The sum of three numbers in HP is 37 and the sum of their reciprocals is `(1)/(4)`. Find the numbers.

Explanation

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`.