Question

If ${{p}^{th}},{{q}^{th}},{{r}^{th}}$ and ${{s}^{th}}$ terms of an A.P are in G.P, then prove that p-q, q-r and r-s are also in G.P.

Answer 1

Hint: Assume that the first term of the A.P is “a” and the common difference is “d”. Use the fact that ${{n}^{th}}$term of an A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$. Use the fact that if a, b, c and d are in G.P, then $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$. Hence prove that p-q, q-r and r-s are in G.P

Complete step-by-step answer:
Let the first term of the A.P be “a” and let the common difference of the A.P be “d”.
We know that the ${{n}^{th}}$ term of the A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$.
Hence, we have
$\begin{align}
  & {{a}_{p}}=a+\left( p-1 \right)d \\
 & {{a}_{q}}=a+\left( q-1 \right)d \\
 & {{a}_{r}}=a+\left( r-1 \right)d \\
 & {{a}_{s}}=a+\left( s-1 \right)d \\
\end{align}$
Since the ${{p}^{th}},{{q}^{th}},{{r}^{th}}$ and ${{s}^{th}}$ term of the A.P are in G.P, we have
$\dfrac{{{a}_{q}}}{{{a}_{p}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}=\dfrac{{{a}_{s}}}{{{a}_{r}}}$
From the first equality, we have
$\dfrac{{{a}_{q}}}{{{a}_{p}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}$
Substituting the value of ${{a}_{p}},{{a}_{q}}$ and ${{a}_{r}}$, we get
$\dfrac{a+\left( q-1 \right)d}{a+\left( p-1 \right)d}=\dfrac{a+\left( r-1 \right)d}{a+\left( q-1 \right)d}$
We know that if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a-b}{b}=\dfrac{c-d}{d}$
Hence, we have
$\begin{align}
  & \dfrac{a+\left( q-1 \right)d-\left( a+\left( p-1 \right)d \right)}{a+\left( p-1 \right)d}=\dfrac{a+\left( r-1 \right)d-\left( a+\left( q-1 \right)d \right)}{a+\left( q-1 \right)d} \\
 & \Rightarrow \dfrac{\left( q-p \right)d}{a+\left( p-1 \right)d}=\dfrac{\left( r-q \right)d}{a+\left( q-1 \right)d} \\
 & \Rightarrow \dfrac{q-p}{a+\left( p-1 \right)d}=\dfrac{r-q}{a+\left( q-1 \right)d} \\
\end{align}$
Multiplying both sides by $\dfrac{a+\left( p-1 \right)d}{r-q}$, we get
$\begin{align}
  & \dfrac{q-p}{r-q}=\dfrac{a+\left( p-1 \right)d}{a+\left( q-1 \right)d}=\dfrac{{{a}_{p}}}{{{a}_{q}}} \\
 & \Rightarrow \dfrac{q-p}{r-q}=\dfrac{{{a}_{p}}}{{{a}_{q}}}\ \ \ \ \left( i \right) \\
\end{align}$
From second equality, we have
$\dfrac{{{a}_{s}}}{{{a}_{r}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}$
Substituting the value of ${{a}_{r}},{{a}_{q}}$ and ${{a}_{s}}$, we get
$\dfrac{a+\left( s-1 \right)d}{a+\left( r-1 \right)d}=\dfrac{a+\left( r-1 \right)d}{a+\left( q-1 \right)d}$
We know that if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a-b}{b}=\dfrac{c-d}{d}$
Hence, we have
$\begin{align}
  & \dfrac{a+\left( s-1 \right)d-\left( a+\left( r-1 \right)d \right)}{a+\left( r-1 \right)d}=\dfrac{a+\left( r-1 \right)d-\left( a+\left( q-1 \right)d \right)}{a+\left( q-1 \right)d} \\
 & \Rightarrow \dfrac{\left( s-r \right)d}{a+\left( r-1 \right)d}=\dfrac{\left( r-q \right)d}{a+\left( q-1 \right)d} \\
 & \Rightarrow \dfrac{s-r}{a+\left( r-1 \right)d}=\dfrac{r-q}{a+\left( q-1 \right)d} \\
\end{align}$
Multiplying both sides by $\dfrac{a+\left( r-1 \right)d}{r-q}$, we get
$\dfrac{s-r}{r-q}=\dfrac{a+\left( r-1 \right)d}{a+\left( q-1 \right)d}=\dfrac{{{a}_{r}}}{{{a}_{q}}}$
We know that $\dfrac{{{a}_{r}}}{{{a}_{q}}}=\dfrac{{{a}_{q}}}{{{a}_{p}}}$
Hence, we have
$\dfrac{s-r}{r-q}=\dfrac{{{a}_{q}}}{{{a}_{p}}}\ \ \ \ \left( ii \right)$
Multiplying equation (i) and equation (ii), we get
$\begin{align}
  & \dfrac{q-p}{r-q}\times \dfrac{s-r}{r-q}=\dfrac{{{a}_{q}}}{{{a}_{p}}}\times \dfrac{{{a}_{q}}}{{{a}_{p}}}=1 \\
 & \Rightarrow \left( q-p \right)\left( s-r \right)={{\left( r-q \right)}^{2}} \\
 & \Rightarrow \left( p-q \right)\left( r-s \right)={{\left( q-r \right)}^{2}} \\
\end{align}$
Hence, we have p-q, q-r and r-s are in G.P
Q.E.D

Note: [1] A common mistake done by students in these types of questions is that they cross multiply the expressions and then simplify which makes the calculations difficult and prone to mistakes. A better strategy in these types of questions is make use of the properties of proportions and simplify as done above.