Question

k = ? if k² + 4k +8, 2k²+ 3k +4, 3k² + 4k +4 are three consecutive terms of an AP​

Answer 1

According to question:

difference=2k²+3k+4-k²+4k+8=k²+7k+12,

Also,

difference=3k²+4k+4-2k²+3k+4=k²+7k+8

then,

$= > k {}^{2} + 7k + 12 = 0$

$k {}^{2} + 3k + 4k + 12 = 0$

$k(k + 3) +4 (k + 3) = 0$

$k = - 3 \: or \: k = - 4$

Now,

$k {}^{2} + 7k - 8 = 0$

$k {}^{2} + 8k - k - 8 = 0$

$= > k(k + 8) - 1(k + 8) = 0$

$k = - 8 \: or \: k = 1$