Solve this:

Q.9. Determine K so that $K^2+4K+8, 2K^2+3K+6, 3K^2+4K+4$ are three consecutive terms of an A.P.

Dear Student,

Please find below the solution to the asked query:

Given :  k² + 4 k + 8 , 2 k² + 3 k + 6 and 3 k² + 4 k + 4 are in AP.

We know In A.P. common difference is equal , So

( 2 k² + 3 k + 6 ) - ( k² + 4 k + 8 ) = ( 3 k² + 4 k + 4 ) - ( 2 k² + 3 k + 6 )

2 k² + 3 k + 6 -  k² - 4 k - 8 = 3 k² + 4 k + 4  -   2 k² - 3 k - 6

k² -  k   - 2 =  k² + k  - 2

2 k = 0

So,

k = 0                                                                              ( Ans )

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