Question

Determine the value of k for which (k2 + 4 K + 8), (2 k2+3 k+6) and (3 k2+4 k + 4) are in A.p
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Answer 1

Answer:

If a,b,c are three consecutive terms of an A.P then 2b=a+c

So, 2(2k

2

+3k+6)=k

2

+4k+8+3k

2

+4k+4

4k

2

+6k+12=4k

2

+8k+12

6k=8k

∴k=0

Step-by-step explanation:

i hope it is a correct answer