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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
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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.
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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