Question

for what value of k; k+2,4k-6,3k-2 are consecutive terms of an AP

Answer 1

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

solving we get k =3 and the numbers are 5, 6, 7.

Answer 2

We know in A.P : a , a+d ,a+2d ...

a = k+2 a+d = 4k-6 a+2d = 3k-2

(a+d) - a = d

4k-6 -(k+2) =d => 3k - 8 = d ......(i)

(a+2d)-(a+d) = d

(3k - 2)-(4k - 6) = d => -k+4 =d .......(ii)

from (i) and (ii)

-k+4 = 3k -8

12 = 4k

k = 3