Question

If b – c, 2b – x and b – a are in H.P. then \[a - \dfrac{x}{2},b - \dfrac{x}{2},c - \dfrac{x}{2}\] are in which progression?

Answer 1

Hint: Here we have three terms which are in H.P. Hence, their reciprocals are in A.P., by using mean property of three consecutive terms of A.P., find x in terms of a, b, c. Then consider the asked terms in G.P. and apply the geometric mean property. Verify whether the value x obtained in A.P. satisfies the geometric terms or not. If satisfied, the asked terms are in G.P.

Complete step-by-step answer:
Given that b – c, 2b – x and b – a are in H.P. therefore, $\dfrac{1}{{b - c}},\dfrac{1}{{2b - x}},\dfrac{1}{{b - a}}$ are in A.P.
$\dfrac{1}{{2b - x}} - \dfrac{1}{{b - c}} = \dfrac{1}{{b - a}} - \dfrac{1}{{2b - x}}$ [In A.P. difference between second term and first term is equal to difference between third term and second term]
$ \Rightarrow \dfrac{1}{{2b - x}} + \dfrac{1}{{2b - x}} = \dfrac{1}{{b - c}} + \dfrac{1}{{b - a}}$
On simplifying
$ \Rightarrow \dfrac{2}{{2b - x}} = \dfrac{{\left( {b - a} \right) + \left( {b - c} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}}$
Cross multiplying
$ \Rightarrow 2b - x = \dfrac{{2\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}$
Rearranging the terms
$ \Rightarrow x = 2b - \dfrac{{2\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}$ …(i)
Let us consider \[a - \dfrac{x}{2},b - \dfrac{x}{2},c - \dfrac{x}{2}\] are in G.P.
Then \[{\left( {b - \dfrac{x}{2}} \right)^2} = \left( {a - \dfrac{x}{2}} \right)\left( {c - \dfrac{x}{2}} \right)\]
Now, L.H.S. = \[{\left( {b - \dfrac{x}{2}} \right)^2}\]
Putting value of x from equation (i)
\[ = {\left( {b - \dfrac{{2b - \dfrac{{2\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}}}{2}} \right)^2}\]
\[ = {\left( {b - b - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)^2}\]
On simplifying
\[ = {\left( { - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)^2}\]
\[ = \dfrac{{{{\left( {b - c} \right)}^2}{{\left( {b - a} \right)}^2}}}{{{{\left( {(b - a) + (b - c)} \right)}^2}}}\]
And R.H.S. = \[\left( {a - \dfrac{x}{2}} \right)\left( {c - \dfrac{x}{2}} \right)\]
Putting value of x from equation (i)
\[ = \left( {a - \dfrac{{2b - \dfrac{{2\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}}}{2}} \right)\left( {c - \dfrac{{2b - \dfrac{{2\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}}}{2}} \right)\]
Simplifying
\[ = \left( {a - b - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\left( {c - b - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\]
\[ = \left( { - (b - a) - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\left( { - (b - c) - \dfrac{{\left( {b - c} \right)\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\]
Taking (−(b – a)) and (−(b – c)) common from each term
\[ = ( - (b - a))( - (b - c))\left( {1 - \dfrac{{\left( {b - c} \right)}}{{(b - a) + (b - c)}}} \right)\left( {1 - \dfrac{{\left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\]
\[ = (b - a)(b - c)\left( {\dfrac{{(b - a) + (b - c) - \left( {b - c} \right)}}{{(b - a) + (b - c)}}} \right)\left( {\dfrac{{(b - a) + (b - c) - \left( {b - a} \right)}}{{(b - a) + (b - c)}}} \right)\]
On simplifying
\[ = (b - a)(b - c)\left( {\dfrac{{(b - a)}}{{(b - a) + (b - c)}}} \right)\left( {\dfrac{{(b - c)}}{{(b - a) + (b - c)}}} \right)\]
Rearranging the terms
\[ = \dfrac{{{{\left( {b - c} \right)}^2}{{\left( {b - a} \right)}^2}}}{{{{\left( {(b - a) + (b - c)} \right)}^2}}} = {\left( {b - \dfrac{x}{2}} \right)^2}\]
L.H.S. = R.H.S.

Hence, \[a - \dfrac{x}{2},b - \dfrac{x}{2},c - \dfrac{x}{2}\] are in G.P.

Note: In these types of questions, always use hit and trial method i.e., first consider the asked terms are in A.P., then G.P., then H.P. you can choose by your own choice but conditions must be fulfilled by the terms. Try to satisfy the same value for A.P and G.P. in this question, which is the easiest method.
In this question you can directly use the mean property of given H.P without changing it into A.P. and can find the value of x in terms of a, b, c.