Question

For what value of k will the consecutive terms 2k+1, 3k+3, 5k-1 from an AP?

Answer 1

2k + 1 = a
3k + 3 = b
5k - 1 = c

To form an AP,
a + c = 2b
2k + 1 + 5k - 1 = 2( 3k + 3)
7k = 6k + 6
k = 6

So, k = 6

Hope This Helps You!

Answer 2

6 is the value of k for the consecutive terms 2k+1, 3k+3, 5k-1 which forms an AP.

Given:

Consecutive terms = 2k + 1, 3k + 3, 5k-1

To find:

The value of k which forms AP in the given consecutive terms  

Solution:

Consecutive terms = 2k + 1, 3k + 3, 5k-1

Therefore, by following the rule of AP that the difference of any two consecutive terms of an AP is same, we can write:

The formula for the finding the common difference is  

$t_{2}-t_{1}=t_{3}-t_{2}$

3k + 3 - 2k - 1 = 5k - 1 - 3k - 3

 k + 2 = 2k - 4

By separating the common terms

We get,

2k - k = 4 + 2  

      k = 6.  

Therefore, 6 is the value of k for the given consecutive terms.