Question

let (x-1)(,x+2)(x+7) are in AP find the value of x​

Answer 1

heya!!!

a,b and c are in A.P iff b - a = c - d

EXP:-

( x + 2 ) - ( x - 1 ) = ( x + 7 ) - ( x + 2 )

3 = 5 ( Absurd )

So, given question is wrong.

Answer 2

Answer:

There is no real value of x for which the numbers are in AP.

Step-by-step explanation:

According to question,

The numbers x-1 , x+2 , x+7 are in A.P.

We know  in case of arithmetic sequence, the difference between two consecutive terms should be equal

Since, the given numbers are in A.P. so the difference between these terms will be equal, that is,

( x + 2 ) - ( x - 1 ) = ( x + 7 ) - ( x + 2 )

x + 2 - x + 1 = x + 7 - x - 2

3 = 5

Since,

This statement is false because 3 is not equal to 5 [ LHS ≠ RHS ]

Therefore, there is no solution for the equation.

Hence,

There is no real value of x for which the numbers are in AP.