Question

For what value of k are 2k+1, 6k-1, 5k+7 the first three terms of an AP?

Answer 1

For an ap, we know that
2b=a+c........(i)
here a=2k+1; b=6k-1 ;c=5k+7
putting the value in eq (i) we get
2 (6k-1)=(2k+1)+(5k+7)
on solving ,
k=2.........Ans

Answer 2

Answer:

k=3

solution =

Step-by-step explanation:

given k,2k+1,6k+1 and 5k+7.

we now that simply we have to find the first three terms of an A.P.

Given = a = k

d = (2k-1)-(k)

= k+1.

An = a+(n-1)d

= k+(n-1)(k+1)

=k+kn-n-k-1

=kn-n-1

n-1=kn

k=1.

hope it will help you.