Question

determine K so that K square + 4 k + 8, 2k square+ 3 k + 6,3k square + 4 k + 4 are three
consecutive terms of an ap​

Answer 1

Given that k²+4k+8,2k²+3k+6 and 3k²+4k+4 are in A.P.

Let d be the common difference of the given A.P

We know that,

Common Difference=Third term - Second term= Second term - First term

→(k²+4k+8)-(2k²+3k+6)=(2k²+3k+6)-(3k²+4k+4)

→ -2k²+k²+4k-3k+8-6=2k²-3k²+3k-4k+6-4

→-k²+k+2= -k²-k+2

→k+2=2-k

→2k=0

→k=0

If the given terms are in A.P,then k would be equal to zero

Verification:

Putting k=0 in all the terms,

First term:k²+4k+8=0²+4(0)+8=8

Second term:2k²+3k+6=2(0)²+3(0)+6=6

Third term:3k²+4k+4=3(0)²+4(0)+4=4

Now,

Third term- Second term= Second term- First term

Hence, verified

Answer 2

GIVEN :

Three consecutive terms of an AP are :

k² + 4k + 8, 2k² + 3k + 6 , 3k² + 4k + 4

We know that,

The common difference between the terms is same.

Thus,

(2k² + 3k + 6) - (k² + 4k + 8) = (3k² + 4k + 4) - (2k² + 3k + 6)

2k² + 3k + 6 - k² - 4k - 8 = 3k² + 4k + 4 - 2k² - 3k - 6

k² - k - 2 = k² + k - 2

k² - k² - 2 + 2 = k + k

0 = 2k

k = 0/2

k = 0

Therefore, the value of k is 0.