Question

The given numbers x+3, 2x+1 and x-7 are in AP. Find the value of x.

Answer 1

Hint: We know when numbers are in Arithmetic progression that is AP then the difference successive terms is constant. Hence here also if we consider $t_1,t_2,t_3$be the first three terms of the sequence then we know $t_2-t_1=t_3-t_2$ this will give us one equation with one variable. Hence we can easily find the value of x.

Complete step by step answer:
Now we are given that the given numbers x+3, 2x+1 and x-7 are in AP.
We know that difference between successive terms in AP is constant
Hence we can write 2x+1 – (x+3) = (x-7) – (2x+1)
$\Rightarrow 2x+1-x-3=x-7-2x-1$
Now we know 2x – x = x and x – 2x = - x Substituting this in the above equation we get
$x+1-3=-x-7-1$
Now 1 – 3 = -2 and -7 – 1 = -8, hence we get
$x-2=-x-8$
Now taking –x from RHS to LHS and -2 from LHS to RHS we get
$x+x=2-8$
2x = -6
$x={\displaystyle \frac{-6}{2}}=-3$
X = -3

Hence the value of x is -3.

Note: now here we can also solve this question with another interesting method since x+3, 2x+1 and x-7 are in AP. 2x+1 is the Arithmetic mean of x+3 and x-7.
This means $2x+1={\displaystyle \frac{(x+3)+(x-7)}{2}}$. This is also one equation in one variable.
Hence we can also find the solution by solving this equation.