Question

Three positive numbers form an increasing G.P. If the first term is doubled and the second term is tripled, the new numbers are in A.P, then the common ratio of the G.P is:
(This question has multiple correct options)
(a) $3-\sqrt{7}$
(b) \[3+\sqrt{7}\]
(c) 2
(d) \[\sqrt{7}\]

Answer 1

Hint: First of all consider three terms in G.P \[a,ar,a{{r}^{2}}\]. Now double the first term and triple the third. Get the resulting 3 terms and apply 2b = a + c in it where a, b and c are the first, second and third terms of A.P or the new sequence. Solve the equation to find the common ratio.

Complete step-by-step answer:
Here, we are given three positive numbers in G.P. Now if we double the first term and triple the second term, the new sequence formed will be in A.P. Then we have to find the common ratio of this G.P. Let us consider our question. We know that G.P or geometric progression is a sequence of numbers with a constant ratio between each number and the one before it. Now, we also know that the terms of the G.P are written as \[a,ar,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}}....\] where ‘a’ is the first term and ‘r’ is the common ratio of G.P.
So, let us consider the three terms given in the question in G.P as \[a,ar,a{{r}^{2}}\].
Now, we are given that the first term of G.P is doubled. So, we get,
First term = 2(a) = 2a.
Also, we are given that the second term of the G.P is tripled. So, we get,
Second term = 3(ar) = 3ar
So, we get the new sequence of the terms as:
\[2a,3ar,a{{r}^{2}}\]
Now, we are given that the above sequence is in A.P. We know that when three terms, say a, b and c are in A.P, then 2b = a + c.
So, by substituting a = 2a, b = 3ar and \[c=a{{r}^{2}}\] in the above equation, we get,
\[2\left( 3ar \right)=2a+a{{r}^{2}}\]
\[\Rightarrow 6ar=2a+a{{r}^{2}}\]
By taking out ‘a’ common from both the sides, we get,
\[a\left( 4r \right)=a\left( 2+{{r}^{2}} \right)\]
By canceling ‘a’ from both the sides of the above equation, we get,
\[6r=2+{{r}^{2}}\]
\[\Rightarrow {{r}^{2}}-6r+2=0\]
We know that for a quadratic equation, \[a{{x}^{2}}+bx+c\], we get, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
So, for the above quadratic equation in ‘r’, we get,
\[r=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times 2}}{2\times 1}\]
\[r=\dfrac{6\pm \sqrt{36-8}}{2}\]
\[r=\dfrac{6\pm 2\sqrt{7}}{2}\]
By canceling 2 from RHS of the above equation, we get,
\[r=3\pm \sqrt{7}\]
So, we get two common ratios as \[3+\sqrt{7}\] and \[3-\sqrt{7}\].
Hence, option (a) and (b) are the correct answers.

Note: Here, students must note that as nothing is given about the third term, we will keep the third term as it is in the new sequence that is in AP. Also, it is given that G.P is increasing G.P. So, ‘r’, that is the common ratio of the G.P should be greater than or equal to 1. If our ‘r’ comes out to be less than 1, then it would be rejected because it would form a decreasing G.P.