find value of k so that 8k+4,6k-2and2k-7 will form A.P

Question

Its about sequences and series on arithmetic progression

Answer 1

The common difference is 6k-2-(8k+4) and 2k-7-(6k-2).

These evaluate to -2k-6 and -4k-5, so -2k-6=-4k-5, 2k=1, k=½.

The terms are 8k+4=8, 6k-2=1, 2k-7=-6; d, common difference=-2k-6=-4k-5=-7.

This is k if the terms are sequential.

But there are 6 ways to arrange the terms. If we call the terms a, b, c then we can have (a=8k+4, b=6k-2, c=2k-7):

1. a, b, c: solution is k=½, d=-7; AP=8,1,-6
2. a, c, b: c-a=-6k-11=b-c=4k+5, k=-1.6, d=-1.4; AP=-8.8,-10.2,-11.6
3. b, a, c: k=-2.125, d=1.75; AP=-14.75,-13,-11.25
4. b, c, a: k=-1.6,d=1.4; AP=-11.6,-10.2,-8.8
5. c, a, b: k=-2.125, d=-1.75; AP=-11.25,-13,-14.75
6. c, b, a: k=½, d=7; AP=-6,1,8