====== Question 8 Exercise 4.2 ======
Solutions of Question 8 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 8=====
If $\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P, then prove $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in A.P. GOOD
====Solution====
Since
$\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P, thus\\
\begin{align}\therefore \dfrac{c+a-b}{b}-\dfrac{b+c-a}{a}&=\dfrac{a+b-c}{c}-\dfrac{c+a-b}{b} \\
\text{Let}\quad S&=\dfrac{a+b+c}{2} \\
\Rightarrow a+b+c&=2 S\\
\text{then}
\Rightarrow a+b-c&=2(S-c) \text {, }\\
a+c-b&=2(S-b), \quad\text{and}\\
b+c-a&=2(S-a)\end{align}
then (1), becomes\\
\begin{align}\dfrac{2(S-b)}{b}-\dfrac{2(S-a)}{a}&=\dfrac{2(S-c)}{c} -\dfrac{2(S-b)}{b}\end{align}
Dividing both sides by $2$\\
\begin{align}\dfrac{S-b}{b}-\dfrac{S-a}{a}&=\dfrac{S-c}{c}-\dfrac{S-b}{b} \\
\dfrac{a(S-b)-b(S-a)}{a b}&=\dfrac{b(S-c)-c(S-b)}{b c} \\
\dfrac{a S-a b-b S+a b}{a b}&=\dfrac{b S-b c-c S+b c}{b c} \\
\Rightarrow \dfrac{(a-b) S}{a b}&=\dfrac{(b-c) S}{b c}\end{align}
Dividing both sides by $S$\\
\begin{align}\Rightarrow \dfrac{a-b}{a b}&=\dfrac{b-c}{b c} \\
\Rightarrow \dfrac{a}{a b}-\dfrac{b}{a b}&=\dfrac{b}{b c}-\dfrac{c}{b c} \\
\Rightarrow \dfrac{1}{b}-\dfrac{1}{a}&=\dfrac{1}{c}-\dfrac{1}{b} \\
\Rightarrow\dfrac{1}{a}, \dfrac{1}{b},& \dfrac{1}{c} \text { are in A.P. }\end{align}


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