Bro plzz give me the answer
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the difference of consecutive terms of an AP are equal.
therefore;
( 2k^2+3k+6)-(k^2+4k+8)=
(3k^2+4k+4)-(2k^2+3k+6)
2k^2+3k+6-k^2-4k+8 = 3k^2+4k+4-2k^2-3k-6
k^2-k+14 = k^2+k-2
2k=6
k=8
therefore;
( 2k^2+3k+6)-(k^2+4k+8)=
(3k^2+4k+4)-(2k^2+3k+6)
2k^2+3k+6-k^2-4k+8 = 3k^2+4k+4-2k^2-3k-6
k^2-k+14 = k^2+k-2
2k=6
k=8
abhishek474067:
But I think this should not be the answer
hii behna
I am not girl
I am a boy
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