Question

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

Answer 1

Answer:

Step-by-step explanation:

the answer is:

or a consecutive ap

d1=d2

so d1= 2k^2+3k+6-[4k+8]

        = 2k^2+3k+6-4k-8

         =2k^2+3k-4k+6-8

so,      d1  =2k^2-k-2

d2 = 3k^2+4k+4-[2k^2+3k+6]

    = 3k^2+4k+4-2k^2-3k-6

    = 3k^2-2k^2+4k-3k+4-6

   = k^2-k-2

now equate d1 &d2

   2k^2-k-2= k^2-k-2

  2k^2-k+k-2+2=0

  k^2=0

    k=0

therefore k=0 is the answer

where the common difference is 2

the ap is 8,6,4.......