Question

insert two numbers between 1/4 and 1/3 so that the resulting sequence is a H.P

Answer 1

The attachment is required answer.

Answer 2

Answer:

HELLO MATE.

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.

Hope it helps you mate.

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sorry if it is confusing.......

@ ANUSHA ❤✌