Question

the difference of two natural numbers is 3 and the difference of their reciprocals is 3/28. find the numbers if a, b, c are in A.P., prove that a 2 ​ + c 2 - 2 bc= 2a(b-a)

Answer 1

Hi ,

i ) let the two numbers are x and y

x - y = 3

x = y + 3 ------( 1 )

Difference of Reciprocals of x and y

equals to 3 / 28

1 / y - 1 / x = 3/28

( x - y ) / xy = 3/ 28

28 ( x - y ) = 3xy

28[ ( y+ 3 ) - y ] = 3 ( y+ 3 ) y

28 ( 3 ) = 3 ( y^2 + 3y )

28 = y^2 + 3y

0 = y^2 +3y - 28

Therefore

y^2 + 7y - 4y - 28 = 0

y( y + 7 ) - 4 ( y + 7 ) = 0

( y + 7 ) ( y - 4 ) = 0

y + 7 = 0 or y - 4 = 0

y = -7 or y = 4

Put y = 4 values in ( 1 ) we get

x = 7 ( since x and y are natural )

Therefore ,
Required two natural numbers are

7 and 4

ii ) a , b , c are in A . P

b - a = c - b

2b = c + a

Multipliing ( c + a ) bothsides

2b ( c + a) = ( c + a ) ( c + a )

2bc + 2ab = c^2 + a^2 + 2ac

Rearrenging the terms

2ab - 2ac = c^2 + a^2 - 2bc

2a( b - c ) = c^2 + a^2 - 2bc

Plz , check your answer.

I hope this helps you.

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