Question 1 plz fast..../


Q1. If a , b, c are in AP and b , c, d be in HP, then 

       (a) ab = cd                    (b) ad = bc                        (c) ac = bd         
                  
 

 

Dear Student,

The solution is as under:

$$\begin{gathered} \sin ce, a,b and c are in AP, thus, \\ 2b=a+c.....\left(i\right) \left\{arithmetic mean\right\} \\ and, b,c and d are in HP, thus, \\ \frac{2}{c}=\frac{1}{b}+\frac{1}{d}.....\left(ii\right) \left\{harmonic mean\right\} \\ or, \frac{1}{b}=\frac{2}{c}-\frac{1}{d} \\ putting the value of b from \left(i\right), we get, \\ \frac{2}{a+c}=\frac{2}{c}-\frac{1}{d} \\ or,\frac{2}{a+c}=\frac{2d-c}{dc} \\ or, 2dc=\left(a+c\right)\left(2d-c\right) \\ or, \cancel{2dc}=2ad-ac+\cancel{2dc}-c^2 \\ or, 2ad=c^2+ac \\ or, 2ad=c\left(c+a\right) \\ from \left(i\right), as a+c=2b \\ or, 2ad=c2b \\ or, 2ad=2bc \\ or, ad=bc \end{gathered}$$

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