Question

Find k, so that 4k+8, 2k²+3k+6, and 3k²+4k+4 are three consecutive terms of an A.P.

Answer 1

the value of k is 0 or k is 2

Answer 2

k = 0 or 2

Step-by-step explanation:

The given $4k+8,2k^2+3k+6$ and $3k^2+4k+4$are three consecutive terms of an A.P.

To find, the value of k = ?

∴ Common difference(d) = Second term - First term =  Third term - Second term

∴ $2k^2+3k+6-(4k+8)=3k^2+4k+4-(2k^2+3k+6)$

⇒ $2k^2+3k+6-4k-8=3k^2+4k+4-2k^2-3k-6$

⇒ $2k^2-k-2=k^2+k-2$

⇒ $2k^2-k^2-k-k-2+2=0$

⇒ $k^2-2k=0$

⇒ $k(k-2)=0$

⇒ k = 0 or 2

Put k = 0, we get

The given 8, 6 and 4 are three consecutive terms of an A.P.

Put k = 2, we get

The given $4(2)+8,2(2)^2+3(2)+6$ and $3(2)^2+4(2)+4$ i.e.,16, 20, 24 are three consecutive terms of an A.P.

Hence, k = 0 or 2