Question

Let \[x,|x - 1|,|x + 1|\] be first three terms of a G.P, then sum of the series
\[\dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + ........\infty \] is
a) \[9\]
b) \[\dfrac{9}{2}\]
c) \[3\]
d) \[6\]

Answer 1

Hint: We will use the property of logarithms in this question and form three cases. Then using the property of geometric progression, we will solve those three cases. Then we will select the most appropriate case of the three and find the sum of the given series by the formula for sum of the infinite geometric series.
Formula used:
Sum of the infinite geometric series with the first term \[a\] and common difference \[r\]\[ = \dfrac{a}{{1 - r}}\]
If three terms like \[a,b,c\] are in G.P then, \[{b^2} = ac\]

Complete answer: Given three terms of G.P are \[x,|x - 1|,|x + 1|\]
We will take three cases.
Case A: \[x \leqslant - 1\]
Now we will evaluate the three terms using the restriction that \[x \leqslant - 1\].
We know that when \[x\]is negative then
\[|x| = - x\]
Since \[x \leqslant - 1\], so \[\left( {x - 1} \right)\] is negative. Hence, \[|x - 1| = - \left( {x - 1} \right)\]
And similarly, \[|x + 1| = - \left( {x + 1} \right)\]
So, we get the three terms as:
\[x, - \left( {x - 1} \right), - \left( {x + 1} \right)\]
Now we use the property of G.P that, if three terms like \[a,b,c\] are in G.P then, \[{b^2} = ac\]
So, we get,
\[ \Rightarrow {\left\{ { - \left( {x - 1} \right)} \right\}^2} = - x\left( {x + 1} \right)\]
\[ \Rightarrow {x^2} + 1 - 2x = - {x^2} - x\]
On further simplification we get,
\[ \Rightarrow 2{x^2} - x + 1 = 0\]
Now to find the value of \[x\], we will use the quadratic formula.
\[\therefore x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
So, we get,
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - 4 \times 2} }}{4}\]
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt { - 7} }}{4}\]
So, in this case \[x\] is an imaginary number.

Case B: \[ - 1 \leqslant x \leqslant 1\]
Since \[ - 1 \leqslant x \leqslant 1\], so \[\left( {x - 1} \right)\] is either zero or negative.
\[\therefore |x - 1| = - \left( {x - 1} \right)\]
Similarly, \[\left( {x + 1} \right)\] is either zero or positive.
\[\therefore |x + 1| = \left( {x + 1} \right)\]
So, we get the three terms as:
\[x, - \left( {x - 1} \right),\left( {x + 1} \right)\]
Same as in case A, we will use the property of G.P.
\[ \Rightarrow {\left\{ { - \left( {x - 1} \right)} \right\}^2} = x\left( {x + 1} \right)\]
On further simplification we get,
\[ \Rightarrow {x^2} + 1 - 2x = {x^2} + x\]
\[ \Rightarrow 3x = 1\]
Or we get,
\[ \Rightarrow x = \dfrac{1}{3}\]
Case C: \[x > 1\]
Since \[x > 1\], so \[\left( {x - 1} \right)\] as well as \[\left( {x - 1} \right)\] are positive. Hence,
\[|x - 1| = x - 1\]
And,
\[|x + 1| = x + 1\]
So, the three terms of the G.P for this case becomes;
\[x,\left( {x - 1} \right),\left( {x + 1} \right)\]
Now we will use the property of G.P as used in the above cases, we get,
\[ \Rightarrow {\left( {x - 1} \right)^2} = x\left( {x + 1} \right)\]
On simplification we get;
\[ \Rightarrow {x^2} + 1 - 2x = {x^2} + x\]
\[ \Rightarrow 3x = 1\]
Or we get,
\[ \Rightarrow x = \dfrac{1}{3}\]
But we assumed that \[x > 1\], so this case is not valid.
Now out of the three cases, Case B is the most appropriate case to form the series given in the question.
For case B; \[x = \dfrac{1}{3}\]
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = \dfrac{1}{{\left( {\dfrac{1}{3}} \right)}} + \dfrac{1}{{|\dfrac{1}{3} - 1|}} + \dfrac{1}{{|\dfrac{1}{3} + 1|}} + ......\infty \]
On solving further, we get;
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = \dfrac{1}{{\left( {\dfrac{1}{3}} \right)}} + \dfrac{1}{{\left( {\dfrac{2}{3}} \right)}} + \dfrac{1}{{\left( {\dfrac{4}{3}} \right)}} + ......\infty \]
Solving further we get;
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = \dfrac{3}{1} + \dfrac{3}{2} + \dfrac{3}{4} + ......\infty \]
Now we will take \[3\] as common;
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = 3\left( {\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{4} + ......\infty } \right)\]
Now we can see that the right-hand side of the above equation is a G.P with first term \[a = 1\] and common difference \[r = \dfrac{1}{2}\].
We apply the formula that Sum of the infinite geometric series with the first term \[a\] and common difference \[r\]\[ = \dfrac{a}{{1 - r}}\].
So, we get;
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = 3\left( {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right)\]
On solving further, we get;
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = 3\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right)\]
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = 3 \times 2\]
So, now we have,
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{|x - 1|}} + \dfrac{1}{{|x + 1|}} + .....\infty = 6\]
So, option d is correct.

Note:
One point to note here is that to form the three cases we have used \[1, - 1\] as the parameter. This is because these are the roots of the terms given in modulus. Suppose we had \[x,|x - 2|,|x + 2|\], as the three terms then we need to form the three cases as; \[x \leqslant - 2, - 2 \leqslant x \leqslant 2,{\text{ }}x \geqslant 2\]. Because here the roots are \[2, - 2\].