Question

If \[a,\,\,b,\,\,c\] are in A.P and \[a,\,\,b,\,\,d\] are in G.P \[~a,\text{ }a\text{ }-\text{ }b,\text{ }d\text{ }-\text{ }c~\] will be in
A.A.P
B.G.P
C.H.P
D.None of these

Answer 1

Hint: We are provided some words of Arithmetic and some terms of Geometric progression in the supplied question, and we must determine what sort of series they belong to or any type of progression for which we must utilize the definitions of each term.

Complete step-by-step answer:
According to the question, we are given that \[a,\,\,b,\,\,c\] are in A.P and we know in any Arithmetic progression the common difference between consecutive terms is the same. Therefore, we can write as
 \[b-a=c-b\]
By rearranging the term we get:
 \[2b=c+a--(1)\]
Multiply equation (1) by a
 \[2ba=ac+{{a}^{2}}\]
By again rearranging the term we get:
 \[-ac={{a}^{2}}-2ab\]
Now, add \[{{b}^{2}}\] on both sides
 \[{{b}^{2}}-ac={{a}^{2}}-2ab+{{b}^{2}}\] .
By using the property of \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get:
 \[{{b}^{2}}-ac={{(a-b)}^{2}}---(2)\]
Now, we know that in G.P the common ratio between the consecutive terms is same therefore, we get:
 \[\dfrac{b}{a}=\dfrac{d}{b}\]
By cross multiply we get:
 \[{{b}^{2}}=ad--(3)\] .
By substituting the equation (3) on equation (2) we get:
 \[ad-ac={{(a-b)}^{2}}\]
By taking common \[a\] on LHS we get:
 \[a(d-c)={{(a-b)}^{2}}\]
Above equation can also be written in the form of \[\dfrac{(a-b)}{a}=\dfrac{(d-c)}{(a-b)}\]
If you observe this above equation you can see that the above equation in the form of G.P
Therefore, \[~a,\text{ }a\text{ }-\text{ }b,\text{ }d\text{ }-\text{ }c~\] are in G.P
So, the correct answer is “Option B”.

Note: When answering such a question, we must be careful to define each sort of progression series involved. Also, we must be aware of the key terminology involved and move immediately as directed by the definition, always attempting to provide the simplest response and working step by step to minimise the risk of making a mistake.