Question

If \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of an A.P. are in G.P and \[m\], \[n\], \[r\]are in H.P. Then find the common difference of A.P.
A. \[ - \dfrac{a}{n}\]
B. \[ - \dfrac{n}{{2a}}\]
C. \[\dfrac{{2a}}{n}\]
D. \[ - \dfrac{{2a}}{n}\]

Answer 1

Hint: First we will assume the first term and common difference of the A.P. Since \[m\], \[n\], \[r\]are in H.P, so apply the condition of AP on \[\dfrac{1}{m}\], \[\dfrac{1}{n}\], \[\dfrac{1}{r}\] to find the relation between \[m\], \[n\], \[r\]. Apply \[{n^{th}}\] term of A.P to calculate \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of the A.P. Since \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of the A.P are in G.P, equate the ratio of \[{\left( {n + 1} \right)^{th}}\] term and \[{\left( {m + 1} \right)^{th}}\] term to the ratio of \[{\left( {r + 1} \right)^{th}}\] term and \[{\left( {n + 1} \right)^{th}}\]. Solve the equation to calculate the common difference.

Formula used
Condition of A.P: If \[a,b,c\] are in A.P, then \[2b = a + c\].
Condition of G.P: If \[a,b,c\] are in G.P, then \[\dfrac{b}{a} = \dfrac{b}{c}\].
\[{n^{th}}\] term of an A.P is \[{t_n} = a + \left( {n - 1} \right)d\], \[a\] is the first term of the A.P. and \[d\] is the common difference.

Complete step by step solution:
Given that, \[m\], \[n\], \[r\]are in H.P, then \[\dfrac{1}{m},\dfrac{1}{n},\dfrac{1}{r}\] are in AP.
Apply the condition A.P on \[\dfrac{1}{m},\dfrac{1}{n},\dfrac{1}{r}\]
\[\dfrac{2}{n} = \dfrac{1}{r} + \dfrac{1}{m}\]
\[ \Rightarrow \dfrac{2}{n} = \dfrac{{r + m}}{{rm}}\]
Multiply both sides by \[rm\].
\[ \Rightarrow \dfrac{{2rm}}{n} = r + m\] ………(i)
Let \[a\] be the first term of the A.P. and \[d\] be the common difference.
Now apply \[{n^{th}}\] term formula of AP to calculate \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of the A.P.
The \[{\left( {m + 1} \right)^{th}}\] term of the A.P is \[{t_{m + 1}} = a + \left( {m + 1 - 1} \right)d\]
\[ = a + md\]
The \[{\left( {n + 1} \right)^{th}}\] term of the A.P is \[{t_{n + 1}} = a + \left( {n + 1 - 1} \right)d\]
 \[ = a + nd\]
The \[{\left( {r + 1} \right)^{th}}\] term of the A.P is \[{t_{r + 1}} = a + \left( {r + 1 - 1} \right)d\]
\[ = a + rd\]
Given that \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of the A.P. are in G.P.
Apply the condition of G.P. on \[{\left( {m + 1} \right)^{th}}\], \[{\left( {n + 1} \right)^{th}}\], and \[{\left( {r + 1} \right)^{th}}\] terms of the A.P.
\[\dfrac{{a + nd}}{{a + md}} = \dfrac{{a + rd}}{{a + nd}}\]
Apply cross multiplication
\[ \Rightarrow {\left( {a + nd} \right)^2} = \left( {a + md} \right)\left( {a + rd} \right)\]
\[ \Rightarrow {a^2} + 2and + {n^2}{d^2} = {a^2} + ard + amd + mr{d^2}\]
Cancel out \[{a^2}\] from both sides of equation.
\[ \Rightarrow 2and + {n^2}{d^2} - ard - amd - mr{d^2} = 0\]
\[ \Rightarrow 2and + {n^2}{d^2} - ad\left( {r + m} \right) - mr{d^2} = 0\]
Now putting \[r + m = \dfrac{{2rm}}{n}\]
\[ \Rightarrow 2and + {n^2}{d^2} - ad \cdot \dfrac{{2rm}}{n} - mr{d^2} = 0\]
\[ \Rightarrow {n^2}\left( {\dfrac{{2ad}}{n} + {d^2}} \right) - \dfrac{{2adrm}}{n} - mr{d^2} = 0\]
\[ \Rightarrow {n^2}\left( {\dfrac{{2ad}}{n} + {d^2}} \right) - rm\left( {\dfrac{{2ad}}{n} + {d^2}} \right) = 0\]
Take common \[\left( {\dfrac{{2ad}}{n} + {d^2}} \right)\]
\[ \Rightarrow \left( {\dfrac{{2ad}}{n} + {d^2}} \right)\left( {{n^2} - rm} \right) = 0\]
Equate each factor with zero
\[\left( {\dfrac{{2ad}}{n} + {d^2}} \right) = 0\] \[\left( {{n^2} - rm} \right) = 0\]
\[ \Rightarrow d\left( {\dfrac{{2a}}{n} + d} \right) = 0\]
Either \[ \Rightarrow d = 0\] or, \[2a = - dn\]
 \[ \Rightarrow - \dfrac{{2a}}{n} = d\]
It is not possible \[d = 0\]. Thus \[ - \dfrac{{2a}}{n} = d\].
Hence option D is the correct option.

Note: Sometimes students take \[d = 0\]. But it is the wrong solution. If \[d = 0\], then the AP is a GP and as well as HP. So, the question becomes invalid. Thus, the correct answer is \[ - \dfrac{{2a}}{n} = d\].