Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $H_1,H_2,\ldots,H_n$ are $n$ harmonic means between $a$ and $b$, $a\ne b$, then the value of
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b}$
is equal to
Solution
Step 1: Convert HP to APSince $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.
Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$
Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$
$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$
Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$
Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $
$= \dfrac{a+b(2n+1)}{b-a} \quad \cdots(1)$
Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$
Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $
Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$
Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $
$= \dfrac{b+a(2n+1)}{a-b} \quad \cdots(2)$
Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $
Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $
$= \dfrac{a+b(2n+1)}{b-a} + \dfrac{b+a(2n+1)}{a-b}$
$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$
$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$
$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$
$= (2n+1) - 1$
$= 2n$
Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$
$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$
$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$
$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$
$= (2n+1) - 1$
$= 2n$
Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
