Question

If the ${{p}^{th}},{{q}^{th}}$ and ${{r}^{th}}$ terms of an A.P. are in G.P. . Then find the common ratio of the GP

Answer 1

Hint: We will first write the given terms in the form of $a$ and $d$ where, $a$ is the first term of the sequence and $d$ is the common difference between terms. This we will do by applying the formula for the ${{n}^{th}}$ term of an arithmetic sequence that is ${{a}_{n}}=a+\left( n-1 \right)d$ . Then we will apply the definition of geometric sequences and we will write the obtained terms as $\left\{ A,AR,A{{R}^{2}},A{{R}^{3}}..... \right\}$, finally we will get two equations , we will divide them and get our common ratio.

Complete step-by-step answer:
We know that the formula for the ${{n}^{th}}$ term of an arithmetic sequence is as follows:
${{a}_{n}}=a+\left( n-1 \right)d$
Where, ${{a}_{n}}$ is the ${{n}^{th}}$ term, $a$ is the first term of the sequence and $d$ is the common difference between terms.
Now, we will find the ${{p}^{th}}$ term , by applying the above formula. We will take $n=p$. Therefore, ${{a}_{p}}=a+\left( p-1 \right)d\text{ }.........\text{Equation 1}\text{.}$
Similarly we will find the ${{q}^{th}}$ term and we will now take $n=q$. So,
${{a}_{q}}=a+\left( q-1 \right)d$
Finally, we will find the ${{r}^{th}}$ term and we will be taking $n=r$. So,
${{a}_{r}}=a+\left( r-1 \right)d$

We now know that in a geometric sequence each term is found by multiplying the previous term by a constant. In general, a sequence say $\left\{ A,AR,A{{R}^{2}},A{{R}^{3}}..... \right\}$ is in geometric progression, where, $A$ is the first term and $R$ is the common ratio.
Since, it is given in the question that ${{p}^{th}},{{q}^{th}}$ and ${{r}^{th}}$ terms of the arithmetic progression are in geometric progression,
Now, the ${{p}^{th}}$ term of the A.P. that is $a+\left( p-1 \right)d$ is the first term of the GP, Therefore: $a+\left( p-1 \right)d=A\text{ }..........\text{Equation 1}\text{.}$

Similarly the ${{q}^{th}}$ term that is $a+\left( q-1 \right)d$ will be the second term of the GP: $a+\left( q-1 \right)d=AR\text{ }........\text{Equation 2}\text{.}$
And finally the ${{r}^{th}}$ term which is $a+\left( r-1 \right)d$ will be the third term of the GP: $a+\left( r-1 \right)d=A{{R}^{2}}\text{ }............\text{Equation 3}\text{.}$
Now we will subtract equation 1 from equation 2 , we will get: $\begin{align}
  & \left[ a+\left( q-1 \right)d \right]-\left[ a+\left( p-1 \right)d \right]=AR-A \\
 & a+\left( q-1 \right)d-a-\left( p-1 \right)d=A\left( R-1 \right) \\
 & \left[ \left( q-1 \right)-\left( p-1 \right) \right]d=A\left( R-1 \right) \\
 & \Rightarrow \left( q-p \right)d=A\left( R-1 \right)\text{ }.........\text{ Equation 4}\text{.} \\
\end{align}$
Similarly we will subtract equation 2 from equation 3:
$\begin{align}
  & \left[ a+\left( r-1 \right)d \right]-\left[ a+\left( q-1 \right)d \right]=A{{R}^{2}}-AR \\
 & a+\left( r-1 \right)d-a-\left( q-1 \right)d=AR\left( R-1 \right) \\
 & \left[ \left( r-1 \right)-\left( q-1 \right) \right]d=AR\left( R-1 \right) \\
 & \Rightarrow \left( r-q \right)d=AR\left( R-1 \right)\text{ }.........\text{ Equation 5}\text{.} \\
\end{align}$
We will now divide equation 5 by equation 4:
$\begin{align}
  & \Rightarrow \dfrac{\left( r-q \right)d}{\left( q-p \right)d}=\dfrac{AR\left( R-1 \right)}{A\left( R-1 \right)}\Rightarrow \dfrac{\left( r-q \right)}{\left( q-p \right)}=R \\
 & \Rightarrow R=\dfrac{-\left( q-r \right)}{-\left( p-q \right)} \\
 & \Rightarrow R=\dfrac{q-r}{p-q} \\
\end{align}$
Hence, the common ratio for the G.P. is $\dfrac{q-r}{p-q}$ .

Note: The resultant common ratio can be applied to general terms of an arithmetic progression which are in G.P. . While solving such types of questions, remember to proceed in such a way that we eliminate the terms which we do not require or we are not given the value of it. For example, here we eliminated the variable $A$ .