Question

Find the value of k. If 2k, k+10 and 3k +2 are in Ap?

Answer 1

Here is your solution

$Let \\ a1=2k \\ a2=k+ 10 \\ a3=3k+2 \\ for \\ finding \: value \: of \: k \\ a2-a1 \: = a3-a2\\ k+ 10 -(2k) \: = \: 3k+2 - (k+10) \\ k+10 -2k = 3k + 2 -k - 10 \\ k -3k -2k +k = -10 -10 +2 \\ -3k = - 18 \\ k = 6$

hope it helps you

Answer 2

given 2k, k+10,3k+2 are in ap
then we know
t2 - t1 = t3 -t2
k+10 -(2k) = 3k+2 -(k+10)
k-2k+10 =3k+2-k-10
-k+10 = 2k-8
2k+k =10+8
3k=18
k=18/3
k=6 is answer obtained