If a, b, c are in H.P., then which one of the following is true? (A) $\dfrac{1}{\text{b}-\text{a}}+\dfrac{1}{\text{b}-\text{c}}=\dfrac{1}{\text{b}}$ (B) $\dfrac{\text{ac}}{\text{a}+\text{c}}=\text{b}$ (C) $\dfrac{\text{b}+\text{a}}{\text{b}-\text{a}}+\dfrac{\text{b}+\text{c}}{\text{b}-\text{c}}=1$ (D) None of these.
Answer
Verified
Hint: Here, we know that, H.P is the opposite of A.P. then, a, b, c are in H.P, $\dfrac{2}{\text{b}}=\dfrac{1}{\text{a}}+\dfrac{1}{\text{c}}\Rightarrow \dfrac{2\,\text{ac}}{\text{a}+\text{c}}$ and check all option which will be true.
Complete step by step solution: Given, a, b, c are in H.P. ⇒ $\dfrac{1}{\text{b}}-\dfrac{1}{\text{a}}=\dfrac{1}{\text{c}}-\dfrac{1}{\text{b}}$ ⇒ $\dfrac{1}{\text{b}}+\dfrac{1}{\text{b}}=\dfrac{1}{\text{a}}+\dfrac{1}{\text{c}}$ ⇒ After addition, we get ⇒ $\dfrac{2}{\text{b}}=\dfrac{\text{a}+\text{c}}{\text{ac}}$ After cross-multiplication, we get, ⇒ $\dfrac{\text{b}}{2}=\dfrac{\text{ac}}{\text{a}+\text{c}}$ ⇒ $\text{b}=\dfrac{\text{2 ac}}{\text{a}+\text{c}}$ Now, option (1) ⇒ $\dfrac{1}{\text{b}-\text{a}}+\dfrac{1}{\text{b}-\text{c}}=\dfrac{1}{\text{b}}$ ⇒ After adding, we get ⇒ $\dfrac{\text{b}-\text{c}+\text{b}-\text{a}}{\text{(b}-\text{c)(b}-\text{a)}}=\dfrac{1}{\text{b}}$ \[\Rightarrow (2\text{b}-\text{a}-\text{c})\,\text{b}={{\text{b}}^{\text{2}}}\,-\,\text{ab}\,-\,\text{cb}\,+\,\text{ac}\] \[\Rightarrow 2{{\text{b}}^{2}}-\text{ab}-\text{bc}={{\text{b}}^{\text{2}}}-\text{ab}-{{\text{b}}^{2}}\text{c}+\text{ac}\] \[\Rightarrow {{\text{b}}^{2}}=\text{ac}\] That is wrong. Now, option (2) $\Rightarrow \dfrac{\text{ac}}{\text{a}+\text{c}}=\text{b}$ That is wrong, Now, option (3). $\Rightarrow \dfrac{\text{b}+\text{a}}{\text{b}-\text{a}}+\dfrac{\text{b}+\text{c}}{\text{b}-\text{c}}=1$. ⇒ After cross-multiplication, we get, \[\Rightarrow (\text{b}+\text{a})(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{a})=(\text{b}-\text{a})(\text{b}-\text{c})\] Now, multiplying \[\Rightarrow {{\text{b}}^{\text{2}}}-\text{bc}+\text{ab}-\text{ac}+{{\text{b}}^{\text{2}}}-\text{ab}+\text{bc}-\text{ac}={{\text{b}}^{\text{2}}}-\text{ab}-\text{bc}+\text{ac}\] \[\Rightarrow {{\text{b}}^{\text{2}}}-2\text{ac}=-\text{ab}-\text{bc}+\text{ac}\] \[\Rightarrow {{\text{b}}^{\text{2}}}=\dfrac{-\text{ab}-\text{bc}+\text{ac}}{2\text{ac}}\] 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 = $\dfrac{1}{\text{b}}-\dfrac{1}{\text{a}}=\dfrac{1}{\text{c}}-\dfrac{1}{\text{b}}$ ,where, a,b.c and d is the number of a series.then, check all option.
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