There are three consecutive integers. if the sum of the reciprocals of the second and third integer is (7/12), what are the three integers?

1 Answer
Mar 23, 2018

#2 , 3, 4#

Explanation:

Let #n# be the first integer. Then the three consecutive integers are:

#n , n+1,n+2#

Sum of the reciprocals of 2nd and 3rd:

#1/(n+1)+1/(n+2)=7/12#

Adding the fractions:

#((n+2)+(n+1))/((n+1)(n+2))=7/12#

Multiply by 12:

#(12((n+2)+(n+1)))/((n+1)(n+2))=7#

Multiply by #((n+1)(n+2))#

#(12((n+2)+(n+1)))=7((n+1)(n+2))#

Expanding:

#12n+24+12n+12=7n^2+21n+14#

Collecting like terms and simplifying:

#7n^2-3n-22=0#

Factor:

#(7n+11)(n-2)=0=>n=-11/7 and n=2#

Only #n=2# is valid since we require integers.

So the numbers are:

#2 , 3, 4#