Question

please help me out with this question of AP

Answer 1

Hi friend,

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Given 4k+8,2k²+3k+6,3k²+4k+4 are in A.p

→we know that if a,b,c are in A.p

→ then b-a=c-b

→2b=a+c

→2(2k²+3k+6)=4k+8+3k²+4k+4

=4k²+6k+12=3k²+8k+12

→k²-2k=0

→k(k-2)=0

Therefore ,the values of k are 0,2

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I hope this will help u ;)

Answer 2

we know that if a,b,c are in A.p

 then b-a=c-b

2b=a+c 

2(2k²+3k+6)=4k+8+3k²+4k+4

=4k²+6k+12=3k²+8k+12

k²-2k=0

k(k-2)=0

Therefore ,the values of k are 0,2