Question

Determine k so that 8k+4, 6k-2 and 2k-7 are the three consecutive terms of an AP

Answer 1

6k-4-8k-4 = 2k-7-6k-4
-2k-8 =3k-11

Answer 2

Given,

8k+4, 6k-2, and 2k-7 are the three consecutive terms of an AP.

To find,

The value of k.

Solution,

• The term a, b, and c  are said to be in Arithmetic Progression when the common difference between the term are equal i.e. b -a = c -b.

• It is given that 8k+4, 6k-2, and 2k-7 are in  Arithmetic Progression.

Hence,

⇒   (6k - 2) - (8k + 4) = (2k - 7) - (6k - 2)

⇒   Solving for k-

⇒   6k - 2 - 8k -4 = 2k - 7 - 6k +2.

⇒   -2k - 6 = -4k - 5

Shifting the k terms to LHS and the constant terms to RHS.

⇒   -2k + 4k = -5 + 6.

⇒   2k = 1

⇒   k = 1/2.

Hence, the value of k = 1/2 so that 8k+4, 6k-2, and 2k-7 are the three consecutive terms of an AP.

And after substituting the value of k in the terms we get-

⇒   8(1/2)+4, 6(1/2)-2 and 2(1/2)-7

⇒  8, 1, -6.

Hence the numbers are 8, 1, and -6 are in arithmetic progression with a common difference of (1-8) = -7.