Question

For what value of k, the numbers (k – 2), (4k – 1) and (5k +2) will be the consecutive terms of an A.P.?

Answer 1

Answer:

We know from formula

a,b,c are in A.P.

then, 2b=a+c

so, put

a=(k – 2)

b=(4k – 1)

c=(5k +2)

Step-by-step explanation:

now, 2*(4k – 1)=(k – 2)+(5k +2)

=> 8k-2=6k-2+2

=> k= 1.

Answer 2

Given:-

• Consecutive terms of an A.P. = (k - 2), (4k - 1) and (5k - 2)

To Find:-

The value of k.

Solution:-

We know,

The common difference between all the terms in an A.P. is always equal.

Hence,

Let (k - 2) = 1st term = $\sf{t_1}$

Let (4k - 1) = 2nd term = $\sf{t_2}$

Let (5k + 2) = 3rd term = $\sf{t_3}$

So,

Difference between the 1st term and 2nd term will always be equal to the difference between 2nd term and 3rd term

Therefore,

$\sf{(4k-1) - (k-2) = (5k + 2) - (4k - 1)}$

= $\sf{4k - 1 - k + 2 = 5k + 2 - 4k + 1}$

= $\sf{3k + 1 = k + 3}$

= $\sf{3k - k = 3-1}$

= $\sf{2k = 2}$

= $\sf{k = \dfrac{2}{2}}$

= $\sf{k = 1}$

Therefore the value of k is 1.

______________________________________

How to do?

Firstly we need to understand the concept of terms in an A.P.

So in an A.P. all the terms are in such a sequence that their difference is equal. In other words the difference between any two terms of an A.P. is always equal.

So here as we were given three terms we equated the difference of 1st and 2nd term with the difference of 2nd and 3rd term. And after equating these we get the value of k as 1.

______________________________________