Question

For what value of k are 2k, k+10 and 3k+2 in AP ?

Answer 1

Solution:

Let a1= 2k, a2= k+10, a3= 3k+2 are in AP, then

Second term - first term = third term - second term

K+10 -(2k) = 3k+2-(k+10)

K+10-2k =3k+2-k-10

k -2k +10 = 3k-k +2-10

-k +10= 2k -8

-k-2k= -8-10

-3k = -18

k =18/3= 6

k= 6

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Hope this will help you....

Answer 2

By definition of Arithmetic progression, the difference between two consecutive numbers must remain constant.
Given numbers are,
2k, k+10 and 3k+2
Let x = 2k
      y = k + 10
      z = 3k + 2
Hence, by definition, the difference of "y" to "x" must be equal to the difference of "z" to "y" 
So, (k + 10) - 2k = (3k + 2) - (k + 10)
k + 10 - 2k = 3k + 2 - k - 10
10 - k = 2k -8
10 + 8 = 2k + k
18 = 3k
On further simplification, we get:
k = 6
which is the required value of "k"
Hope this will help you.