Question

determine k so that k²+4k+8,2k²+3k+6,3k²+4k+4 are the three consecutive terms of an AP

​

Answer 1

Answer:

hope it helps you

please mark it as brainliest and follow me

Answer 2

$\huge \bold \blue {heya \: mate}$

Given that,

k²+4k+8, 2k²+3k+6 and 3k²+4k+4 are the 3 consecutive terms of an AP,

Now,

We know that,

a2-a1=a3-a2

Here,

a1=k²+4k+8, a2=2k²+3k+6, a3=3k²+4k+4

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-k-k-2+2=0

-2k=0

k=0

Thus, the value of k is 0.