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If `a,b,c` are in HP,b,c,d are in GP and `c,d,e` are in AP, than show that `e=(ab^(2))/(2a-b)^(2)`. |
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Answer» Given, `a,b,c` are in HP. `therefore b=(2ac)/(a+c)" or "c=(ab)/(2a-b) "……..(i)"` Given,` b,c,d` are in GP. `therefore " " c^(2)=bd " " "……..(ii)"` and given, `c,d,e` are in AP. `therefore " " d=(c+e)/(2)` `implies e=2d-c` `e=((2c^(2))/(b)-c) " " [" from Eq. (ii) "] "........(iii)"` From Eqs. (i) and (iii), `e=(2)/(b)((ab)/(2a-b))^(2)-((ab)/(2a-b))` `=(ab)/(2a-b)^(2){(2a-(2a-b)}` `=(ab^(2))/(2a-b)^(2)` |
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