Question

Prove that a,b,c are in A.P., G.P. or H.P. according as the value of $\dfrac{a-b}{b-c}$ is equal to $\dfrac{a}{a},\dfrac{a}{b},\dfrac{a}{c}$ respectively. \[\]

Answer 1

Hint: We recall the definitions of arithmetic progression (AP), Geometric progression (GP), Harmonic progression (HP). We find the standard relation of three consecutive terms $a,b,c$ using the definitions of AP, GP, HP. We equate $\dfrac{a-b}{b-c}$ one by one with $\dfrac{a}{a},\dfrac{a}{b},\dfrac{a}{c}$ and simplify to obtain relation among $a,b,c$. We compare them with the standard relation of AP, GP and HP to prove the statement. \[\]

Complete step-by-step solution:
We know that a sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
We also know that Arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP, is a type sequence where the difference between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then
\[{{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}={{x}_{4}}-{{x}_{3}}...\]
We know that a Geometric sequence otherwise known as Geometric progression, abbreviated as GP is a type sequence where the ratio between any two consecutive numbers is constant . If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an GP, then
\[\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{1}}}=\dfrac{{{x}_{4}}}{{{x}_{3}}}...\]
We know that a harmonic sequence otherwise known as Harmonic progression, abbreviated as HP is the sequence of reciprocals of terms in arithmetic sequence which means if ${{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an HP then the sequence $\dfrac{1}{{{x}_{1}}},\dfrac{1}{{{x}_{2}}},\dfrac{1}{{{x}_{3}}},...$ is in AP which means
\[\dfrac{1}{{{x}_{2}}}-\dfrac{1}{{{x}_{1}}}=\dfrac{1}{{{x}_{3}}}-\dfrac{1}{{{x}_{2}}}...\]
If $a,b,c$ are three consecutive terms in AP then we have,
\[\begin{align}
  & b-a=c \\
 & \therefore 2b=a+c........\left( 1 \right) \\
\end{align}\]
If $a,b,c$ are three consecutive terms in GP then we have,
\[\begin{align}
  & \dfrac{b}{a}=\dfrac{c}{b} \\
 & \therefore {{b}^{2}}=ac.......\left( 2 \right) \\
\end{align}\]
If $a,b,c$ are three consecutive terms in HP then we have,
\[\begin{align}
  & \dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b} \\
 & \Rightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\
 & \Rightarrow \dfrac{2}{b}=\dfrac{a+c}{ac} \\
 & \therefore b=\dfrac{2ac}{a+c}......\left( 3 \right) \\
\end{align}\]
We are asked to prove that of $\dfrac{a-b}{b-c}$ is equal to $\dfrac{a}{a},\dfrac{a}{b},\dfrac{a}{c}$ the three terms $a,b,c$are in AP, GP and HP respectively. Let us consider the case for AP. We have,
\[\dfrac{a-b}{b-c}=\dfrac{a}{a}\]
We cross multiply to have,
\[\begin{align}
  & \Rightarrow a\left( a-b \right)=a\left( b-c \right) \\
 & \Rightarrow {{a}^{2}}-ab=ab-ac \\
 & \Rightarrow a\left( a-b \right)=a\left( b-c \right) \\
\end{align}\]
We divide both sides of the above equation by $a$ assuming $a\ne 0$. We have,
\[\begin{align}
  & a-b=b-c \\
 & \therefore 2b=a+c \\
\end{align}\]
We compare the above result with (1) and hence it is proved that $a,b,c$ are in AP if $\dfrac{a-b}{b-c}=\dfrac{a}{a}$.
Let us consider the case for GP. We have,
\[\dfrac{a-b}{b-c}=\dfrac{a}{b}\]
We cross multiply to have,
\[\begin{align}
  & \Rightarrow \left( a-b \right)b=\left( b-c \right)a \\
 & \Rightarrow ab-{{b}^{2}}=ab-ac \\
 & \therefore {{b}^{2}}=ac \\
\end{align}\]
We compare the above result with (2) and hence it is proved that $a,b,c$ are in GP if $\dfrac{a-b}{b-c}=\dfrac{a}{b}$.
Let us consider the case for HP. We have,
\[\begin{align}
  & \dfrac{a-b}{b-c}=\dfrac{a}{c} \\
 & \Rightarrow \left( a-b \right)c=\left( b-c \right)a \\
 & \Rightarrow ac-bc=ab-ca \\
 & \Rightarrow b\left( a+c \right)=2ac \\
 & \therefore b=\dfrac{2ac}{a+c} \\
\end{align}\]
We compare the above result with (3) and hence it is proved that $a,b,c$ are in HP if $\dfrac{a-b}{b-c}=\dfrac{a}{c}$.\[\]

Note: We can alternatively prove by putting the expression for consecutive terms for AP $f-d,f,f+d$ where $f$ the first term is and $d$ is the common difference. We can similarly prove for GP with three for consecutive terms $\dfrac{f}{r},f,fr$ where $r$ is the common ratio and for HP with three for consecutive terms $\dfrac{1}{f-d},\dfrac{1}{f},\dfrac{1}{f+d}$.