Question

If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is
(a) −2
(b) 3
(c) −3
(d) 6

Answer 1

Answer:

The value of k is 3.

Among the given options option (b) 3 is a correct answer.

Step-by-step explanation:

Given :  

k, 2k -1, 2k + 1 are in A.P

Let a1 =  k , a2 = 2k - 1  , a3 = 2k + 1

If terms are in A.P, then common difference (d) of any two consecutive terms is same.  

a2 - a1 = a3 - a2

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

2k - 1- k = 2k + 1 - 2k + 1

k - 1 = 2

k = 2 + 1

k = 3

Hence, the value of k is 3.

HOPE THIS ANSWER WILL HELP YOU….

 

Answer 2

Answer

K = 3

Explanation

Given:

k, 2k-1, 2k+1 are the three consecutive terms of an A.P.

To Find

Value of k

Solution

The series is...

k, 2k-1, 2k+1....

Here

a1 = k

a2 = 2k-1

a3 = 2k+1

since, the series is in A.P. Common difference(d) is equal!

Common difference

a2-a1 = a3-a2

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

k-1 = 2k+1-2k+1

k-1 = 2

$\boxed{k = 3}$