Question

If x, | x + 1 |, |x - 1| are in AP, then find the common difference of an AP series.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: x, \: |x + 1|, \: |x - 1| \: are \: in \: AP$

We know, 3 numbers a, b, c are in AP iff b - a = c - b

So, using this result, we get

$\rm \: |x + 1| - x = |x - 1| - |x + 1|$

$\rm\implies \:2 |x + 1| = |x - 1| + x$

Now, to solve this modulus equation, we have two critical points, i. e. - 1 and 1

So, three cases arises.

Case - 1

$\red{\rm \: When \: x \leqslant - 1}$

We know, from definition of Modulus function

Modulus function is defined as

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, we get

$\rm \: - 2(x + 1) = x - (x - 1)$

$\rm \: - 2x - 2 = x - x + 1$

$\rm \: - 2x - 2 = 1$

$\rm \: - 2x = 1 + 2$

$\rm \: - 2x =3$

$\rm\implies \:x = - \dfrac{3}{2}$

So, three numbers

$\rm \: x, \: |x + 1|, \: |x - 1|$

on substituting the value of x, can be reduced to

$\rm \: - \dfrac{3}{2}, \: \dfrac{1}{2}, \: \dfrac{5}{2}$

So,

$\rm\implies \:Common \: difference = \dfrac{1}{2} + \dfrac{3}{2} = \dfrac{4}{2} = 2$

Case - 2

$\red{\rm \: When \: - 1 < x < 1}$

Now, in this case, equation reduces to

$\rm \: 2(x + 1) = x - (x - 1)$

$\rm \: 2x + 2 = x - x + 1$

$\rm \: 2x + 2 = 1$

$\rm \: 2x = 1 - 2$

$\rm \: 2x = - 1$

$\rm\implies \:x = - \dfrac{1}{2}$

So, three numbers

$\rm \: x, \: |x + 1|, \: |x - 1|$

on substituting the value of x, can be reduced to

$\rm \: - \dfrac{1}{2}, \: \dfrac{1}{2}, \: \dfrac{3}{2}$

So,

$\rm\implies \:Common \: difference = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{2}{2} = 1$

Hence,

Common difference of the series is 2 or 1

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: x, \: |x + 1|, \: |x - 1| \: are \: in \: AP$

We know, 3 numbers a, b, c are in AP iff b - a = c - b

So, using this result, we get

$\rm \: |x + 1| - x = |x - 1| - |x + 1|$

$\rm\implies \:2 |x + 1| = |x - 1| + x$

Now, to solve this modulus equation, we have two critical points, i. e. - 1 and 1

So, three cases arises.

Case - 1

$\red{\rm \: When \: x \leqslant - 1}$

We know, from definition of Modulus function

Modulus function is defined as

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, we get

$\rm \: - 2(x + 1) = x - (x - 1)$

$\rm \: - 2x - 2 = x - x + 1$

$\rm \: - 2x - 2 = 1$

$\rm \: - 2x = 1 + 2$

$\rm \: - 2x =3$

$\rm\implies \:x = - \dfrac{3}{2}$

So, three numbers

$\rm \: x, \: |x + 1|, \: |x - 1|$

on substituting the value of x, can be reduced to

$\rm \: - \dfrac{3}{2}, \: \dfrac{1}{2}, \: \dfrac{5}{2}$

So,

$\rm\implies \:Common \: difference = \dfrac{1}{2} + \dfrac{3}{2} = \dfrac{4}{2} = 2$

Case - 2

$\red{\rm \: When \: - 1 < x < 1}$

Now, in this case, equation reduces to

$\rm \: 2(x + 1) = x - (x - 1)$

$\rm \: 2x + 2 = x - x + 1$

$\rm \: 2x + 2 = 1$

$\rm \: 2x = 1 - 2$

$\rm \: 2x = - 1$

$\rm\implies \:x = - \dfrac{1}{2}$

So, three numbers

$\rm \: x, \: |x + 1|, \: |x - 1|$

on substituting the value of x, can be reduced to

$\rm \: - \dfrac{1}{2}, \: \dfrac{1}{2}, \: \dfrac{3}{2}$

So,

$\rm\implies \:Common \: difference = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{2}{2} = 1$

Hence,

Common difference of the series is 2 or 1