If the reciprocal of the product of the two consecutive integers is 1/30 how do you find the two integers?

Question

Genesis Allen · Accepted Answer

Then the product must be the reciprocal of #1//30# and that is #30# (reciprocal goes both ways).

The reciprocal of the reciprocal is the original number. So we need #n*(n+1)=30# You can try factoring #30# in different ways, but it will be clear that only #5and6# satisfy the condition of being consecutive.

Ethan Adkins · Answer

#5,6 or -6,-5#

*Integers:* The numbers which are not fractional(like #26/9#) are called integers.

Consider two consecutive numbers, A and B.

Let's say A=n.

Since B is the next number after A, it must be "n+1".

2 consecutive numbers multiplied by one is AB.

The reciprocal of the product of the two consecutive integers=#1/(AB)#

However, 1/30 is the reciprocal of the product of the two successive integers.

From the equality rule, in the end,

#1/(AB)=1/30#

Simply replace the assumed A and B values.

#1/(n(n+1))=1/30#

#=>1/(n^2+n)=1/30#

#=>n^2+n-30=0#

#=>n^2+6n-5n-30=0#

#=>n(n+6)-5(n+6)=0#

#=>(n-5)(n+6)=0#

#=>n=5 | n=-6#

If # n=5#;

#A=5,#

#B=6#

If #n=-6:#

#A=-6#

#B=-5#

Jace Anderson · Answer

undefined Let $ n $ be the first integer. Then the next consecutive integer is $ n + 1 $. So the product of these two consecutive integers is $ n(n + 1) $. Given that the reciprocal of this product is $ \frac{1}{30} $, we can write the equation:

$$ \frac{1}{n(n + 1)} = \frac{1}{30} $$

To solve for $ n $, we can cross multiply and solve the resulting quadratic equation.