Question

the sum of three terms of an HP is 33/40 sum of their reciprocals is 15.
The sum of three terms of a H.P. is
Find the three terms.
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Answer 1

Question:

The sum of three terms of an HP is 33/40 and the sum of their reciprocals is 15. Find the three terms.

Answer:

1/2 , 1/5 , 1/8 or 1/8 , 1/5 , 1/2

Note:

• Harmonic progression (HP) : A sequence is said to be in harmonic progressive if the reciprocal of its terms are in authentic progression (AP) .

• If a , a+d , a+2d , a +3d ,......are in AP , then 1/a , 1/(a+d) , 1/(a+2d) , 1/(a+3d) ......are in HP .

• If 1/A , 1/B , 1/C are in HP then A,B,C are in AP and hence , 2B = A+C .

Solution:

Let the required HP be ,

1/(a-d) , 1/a , 1/(a+d) , so that (a-d) , a , (a+d) are in AP.

Now,

According to the question,

The sum of the three terms of HP is 33/40.

=> 1/(a-d) + 1/a + 1/(a+d) = 33/40 ---(1)

Also ,

The sum of the reciprocals of the three terms of HP is 15 .

=> (a-d) + a + (a+d) = 15

=> 3a = 15

=> a = 15/3

=> a = 5

Now,

Putting a = 5 in eq-(1) , we have ;

$= > \frac{1}{5 - d} + \frac{1}{5} + \frac{1}{5 + d} = \frac{33}{40 } \\ = > \frac{1}{5 - d} + \frac{1}{5 + d} = \frac{33}{40} - \frac{1}{5} \\ = > \frac{(5 + d) + (5 - d)}{(5 - d)(5 + d)} = \frac{33 - 8}{40} \\ = > \frac{5 + 5}{ {5}^{2} - {d}^{2} } = \frac{25}{40} \\ = > \frac{10}{25 - {d}^{2} } = \frac{25}{40} \\ = > 25 - {d}^{2} = \frac{10 \times 40}{25} \\ = > 25 - {d}^{2} = 16 \\ = > {d}^{2} = 25 - 16 \\ = > {d}^{2} = 9 \\ = > d = \sqrt{9} \\ = > d = 3 ,- 3$

Here,

Two HP are possible corresponding to two different values of d .

Case:1

When a = 5 and d = 3

Then,

1st term of HP = 1/(a-d) = 1/(5-3) = 1/2

2nd term of HP = 1/a = 1/5

3rd term of HP = 1/(a+d) = 1/(5+3) = 1/8

Case:2

When a = 5 and d = -3

Then,

1st term of HP = 1/(a-d) = 1/[5-(-3)] = 1/8

2nd term of HP = 1/a = 1/5

3rd term of HP = 1/(a+d) = 1/(5-3) = 1/2

Hence,

The required HPs are ;

1/2 , 1/5 , 1/8 or 1/8 , 1/5 , 1/2 .