Determaine k so that k^2+4k+8,2k^2+3k+6,3k^2+4k+4 are three consecutive terms of an ap
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Since they are 3 consecutive terms of an arithmetic progression.
second term - first term = third term - second term
t2 -t1 = t3 -t2
(2k^2 +3k +6) - (k^2+4k +8) = (3k^2 +4k +4) - (2k^2 +3k +6)
k^2 - k -2 = k^2 + k -2
2k = 0
k = 0.
Hope this helps!
second term - first term = third term - second term
t2 -t1 = t3 -t2
(2k^2 +3k +6) - (k^2+4k +8) = (3k^2 +4k +4) - (2k^2 +3k +6)
k^2 - k -2 = k^2 + k -2
2k = 0
k = 0.
Hope this helps!
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