Q 5 6 7 all

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

2k = 0

So, k = 0

For any value of " K " given equations are in A.P. 

Regards