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Question: If a, b, c are in H.P., then which one of the following is true? (A) \(\dfrac{1}{\text{b}-\text{a}...

If a, b, c are in H.P., then which one of the following is true?
(A) 1b−a+1b−c=1b\dfrac{1}{\text{b}-\text{a}}+\dfrac{1}{\text{b}-\text{c}}=\dfrac{1}{\text{b}}
(B) aca+c=b\dfrac{\text{ac}}{\text{a}+\text{c}}=\text{b}
(C) b+ab−a+b+cb−c=1\dfrac{\text{b}+\text{a}}{\text{b}-\text{a}}+\dfrac{\text{b}+\text{c}}{\text{b}-\text{c}}=1
(D) None of these.

Explanation

Solution

Here, we know that, H.P is the opposite of A.P. then,
a, b, c are in H.P,
2b=1a+1c⇒2 aca+c\dfrac{2}{\text{b}}=\dfrac{1}{\text{a}}+\dfrac{1}{\text{c}}\Rightarrow \dfrac{2\,\text{ac}}{\text{a}+\text{c}} and check all option which will be true.

Complete step by step solution: Given,
a, b, c are in H.P.
⇒ 1b−1a=1c−1b\dfrac{1}{\text{b}}-\dfrac{1}{\text{a}}=\dfrac{1}{\text{c}}-\dfrac{1}{\text{b}}
⇒ 1b+1b=1a+1c\dfrac{1}{\text{b}}+\dfrac{1}{\text{b}}=\dfrac{1}{\text{a}}+\dfrac{1}{\text{c}}
⇒ After addition, we get
⇒ 2b=a+cac\dfrac{2}{\text{b}}=\dfrac{\text{a}+\text{c}}{\text{ac}}
After cross-multiplication, we get,
⇒ b2=aca+c\dfrac{\text{b}}{2}=\dfrac{\text{ac}}{\text{a}+\text{c}}
⇒ b=2 aca+c\text{b}=\dfrac{\text{2 ac}}{\text{a}+\text{c}}
Now, option (1)
⇒ 1b−a+1b−c=1b\dfrac{1}{\text{b}-\text{a}}+\dfrac{1}{\text{b}-\text{c}}=\dfrac{1}{\text{b}}
⇒ After adding, we get
⇒ b−c+b−a(b−c)(b−a)=1b\dfrac{\text{b}-\text{c}+\text{b}-\text{a}}{\text{(b}-\text{c)(b}-\text{a)}}=\dfrac{1}{\text{b}}
⇒(2b−a−c) b=b2 − ab − cb + ac\Rightarrow (2\text{b}-\text{a}-\text{c})\,\text{b}={{\text{b}}^{\text{2}}}\,-\,\text{ab}\,-\,\text{cb}\,+\,\text{ac}
⇒2b2−ab−bc=b2−ab−b2c+ac\Rightarrow 2{{\text{b}}^{2}}-\text{ab}-\text{bc}={{\text{b}}^{\text{2}}}-\text{ab}-{{\text{b}}^{2}}\text{c}+\text{ac}
⇒b2=ac\Rightarrow {{\text{b}}^{2}}=\text{ac}
That is wrong.
Now, option (2)
⇒aca+c=b\Rightarrow \dfrac{\text{ac}}{\text{a}+\text{c}}=\text{b}
That is wrong,
Now, option (3).
⇒b+ab−a+b+cb−c=1\Rightarrow \dfrac{\text{b}+\text{a}}{\text{b}-\text{a}}+\dfrac{\text{b}+\text{c}}{\text{b}-\text{c}}=1.
⇒ After cross-multiplication, we get,
⇒(b+a)(b−c)+(b+c)(b−a)=(b−a)(b−c)\Rightarrow (\text{b}+\text{a})(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{a})=(\text{b}-\text{a})(\text{b}-\text{c})
Now, multiplying
⇒b2−bc+ab−ac+b2−ab+bc−ac=b2−ab−bc+ac\Rightarrow {{\text{b}}^{\text{2}}}-\text{bc}+\text{ab}-\text{ac}+{{\text{b}}^{\text{2}}}-\text{ab}+\text{bc}-\text{ac}={{\text{b}}^{\text{2}}}-\text{ab}-\text{bc}+\text{ac}
⇒b2−2ac=−ab−bc+ac\Rightarrow {{\text{b}}^{\text{2}}}-2\text{ac}=-\text{ab}-\text{bc}+\text{ac}
⇒b2=−ab−bc+ac2ac\Rightarrow {{\text{b}}^{\text{2}}}=\dfrac{-\text{ab}-\text{bc}+\text{ac}}{2\text{ac}}
That is False.
Hence, the correct answer is none of the above.

Note: The above question is of arithmetic progression in which A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always same and the h.p is a harmonic progression is a progression formed by taking the reciprocals of an arithmetic progression.
Here, H.P = 1b−1a=1c−1b\dfrac{1}{\text{b}}-\dfrac{1}{\text{a}}=\dfrac{1}{\text{c}}-\dfrac{1}{\text{b}}
,where, a,b.c and d is the number of a series.then, check all option.