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Question: The first term and last term of a GP are a and l respectively where r is the common ratio, then the ...

The first term and last term of a GP are a and l respectively where r is the common ratio, then the number of terms in this GP is
(A) (log⁡l−log⁡alog⁡r)\left( {\dfrac{{\log l - \log a}}{{\log r}}} \right)
(B) 1−(log⁡l−log⁡alog⁡r)1 - \left( {\dfrac{{\log l - \log a}}{{\log r}}} \right)
(C) (log⁡a−log⁡llog⁡r)\left( {\dfrac{{\log a - \log l}}{{\log r}}} \right)
(D) 1+(log⁡l−log⁡alog⁡r)1 + \left( {\dfrac{{\log l - \log a}}{{\log r}}} \right)

Explanation

Solution

Hint : The given problem involves the concepts of geometric progression. We are given the expressions for a few terms of the GP and we have to find the required term of the same. So, we make use of the formula for the general term of a geometric progression an=ar(n−1){a_n} = a{r^{\left( {n - 1} \right)}} to solve the problem. We must have a good grip over transposition and laws of logarithms in order to get to the final answer.

Complete step-by-step answer :
So, we have the first term of a GP as a and last term of GP as l and we have to find the number of terms in the geometric progression.
Let us assume the number of terms to be n.
We are also given the common ratio of GP as r.
So, we have the last nth term of the series as an=ar(n−1){a_n} = a{r^{\left( {n - 1} \right)}} .
So, we get, l=ar(n−1)l = a{r^{\left( {n - 1} \right)}} .
Dividing both the sides of the above equation by a, we get,
⇒la=r(n−1)\Rightarrow \dfrac{l}{a} = {r^{\left( {n - 1} \right)}}
Taking logarithm on both sides of equation, we get,
⇒log⁡(la)=log⁡r(n−1)\Rightarrow \log \left( {\dfrac{l}{a}} \right) = \log {r^{\left( {n - 1} \right)}}
Making use of the logarithmic property log⁡ab=log⁡a−log⁡b\log \dfrac{a}{b} = \log a - \log b , we get,
⇒log⁡l−log⁡a=log⁡r(n−1)\Rightarrow \log l - \log a = \log {r^{\left( {n - 1} \right)}}
Now, we use the logarithmic property log⁡xn=nlog⁡x\log {x^n} = n\log x , we get,
⇒log⁡l−log⁡a=(n−1)log⁡r\Rightarrow \log l - \log a = \left( {n - 1} \right)\log r
Now, we have to find the value of n. So, isolating the term consisting n, we get,
⇒log⁡l−log⁡alog⁡r=(n−1)\Rightarrow \dfrac{{\log l - \log a}}{{\log r}} = \left( {n - 1} \right)
⇒n=1+(log⁡l−log⁡alog⁡r)\Rightarrow n = 1 + \left( {\dfrac{{\log l - \log a}}{{\log r}}} \right)
Hence, the option (D) is correct.
So, the correct answer is “Option D”.

Note : Geometric progression is a series where any two consecutive terms have the same ratio between them. The common ratio of a geometric series can be calculated by dividing any two consecutive terms of the series. Any term of a geometric progression can be calculated if we know the first term and the common ratio of the series as: an=ar(n−1){a_n} = a{r^{\left( {n - 1} \right)}} . Laws of logarithms must be remembered for solving such questions.