Question 1204069
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A, B and C are consecutive natural numbers. If 2/7 of the reciprocal of A is equal to ⅓ of the reciprocal of C, find B.​
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<pre>
Let B = n;  then  A = n-1;  C = n+1.


The equation is

    {{{(2/7)*(1/A)}}} = {{{(1/3)*(1/C)}}},

or

    {{{(2/7)*(1/(n-1))}}} = {{{(1/3)*(1/(n+1))}}}.


To solve, multiply both sides by  7*3*(n-1)*(n+1).  You will get

    2*3*(n+1) = 7*(n-1),

     6(n+1)   = 7(n-1)

     6n + 6   = 7n - 7

     6  + 7   = 7n - 6n

       13    = n.


<U>ANSWER</U>.  The numbers are 12, 13 and 14.
</pre>

Solved.


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If you read the post by @greenestamps, &nbsp;you will see in the third line


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"2/7 of the reciprocal of A is 7/2 times A; 1/3 of the reciprocal of C is 3/1 times C."



This statement is incorrect.  

For the correct setup, &nbsp;read the posts of other tutors.