Question

Insert two numbers between 1/4 and 1/13 so that the resulting sequence is a H.P.

Answer 1

Answer:

1/7 & 1/10 to be inserted

Step-by-step explanation:

1/4  , a  , b  , 1/13  are in HP

=> a  =  (2(1/4)b ) /(1/4 + b)

=> a =  2b/(4b + 1)

=> 4ab + a = 2b    Eq1

b =  (2(1/13)a ) /(1/13 + a)

=> b = 2a /(13a + 1))

=> 13ab + b = 2a    Eq2

4*Eq2 - 13 * eq1

=> 4b - 13a = 8a - 26b

=> 30b = 21a

=> 10b = 7a

4ab + a = 2b

=> 4a (7a/10) + a = 2(7a/10)

=> 28a² + 10a = 14a

=> 28a² = 4a

=> a = 1/7

10b = 7a

=> 10b = 1

=> b = 1/10

so 1/7 & 1/10 to be inserted

Answer 2

Answer:

$\mathbf{\frac{1}{7}\ &\ \frac{1}{10}\ needs\ to\ be\ inserted}$

Step-by-step explanation:

In this question,

We need to insert two numbers so that resulting number is a H.P

Let a, b be the number to be inserted between $\frac{1}{4}\ and\ \frac{1}{13}$

Therefore, $\mathbf{\frac{1}{4}\ ,a\ ,b\ , \frac{1}{13}}$  are in H.P

=> a  =  $\frac{(2(\frac{1}{4})b )}{(\frac{1}{4}\ +\ b)}$

=> a =  $\frac{2b}{(4b + 1)}$

=> 4ab + a = 2b            (i)........

b =  $\frac{2(\frac{1}{13})a)}{(\frac{1}{13}\ +\ a)}$

=> b = $\frac{2a}{(13a + 1))}$

=> 13ab + b = 2a         (ii).........

Now,

$4\times Eq(ii)\ -\ 13\times eq(i)$

=> 4b - 13a = 8a - 26b

=> 30b = 21a

=> 10b = 7a

4ab + a = 2b

=> 4a ($\frac{7a}{10})\ +\ a\ =\ 2(\frac{7a}{10})$

=> 28a² + 10a = 14a

=> 28a² = 4a

=> a = $\frac{1}{7}$

10b = 7a

=> 10b = 1

=> b = 1$\frac{1}{10}$

So $\mathbf{\frac{1}{7}\ &\ \frac{1}{10}\ needs\ to\ be\ inserted}$