Find k so that k sq +4k +8,2k sq+ 3k+6, 3k sq+4k+4 are three consecutive terms of an ap
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Pickeyleo:
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yeah
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(2k^2+3k+6)-(k^2+4k+8)
=2k^2+3k+6-k^2-4k-2
=k^2-k-2 ................1
(3k^2+4k+4)-(2k^2+3k+6)
=3k^2+4k+4-2k^2-3k-6
=k^2+k-2 ..,,.....,.,.....2
From 1 and 2,
k^2-k-2= k^2+k-2
=> -k-k=-2+2
=> k=0
=2k^2+3k+6-k^2-4k-2
=k^2-k-2 ................1
(3k^2+4k+4)-(2k^2+3k+6)
=3k^2+4k+4-2k^2-3k-6
=k^2+k-2 ..,,.....,.,.....2
From 1 and 2,
k^2-k-2= k^2+k-2
=> -k-k=-2+2
=> k=0
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