Question

Three numbers form a H.P. The sum of the numbers is 11 and the sum
of the reciprocals is one. Find the numbers.

Answer 1

let three number are a,b,c 
a+b+c = 11 --------------(1)
1/a + 1/b + 1/c = 1 --------(2)
a,b,c are in H.P. then 1/a,1/b,1/c are in A.P.
1/a + 1/c = 2/b
put in equation (2)
1/b + 2/b = 1
3/b = 1
b = 3
a+c = 8  ----------------(3)
1/a + 1/c = 2/3
(a+c)/ac = 2/3
ac = 12
(a-c)² = (a+c)² - 4ac
         = 64 - 48
         = 16
a-c = 4  ---------------(4)
(3) + (4)
2a = 12
a = 6
c = 2

Answer 2

Given : Three numbers form a H.P. the sum of the numbers is 11 and the sum of their reciprocals is 1

To Find : Find the numbers.

Solution:

Let say three numbers  are

1/a  , 1/b . 1/c

Three numbers form a H.P.

=> a   , b & c are in ap

=> 2b  = a + c

1/a + 1/b  + 1/c  = 11

sum of their reciprocals is 1

=> a + b  + c  =  1

=> 2b + b = 1

=> 3b  = 1

=> 3 = 1/b

1/a  +  3  + 1/c  = 11

=> 1/a  + 1/c = 8

=> c + a  = 8ac

=> 2b = 8ac

=> 2/3 = 8ac

=> 1/12 = ac

a + c  =  2/3

=> x²  - (2/3)x  + 1/12 = 0

=> 12x² - 8x  + 1  = 0

=> 12x² - 6x - 2x + 1 = 0

=> 6x(2x - 1) - 1(2x - 1)  = 0

=> x = 1/6 , 1/2

a & c are 1/6 & 1/2

1/a , 1/c are  6 & 2

Hence numbers are

2 , 3  , 6

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