If a, b, c are in HP, then what is (1)/(b-a)+(1)/(b-c) equal to? 1. (2)/(b) 2. (1)/(a)+(1)/(c) 3. (1)/(2) ((1)/(a)+(1)/(b)+(1)/(c)) Select the correct answer using the code given below:

Correct answer: B. 2 only a, b, c are in H.P. (2)/(b)=(1)/(a)+(1)/(c) aligned & b=(2 a c)/(a+c) & b-a=(2 a c)/(a+c)-a=(2 a c-a^2-a c)/(a+c) & =(a c-a^2)/(a+c)=(a(c-a))/(a+c) & (1)/(b-a)=(a+c)/(-a(a-c)) & Now, b-c=(2 a c)/(a+c)-c=(2 a c-a c-c^2)/(a+c) & =(a c-c^2)/(a+c)=(c(a-c))/(a+c) & (1)/(b-c)=(a+c)/(c(a-c)) & (1)/(b-a)+(1)/(b-c)=(a+c)/(-a(a-c))+(a+c)/(c(a-c)) & =((a+c))/((a-c))[-(1)/(a)+(1)/(c)] & =(a+c)/(a c)=(1)/(a)+(1)/(c) & =(2)/(b)=(1)/(b)+(1)/(b)=(1)/(b)+(1)/(2)((1)/(a)+(1)/(c)) aligned Hence, only statement (1) and (2) are correct. Statement (3) is not correct.

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Sequences and Series

NDA 2024 · Mathematics · Sequences and Series

If a, b, c are in HP, then what is (1)/(b-a)+(1)/(b-c) equal to? 1....

a, b, c

are in HP, then what is

\frac{1}{b-a}+\frac{1}{b-c} equal to? 1. \frac{2}{b} 2. \frac{1}{a}+\frac{1}{c} 3. \frac{1}{2}

\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)

Select the correct answer using the code given below:

Options

A1 only
B2 only
C3 only
D1,2 and 3

Correct answer

B. 2 only

Step-by-step solution

\because a, b, c

are in H.P.

\Rightarrow \quad \frac{2}{b}=\frac{1}{a}+\frac{1}{c}

\begin{aligned} & \Rightarrow \quad b=\frac{2 a c}{a+c} \\ & b-a=\frac{2 a c}{a+c}-a=\frac{2 a c-a^2-a c}{a+c} \\ & =\frac{a c-a^2}{a+c}=\frac{a(c-a)}{a+c} \\ & \Rightarrow \quad \frac{1}{b-a}=\frac{a+c}{-a(a-c)} \\ & \text{ Now, } \quad b-c=\frac{2 a c}{a+c}-c=\frac{2 a c-a c-c^2}{a+c} \\ & =\frac{a c-c^2}{a+c}=\frac{c(a-c)}{a+c} \\ & \Rightarrow \quad \frac{1}{b-c}=\frac{a+c}{c(a-c)} \\ & \therefore \quad \frac{1}{b-a}+\frac{1}{b-c}=\frac{a+c}{-a(a-c)}+\frac{a+c}{c(a-c)} \\ & =\frac{(a+c)}{(a-c)}\left[-\frac{1}{a}+\frac{1}{c}\right] \\ & =\frac{a+c}{a c}=\frac{1}{a}+\frac{1}{c} \\ & =\frac{2}{b}=\frac{1}{b}+\frac{1}{b}=\frac{1}{b}+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right) \end{aligned}

$$ Hence, only statement (1) and (2) are correct. Statement (3) is not correct.

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