Question

 If \[\dfrac{{\left( {a + bx} \right)}}{{\left( {a - bx} \right)}} = \dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}} = \dfrac{{\left( {c + dx} \right)}}{{\left( {c - dx} \right)}}\] \[\left( {x \ne 0} \right)\] ,then which of the following is true for a,b,c,d?
A. AP
B. GP
C. HP
D. None Of These

Answer 1

Hint:
First we will take the first two ratios from the given ratios and simplify to find the relation between a, b, and c. In the same manner, we will take the last two ratios and find the relation between b,c,d.



Complete step-by-step answer:
Given \[\dfrac{{\left( {a + bx} \right)}}{{\left( {a - bx} \right)}} = \dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}} = \dfrac{{\left( {c + dx} \right)}}{{\left( {c - dx} \right)}}\] where \[\left( {x \ne 0} \right)\]
Simplifying \[\dfrac{{\left( {a + bx} \right)}}{{\left( {a - bx} \right)}} = \dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}}\] we will get,
\[\dfrac{{\left( {a + bx} \right)}}{{\left( {a - bx} \right)}} = \dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}}\]
After cross multiplying,
\[\left( {a + bx} \right)\left( {b - cx} \right) = \left( {b + cx} \right)\left( {a - bx} \right)\]
\[ab + {b^2}x - acx - bc{x^2} = ab + acx - {b^2}x - bc{x^2}\]
\[2{b^2}x = 2acx\]
\[{b^2} = ac\]
\[\dfrac{b}{a} = \dfrac{c}{b}\] ----- (i)
After cross multiplying we have successfully calculated the simplest equation which is equation (i).
Now,
Simplifying \[\dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}} = \dfrac{{\left( {c + dx} \right)}}{{\left( {c - dx} \right)}}\] we will get,
\[\dfrac{{\left( {b + cx} \right)}}{{\left( {b - cx} \right)}} = \dfrac{{\left( {c + dx} \right)}}{{\left( {c - dx} \right)}}\]
\[bc + {c^2}x - bdx - cd{x^2} = bc + bdx - {c^2}x - cd{x^2}\]
\[2{c^2}x = 2bdx\]
\[{c^2} = bd\]
\[\dfrac{c}{b} = \dfrac{d}{c}\]----- (ii)
From equation (i) and (ii) we obtain,
\[\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}\]
 Hence, we can say that \[a,b,c,d\] are in G.P
Hence the answer is (B) which is GP .



Note:
Students often confused with the conditions of AP and GP. If a, b, and c are in A.P., then 2b = a+c. If a, b, and c are in G.P., then \[b^{2}=ac\].