Question 1006529
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What to find ?


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<U>Comment from student</U>: simplify it into an expression using n
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OK. (Why do not write it from the very beginning?)


The sum in the numerator is the sum of first 19 terms of an arithmetic progression 1, 2, 3, . . . , 19


S = 1 + 2 + 3 + . . . + 19 = {{{(1 + 19)/2*19}}} = {{{20/2*19}}} = 10*19.


The sum in the numerator is the sum of first 19 terms of an arithmetic progression 21, 22, 23, . . . , 39


T = 21 + 22 + 23 + . . . + 39 = {{{(21 + 39)/2*19}}} = {{{60/2*19}}} = 30*19.


Hence, the ratio you are asking for, is


{{{(S*20)/(R*40)}}} = {{{(10*19)/(30*19*2)}}} = {{{10/(30*2)}}} = {{{1/6}}}.