If p + 1, 2P – 10, 1 – 4p2are three consecutive terms of an arithmetic progression, find the possible values of p
The correct answer is: C
Explanation
2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean)= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -\(\frac{4}{11}\)