Question

Three non-zero real numbers form A.P. and the squares of these numbers taken in the same order form a G.P. Then the number of all possible common ratio of the G.P. is
1. \[1\]
2. \[2\]
3. \[3\]
4. None of these

Answer 1

Hint: Assume the series by considering the condition of AP series. After squaring the terms of the AP series the series we will get is the GP series according to the given question. Then apply the condition of the GP series i.e. the common ratio is the same and this will give the value for \[d\]. After substituting the value of \[d\]we will get the required answer.

Complete step-by-step solution:
As we know, the AP series is the kind of series in which the difference between the two consecutive terms are the same. That same difference is some constant value known as the common difference which is denoted by the symbol \['d'\].
Let us assume that the series of three non- zero real numbers which form an AP series is \[a-d,a,a+d\]where \[d\]is the common difference.
And in the question it is given that the squares of these non-zero real numbers form a GP series or we can say Geometric Progression series.
This means that \[{{(a-d)}^{2}},{{a}^{2}},{{(a+d)}^{2}}\]are the terms of the GP series.
If the sequence is in GP then it means that the ratio between the two consecutive numbers is the same. So after applying the definition of GP, we get
\[\dfrac{{{a}^{2}}}{{{(a-d)}^{2}}}=\dfrac{{{(a+d)}^{2}}}{{{a}^{2}}}\]
\[{{({{a}^{2}})}^{2}}={{(a+d)}^{2}}{{(a-d)}^{2}}\]
\[{{({{a}^{2}})}^{2}}={{((a+d)(a-d))}^{2}}\]
And we know that, \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]. So applying the same identity, we get
\[{{({{a}^{2}})}^{2}}={{({{a}^{2}}-{{d}^{2}})}^{2}}\]
Apply root on both the sides, we will get
\[({{a}^{2}})=({{a}^{2}}-{{d}^{2}})\]
But \[{{a}^{2}}={{a}^{2}}-{{d}^{2}}\] or \[{{a}^{2}}={{d}^{2}}-{{a}^{2}}\]
First case i.e. \[{{a}^{2}}={{a}^{2}}-{{d}^{2}}\]is not possible because it gives \[d=0\]which is not possible because all the terms of the series are different. But \[d=0\]means all the terms of the series are the same.
 So from above we can say that,
\[{{d}^{2}}=2{{a}^{2}}\]
\[d=\pm a\sqrt{2}\]
And we know that common ratio \[r\] between the consecutive term is given as
\[r=\dfrac{{{a}^{2}}}{{{(a-d)}^{2}}}\]
We have just calculated \[d=\pm a\sqrt{2}\]. Substitute this value in the common ratio, we get
\[r=\dfrac{{{a}^{2}}}{{{(a\pm a\sqrt{2})}^{2}}}\]
Taking \[a\]from denominator, we will get
\[r=\dfrac{1}{{{(1\pm \sqrt{2})}^{2}}}\]
This proves that there are two common ratios possible. And they are
\[r=\dfrac{1}{{{(1+\sqrt{2})}^{2}}}\] or \[r=\dfrac{1}{{{(1-\sqrt{2})}^{2}}}\]
Hence after solving the given question we can conclude that option \[(2)\] is the correct one.

Note: The series is a sequence of numbers in which the first, second, third and so on all the numbers follow some pattern that is if we apply that pattern on the first number we will get the second number. Similarly if we apply the same pattern on the second number we will get the third element of the series.