Question

Determine k so that k2+ 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP

Answer 1

Hii frnd.

As we know that in an AP the common difference is equal between terms
i.e 
Consider an AP series:
a,a+1,a+2
Here the common difference is 1

So.
A to Q
(2k2+3k+6)-(k2+4k+8)=(3k2+4k+4)-(2k2+3k+6)
k2-k-2=k2+k-2
2k=0
k=0

Hope it helped u.

Answer 2

 Given, First term, T1 = k^2 + 4k + 8

Given, Second term, T2 = 2k^2 + 3k + 6 

Given,Third term, T3 = 3k^2 + 4k + 4 

Given, They are 3 consecutive terms of an AP.

We know that 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) 

= 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!