Question

find k if k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P.
​

Answer 1

Step-by-step explanation:

k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P.

k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P. 2(2k² + 3k + 6) = k² + 4k + 8 + 3k² + 4k + 4

k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P. 2(2k² + 3k + 6) = k² + 4k + 8 + 3k² + 4k + 44k² + 6k + 12 = 4k² + 8k + 12

k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P. 2(2k² + 3k + 6) = k² + 4k + 8 + 3k² + 4k + 44k² + 6k + 12 = 4k² + 8k + 122k = 0

k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P. 2(2k² + 3k + 6) = k² + 4k + 8 + 3k² + 4k + 44k² + 6k + 12 = 4k² + 8k + 122k = 0k = 0

Hope it helps

Answer 2

GIVEN :

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

The common difference between the terms is same.

Difference = a2 - a1 ; a3 - a2

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

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

k² - k - 2 = k² + k - 2

k² - k² - 2 + 2 = k + k

2k = 0

k = 0

Therefore, the value of k is 0.