Question

Determine k, so that ksquare+4k+8, 2ksquare+3k+6 and 3ksquare+4k+4 are three consecutive terms of AP

Answer 1

Given a1 = k^2 + 4k + 8.

           a2 = 2k^2 + 3k + 6

           a3 = 3k^2 + 4k + 4.


a2 - a1 = a3 - a2

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 - 2 = k^2 + k - 2

-k = k

-2k = 0

k = 0.


Hope this helps!

Answer 2

Answer
...........

The given is ******. a1 = k^2 + 4k + 8.

******* a2 = 2k^2 + 3k + 6

******* a3 = 3k^2 + 4k + 4.


#################
a2 - a1
= a3 - a2
_________________
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 - 2 = k^2 + k - 2

( -k = k )

-2k = 0

k = 0.