If $ A = 1 + {r^a} + {r^{2a}} + {r^{3a}} + ......\infty $ and $ B = 1 + {r^b} + {r^{2b}} + .......\infty $ then find $ \dfrac{a}{b} $
A) $ \dfrac{{\log \left( {A - 1} \right)}}{{\log \left( {B - 1} \right)}} $
B) $ \dfrac{{\log \left( {\dfrac{{A - 1}}{A}} \right)}}{{\log \left( {\dfrac{{B - 1}}{B}} \right)}} $
C) $ \dfrac{{\log A}}{{\log B}} $
D) $ \dfrac{{\log B}}{{\log A}} $
Answer
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Hint: In the given question, A and B are two geometric progressions. Find the common ratio of both the geometric progressions. Use the following formula to find the values of both the progressions and then calculate the values of a and b from them.
Formula used: $ S = \dfrac{a}{{1 - r}} $ , here a is the first term of the series and r is the common ratio.
Complete step-by-step answer:
Given to us are two geometric progressions, $ A = 1 + {r^a} + {r^{2a}} + {r^{3a}} + ......\infty $ and $ B = 1 + {r^b} + {r^{2b}} + .......\infty $
Let us first find the value of A. In order to find this we have to first calculate the common ratio.
Common ratio is given as $ \dfrac{{{r^a}}}{1} = {r^a} $ and we know that the first term is one. By substituting these values in the above formula, we get the value of A.
$ A = \dfrac{1}{{1 - {r^a}}} $
From this we have to calculate the value of a. This can be done as follows.
$ 1 - {r^a} = \dfrac{1}{A} \Rightarrow 1 - \dfrac{1}{A} = {r^a} $
On further solving this, we get $ {r^a} = \dfrac{{A - 1}}{A} $
Taking log on both sides, we get $ a\log r = \log \left( {\dfrac{{A - 1}}{A}} \right) $ and let this equation be X
Similarly we can find the value of the progression B to be $ B = \dfrac{1}{{1 - {r^b}}} $
From this we get the value of b as $ b\log r = \log \left( {\dfrac{{B - 1}}{B}} \right) $ and let this equation be Y.
Now, by dividing X and Y equations, we get $ \dfrac{{a\log r}}{{b\log r}} = \dfrac{{\log \left( {\dfrac{{A - 1}}{A}} \right)}}{{\log \left( {\dfrac{{B - 1}}{B}} \right)}} $
On solving, we get $ \dfrac{a}{b} = \dfrac{{\log \left( {\dfrac{{A - 1}}{A}} \right)}}{{\log \left( {\dfrac{{B - 1}}{B}} \right)}} $
So, the correct answer is “Option B”.
Note: It is to be noted that the above formula of geometric progression is only applicable if the value of common ratio is less than one. This means that in the above progressions A and B, we are assuming that the common ratios $ {r^a} $ and $ {r^b} $ are less than one.
Formula used: $ S = \dfrac{a}{{1 - r}} $ , here a is the first term of the series and r is the common ratio.
Complete step-by-step answer:
Given to us are two geometric progressions, $ A = 1 + {r^a} + {r^{2a}} + {r^{3a}} + ......\infty $ and $ B = 1 + {r^b} + {r^{2b}} + .......\infty $
Let us first find the value of A. In order to find this we have to first calculate the common ratio.
Common ratio is given as $ \dfrac{{{r^a}}}{1} = {r^a} $ and we know that the first term is one. By substituting these values in the above formula, we get the value of A.
$ A = \dfrac{1}{{1 - {r^a}}} $
From this we have to calculate the value of a. This can be done as follows.
$ 1 - {r^a} = \dfrac{1}{A} \Rightarrow 1 - \dfrac{1}{A} = {r^a} $
On further solving this, we get $ {r^a} = \dfrac{{A - 1}}{A} $
Taking log on both sides, we get $ a\log r = \log \left( {\dfrac{{A - 1}}{A}} \right) $ and let this equation be X
Similarly we can find the value of the progression B to be $ B = \dfrac{1}{{1 - {r^b}}} $
From this we get the value of b as $ b\log r = \log \left( {\dfrac{{B - 1}}{B}} \right) $ and let this equation be Y.
Now, by dividing X and Y equations, we get $ \dfrac{{a\log r}}{{b\log r}} = \dfrac{{\log \left( {\dfrac{{A - 1}}{A}} \right)}}{{\log \left( {\dfrac{{B - 1}}{B}} \right)}} $
On solving, we get $ \dfrac{a}{b} = \dfrac{{\log \left( {\dfrac{{A - 1}}{A}} \right)}}{{\log \left( {\dfrac{{B - 1}}{B}} \right)}} $
So, the correct answer is “Option B”.
Note: It is to be noted that the above formula of geometric progression is only applicable if the value of common ratio is less than one. This means that in the above progressions A and B, we are assuming that the common ratios $ {r^a} $ and $ {r^b} $ are less than one.
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