{-3 \pm \sqrt{361}}{8}
\]
\[
p = \frac{-3 \pm 19}{8}
\]
6. This gives us the solutions:
\[
p = \frac{16}{8} = 2 \quad \text{or} \quad p = \frac{-22}{8} = -\frac{11}{4}
\]
Thus, the possible values of \(p\) are:
\[
\boxed{2} \quad \text{and} \quad \boxed{-\frac{11}{4}}
\]
\[
2p - 10 = \frac{(p + 1) + (1 - 4p^2)}{2}
\]
Now, we'll simplify and solve for \(p\).
1. Start by simplifying the right side:
\[
\frac{(p + 1) + (1 - 4p^2)}{2} = \frac{p + 1 + 1 - 4p^2}{2} = \frac{p - 4p^2 + 2}{2}
\]
2. Therefore, the equation becomes:
\[
2p - 10 = \frac{p - 4p^2 + 2}{2}
\]
3. To eliminate the fraction, multiply through by 2:
\[
4p - 20 = p - 4p^2 + 2
\]
4. Rearranging gives us:
\[
4p - p + 4p^2 - 20 - 2 = 0
\]
\[
4p^2 + 3p - 22 = 0
\]
5. Now, we solve this quadratic equation using the quadratic formula:
\[
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 4\), \(b = 3\), and \(c = -22\):
\[
p = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-22)}}{2 \cdot 4}
\]
\[
p = \frac{-3 \pm \sqrt{9 + 352}}{8}
\]
\[
p = \frac{-3 \pm \sqrt{361}}{8}
\]
\[
p = \frac{-3 \pm 19}{8}
\]
6. This gives us the solutions:
\[
p = \frac{16}{8} = 2 \quad \text{or} \quad p = \frac{-22}{8} = -\frac{11}{4}
\]
Thus, the possible values of \(p\) are:
\[
\boxed{2} \quad \text{and} \quad \boxed{-\frac{11}{4}}
\]