Question

if a,b,c are distinct positive real in H.P then the value of the expression b+a/b-a + b+c/b-c is

Answer 1

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Answer 2

Answer: 2

Step-by-step explanation:

• If a,b and c are in H.P. then b=$\frac{2ac}{a+c}$
• So, b+a=$\frac{2ac}{a+c}+a$

                    =$\frac{a^{2}+ 3ac}{a+c}$

       Similarly,b-a=$\frac{-a^{2}+ ac}{a+c}$

       And, b+c=$\frac{2ac}{a+c}+c$

                      =$\frac{c^{2}+ 3ac}{a+c}$

               b-c=$\frac{-c^{2}+ ac}{a+c}$

• Now,putting the values of $\frac{b+a}{b-a} +\frac{b+c}{b-c}$

                 $=\frac{\frac{a^{2}+3ac}{a+c} }{\frac{-a^{2}+ac }{a+c} } +\frac{\frac{c^{2}+3ac}{a+c} }{\frac{-c^{2}+ac }{a+c} }$

                $={\frac{a^{2}+3ac}{-a^{2}+ac } +{\frac{c^{2}+3ac}{-c^{2}+ac }$

                $=\frac{a(a+3c)}{a(c-a)}+\frac{c(c+3a)}{c(a-c)}$

               $=\frac{(a+3c)}{(c-a)}-\frac{(c+3a)}{(c-a)}$

               $=\frac{a+3c-c-3a}{(c-a)}$

               $=\frac{2(c-a)}{(c-a)}$

              $=2$

• Hence,on solving the above problem the result obtained is 2.

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