If p + 1, 2P - 10, 1 - 4p²are three consecutive terms of an arithmetic progression, find the possible values of p

-4, 2

-3, \(\frac{4}{11}\)

-\(\frac{4}{11}\), 2

5, -3

Correct answer is C

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 -\(\frac{4}{11}\)

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If p + 1, 2P - 10, 1 - 4p²are three consecutive terms of an arithmetic progression, find the possible values of p