Question 1204069



{{{ N}}}  = { {{{ 1}}} , {{{ 2}}} , {{{ 3}}} , {{{ 4}}} ,{{{  5}}} , {{{ 6}}} , ……} So consecutive natural numbers are the numbers that continuously follow each other in order from the smallest number to the largest number. 

let

{{{ A=n }}} 
{{{ B=n+1 }}} 
{{{ C=n+2}}} 

 
the reciprocal of {{{  A=1/n}}} 

{{{ 2/7}}}  of the reciprocal of  {{{ A }}} is {{{ (2/7)(1/n)=2/(7n)}}} 

the reciprocal of {{{ C}}}  is {{{ 1/(n+2)}}} 

 {{{ 1/3}}}  of the reciprocal of {{{ C}}}  is {{{ (1/3)(1/(n+2))=1/(3(n+2))}}} 

If {{{ 2/7 }}} of the reciprocal of  {{{ A}}}  is equal to  {{{ 1/3 }}} of the reciprocal of {{{ C}}} , we have


{{{ 2/(7n)=1/(3(n+2))}}} ...cross multiply

{{{ 2(3(n+2))=1(7n)}}} 

{{{ 6(n+2)=7n}}} 

{{{ 6n+12=7n}}} 

{{{ 12=7n-6n}}} 

{{{ n=12}}} 

then

{{{ B=12+1 }}} 

{{{ B=13}}}