Question

Please guys solve this problem​

Answer 1

Answer:

If 4p + 8, 2p² + 3p + 6 and 3p² + 4p + 4 are in A.P

Then, 3p² + 4p + 4 - 2p² + 3p + 6

= 2p² + 3p + 6 - 4p + 8

= p² + p - 2 = 2p² - p - 2

= p - 2 = p² - p -2

Taking p common in RHS we will get,

( p - 2 ) = ( p - 1 ) ( p - 2 )

p - 1 = 1 , p = 1 + 1 , p = 2

Hence the value of P is 2

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