Find the value of x for which the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.

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Sir please give me a detailed solution of this question in which we are to find for which value of X the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.

Given: (5x+2),(4x−1) and (x+2) are terms in AP.
So, d=(4x−1)−(5x+2)=(x+2)−(4x−1) 2(4x−1)=(x+2)+(5x+2) 8x−2=6x+2+2 8x−2=6x+4 8x−6x=4+2
or x=3
The value of x is 3.

Given: (5x+2),(4x−1) and (x+2) are terms in AP.

So, d=(4x−1)−(5x+2)=(x+2)−(4x−1)

2(4x−1)=(x+2)+(5x+2)

8x−2=6x+2+2

8x−2=6x+4

8x−6x=4+2

or x=3

The value of x is 3.