Question

Determine k so that k^2+4k+8, 2k^2+3k+6 and 3k^2+4k+4 are three consecutive terms of an ap

Answer 1

If a, b, c are in A. P. then 2b = a+c; Similarly here, k^2+4k+8, 2k^2+3k+6 and 3k^2+4k+4 are in A.P. Therefore,2(2k^2+3k+6)=(k^2+4k+8) + (3k^2+4k+4) or, 4k^2+6k+12 = 4k^2+8k+12 or, 8k-6k =0 or, 2k =0 or, k =0

Answer 2

Answer:

Step-by-step explanation:

According to the Question

As we know that a,b,c are in AP if b - a = c - b or 2b = a + c

Now

2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4)

4k^2 + 6k + 12 = 4k^2 + 8k + 12

6k = 8k

6k - 8k = 0

-2k = 0

k = 0