Question

Let \[{a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots ..\] be in AP and \[{a_p},{\text{ }}{a_q},{\text{ }}{a_r}\] be in GP. Then \[{a_q}:{\text{ }}{a_p}\] is equal to:
A) $ \dfrac{{r - p}}{{q - p}} $
B) $ \dfrac{{q - p}}{{r - q}} $
C) $ \dfrac{{r - q}}{{q - p}} $
D) $ 1 $

Answer 1

Hint: To solve this problem, we will first apply the formula of getting the value of nth term in an AP, then using that we will find \[{a_p},{\text{ }}{a_q}\] and \[{a_r}.\] Then after that we will use the law of proportion, we will put the solved values of \[{a_p},{\text{ }}{a_q}\] and \[{a_r}.\]Hence, with that we will get our required ratio.

Complete step-by-step answer:
We have been given that, \[{a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots ..\] be in AP and \[{a_p},{\text{ }}{a_q},{\text{ }}{a_r}\] in GP. We need to find the value of \[{a_q}:{\text{ }}{a_p}.\]
We know that, if the first term of an arithmetic progression is \[{a_1}\] and the common difference of successive members is \[d,\] then the nth term of the sequence an is given by, \[{a_n} = {\text{ }}{a_1} + {\text{ }}\left( {n - 1} \right)d \ldots \ldots eq.\left( 1 \right)\]
Now, here we have \[{a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots ..\] in AP and \[{a_p},{\text{ }}{a_q},{\text{ }}{a_r}\] in GP.
So, on putting the values in the \[eq.\left( 1 \right),\] we get
\[{a_p} = {\text{ }}{a_1} + {\text{ }}\left( {p{\text{ }} - {\text{ }}1} \right)d,{\text{ }}{a_q} = {\text{ }}{a_1} + {\text{ }}\left( {q{\text{ }} - {\text{ }}1} \right)d{\text{ }}\] and \[{a_r} = {\text{ }}{a_1} + {\text{ }}\left( {r{\text{ }} - {\text{ }}1} \right)d.\]
Since, \[{a_p},{\text{ }}{a_q},{\text{ }}{a_r}\] in GP, so by law of proportion, we get, \[\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{{a_r}}}{{{a_q}}}\]
\[\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{{a_q} - {a_r}}}{{{a_p} - {a_q}}}\]
Now, on putting the values of \[{a_p} = {\text{ }}{a_1} + {\text{ }}\left( {p{\text{ }} - {\text{ }}1} \right)d,{\text{ }}{a_q} = {\text{ }}{a_1} + {\text{ }}\left( {q{\text{ }} - {\text{ }}1} \right)d{\text{ }}\] and \[{a_r} = {\text{ }}{a_1} + {\text{ }}\left( {r{\text{ }} - {\text{ }}1} \right)d\] in the above equation, we get
$
\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{{a_1} + {\text{ }}\left( {q{\text{ }} - {\text{ }}1} \right)d - {a_1} + {\text{ }}\left( {r{\text{ }} - {\text{ }}1} \right)d}}{{{a_1} + {\text{ }}\left( {p{\text{ }} - {\text{ }}1} \right)d - {a_1} + {\text{ }}\left( {{\text{q }} - {\text{ }}1} \right)d}} \\
\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{{a_1} + {\text{ q}}d - d - {a_1} - {\text{ r}}d + d}}{{{a_1} + {\text{ p}}d - d - {a_1} - {\text{ q}}d + d}} \\
\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{{\text{q}}d - rd}}{{pd - qd}} \\
\dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{q - r}}{{p - q}} \\
\Rightarrow \dfrac{{{a_q}}}{{{a_p}}} = \dfrac{{r - q}}{{q - p}} \\
 $
So, \[{a_q}:{a_p}\] is \[\left( {r - q} \right):\left( {q - p} \right).\] Thus, option (C) $ \dfrac{{r - q}}{{q - p}}, $ is correct.
So, the correct answer is “Option C”.

Note: In the question we were given two sets one is in AP and while other is in GP. So, the basic difference between AP and GP is that, in Arithmetic progression, sequence is described as a list of numbers, in which each new term differs from a preceding term by a constant quantity, while in geometric progression, sequence is a set of numbers where each element after the first is obtained by multiplying the preceding number by a constant factor.