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Question: If \({{p}^{th}},{{q}^{th}},{{r}^{th}}\) and \({{s}^{th}}\) terms of an A.P are in G.P, then prove th...

If pth,qth,rth{{p}^{th}},{{q}^{th}},{{r}^{th}} and sth{{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.

Explanation

Solution

Assume that the first term of the A.P is “a” and the common difference is “d”. Use the fact that nth{{n}^{th}}term of an A.P is given by an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d. Use the fact that if a, b, c and d are in G.P, then ab=bc=cd\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 nth{{n}^{th}} term of the A.P is given by an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d.
Hence, we have
ap=a+(p−1)d aq=a+(q−1)d ar=a+(r−1)d as=a+(s−1)d \begin{aligned} & {{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{aligned}
Since the pth,qth,rth{{p}^{th}},{{q}^{th}},{{r}^{th}} and sth{{s}^{th}} term of the A.P are in G.P, we have
aqap=araq=asar\dfrac{{{a}_{q}}}{{{a}_{p}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}=\dfrac{{{a}_{s}}}{{{a}_{r}}}
From the first equality, we have
aqap=araq\dfrac{{{a}_{q}}}{{{a}_{p}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}
Substituting the value of ap,aq{{a}_{p}},{{a}_{q}} and ar{{a}_{r}}, we get
a+(q−1)da+(p−1)d=a+(r−1)da+(q−1)d\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 ab=cd\dfrac{a}{b}=\dfrac{c}{d}, then a−bb=c−dd\dfrac{a-b}{b}=\dfrac{c-d}{d}
Hence, we have
a+(q−1)d−(a+(p−1)d)a+(p−1)d=a+(r−1)d−(a+(q−1)d)a+(q−1)d ⇒(q−p)da+(p−1)d=(r−q)da+(q−1)d ⇒q−pa+(p−1)d=r−qa+(q−1)d \begin{aligned} & \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{aligned}
Multiplying both sides by a+(p−1)dr−q\dfrac{a+\left( p-1 \right)d}{r-q}, we get
q−pr−q=a+(p−1)da+(q−1)d=apaq ⇒q−pr−q=apaq    (i) \begin{aligned} & \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{aligned}
From second equality, we have
asar=araq\dfrac{{{a}_{s}}}{{{a}_{r}}}=\dfrac{{{a}_{r}}}{{{a}_{q}}}
Substituting the value of ar,aq{{a}_{r}},{{a}_{q}} and as{{a}_{s}}, we get
a+(s−1)da+(r−1)d=a+(r−1)da+(q−1)d\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 ab=cd\dfrac{a}{b}=\dfrac{c}{d}, then a−bb=c−dd\dfrac{a-b}{b}=\dfrac{c-d}{d}
Hence, we have
a+(s−1)d−(a+(r−1)d)a+(r−1)d=a+(r−1)d−(a+(q−1)d)a+(q−1)d ⇒(s−r)da+(r−1)d=(r−q)da+(q−1)d ⇒s−ra+(r−1)d=r−qa+(q−1)d \begin{aligned} & \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{aligned}
Multiplying both sides by a+(r−1)dr−q\dfrac{a+\left( r-1 \right)d}{r-q}, we get
s−rr−q=a+(r−1)da+(q−1)d=araq\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 araq=aqap\dfrac{{{a}_{r}}}{{{a}_{q}}}=\dfrac{{{a}_{q}}}{{{a}_{p}}}
Hence, we have
s−rr−q=aqap    (ii)\dfrac{s-r}{r-q}=\dfrac{{{a}_{q}}}{{{a}_{p}}}\ \ \ \ \left( ii \right)
Multiplying equation (i) and equation (ii), we get
q−pr−q×s−rr−q=aqap×aqap=1 ⇒(q−p)(s−r)=(r−q)2 ⇒(p−q)(r−s)=(q−r)2 \begin{aligned} & \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{aligned}
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.