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The sum of three numbers in HP is 37 and the sum of their reciprocals is `(1)/(4)`. Find the numbers. |
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Answer» Three numbers in HP can be taken as ` (1)/(a-d),(1)/(a),(1)/(a+d)`. Then, ` (1)/(a-d)+(1)/(a)+(1)/(a+d)=37 "…….(i)"` and `a-d+a+a+d=(1)/(4)` `therefore " " a=(1)/(12)` From Eq.(i), `(12)/(1-12d)+12+(12)/(1+12d)=37` `implies (12)/(1-12d)+(12)/(1+12d)=25 implies (24)/(1-144d^(2))=25` `1-144d^(2)=(24)/(25)` or `d^(2)=(1)/(25xx144)` `therefore " " d= pm (1)/(60)` `therefore a-d,a,a+d, " are " (1)/(15),(1)/(12),(1)/(10), " or " (1)/(10),(1)/(12),(1)/(15).` Hence, three numbers in HP are `15,12,10, " or " 10,12,15`. |
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