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

Answer
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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 t1,t2,t3be the first three terms of the sequence then we know t2t1=t3t2 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)
2x+1x3=x72x1
Now we know 2x – x = x and x – 2x = - x Substituting this in the above equation we get
x+13=x71
Now 1 – 3 = -2 and -7 – 1 = -8, hence we get
x2=x8
Now taking –x from RHS to LHS and -2 from LHS to RHS we get
x+x=28
2x = -6
x=62=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=(x+3)+(x7)2. This is also one equation in one variable.
Hence we can also find the solution by solving this equation.
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