12(1) The sum of the reciprocals of two consecutive odd natural number

1.	12(1) The sum of the reciprocals of two consecutive odd natural numbers isnumbers
Answer» Let the two consecutive odd natural numbers are x and x+2. As given sum of reciprocal of two consecutive odd natural number is 12/35 . Writing the statement in terms of equation: →1 /x + 1/(x+2) = 12/35 →35× (2 x +2) = 12× (x² +2 x) → 70 x + 70 = 12 x² + 24 x → [ Using Distributive property: a× (b+c) = a ×b + a× c] Taking variable and constant on one side of equation → 12 x² + 24 x - 70 x - 70 =0 → 12 x² - 46 x - 70=0 → 2 × (6 x² - 23 x -35) =0 → 6 x² - 23 x -35=0 Splitting the middle term → 6 x² - 30 x + 7 x - 35=0 → 6 x × ( x -5) + 7 × (x -5) =0 → (6 x +7)(x-5) =0 → 6 x +7 =0 ∧ x -5 =0 x ≠ can't be the solution as x is a natural number. and x = 5. So, x = 5 and 5+2 =7 are those two numbers.

12(1) The sum of the reciprocals of two consecutive odd natural numbers isnumbers

Answer»

Let the two consecutive odd natural numbers are x and x+2.

As given sum of reciprocal of two consecutive odd natural number is 12/35 .

Writing the statement in terms of equation:

→1 /x + 1/(x+2) = 12/35

→35× (2 x +2) = 12× (x² +2 x)

→ 70 x + 70 = 12 x² + 24 x → [ Using Distributive property: a× (b+c) = a ×b + a× c]

Taking variable and constant on one side of equation

→ 12 x² + 24 x - 70 x - 70 =0

→ 12 x² - 46 x - 70=0

→ 2 × (6 x² - 23 x -35) =0

→ 6 x² - 23 x -35=0

Splitting the middle term

→ 6 x² - 30 x + 7 x - 35=0

→ 6 x × ( x -5) + 7 × (x -5) =0

→ (6 x +7)(x-5) =0

→ 6 x +7 =0 ∧ x -5 =0

x ≠ can't be the solution as x is a natural number. and x = 5.

So, x = 5 and 5+2 =7 are those two numbers.