Question

Determine K so that k 2 +4k+8,2k 2 +3k+6 and 3k 2 ​+4k+4 are three consecutive terms of AP.

Answer 1

i hope this will usful to u

Answer 2

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

Therefore for k = 0 the given 3 terms are in AP