Question

If k + 1, 3k , 4k + 2 be any three consecutive terms of A.P then find value of k.

Answer 1

1st term t1 = (k + 1).

2nd term t2 = 3k.

3rd term t3 = (4k + 2).

Given that t1,t2,t3 are in AP.


We know that when they are in AP, their common difference will be:

t2 - t1 = t3 - t2

3k - (k + 1) = (4k + 2) - 3k

3k - k - 1 = 4k + 2 - 3k

2k - 1 = k + 2

2k = k + 2 + 1

2k = k + 3

2k - k = 3

k = 3.


Therefore the value of k = 3.


Hope this helps!

Answer 2

Solution:

Since,
k + 1 , 3k , 4k + 2 is three consecutive terms of A.P.

Therefore,

2×(3k) = (k+1)+(4k+2)

6k = 5k+3

=> k = 3 Ans.