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GAŅIITASĀRASAŃGRAHA.

which, (while having the given denominators), have one for the numerators and (are then reduced so as to) have a common denominator becomes the (first required) numerator among those which (successively) rise in value by one (and are to be found out). On the remainder (obtained in this division) being divided by the sum of the other numerators (having the common denominator as above), it, (i.e., the resulting quotient), becomes another (viz., tho second required ) numerator (if added to the first one already obtained). In this manner (the problem has to be worked out) to the end.

An example in illustration thereof.

7. The Sum of (certain numbers which are divided (respectively) by 9, 10 and 11 is 877 as divided by 990. Give out what the numerators are (in this operation of adding fractions).

The rule for arriving at the (required) denominators (is as follows):--

75. When the sum of the (different fractional) quantities having one for their numerators is one the (required) denominators are such as, beginning with one, are in order multiplied (successively) by

obtained in this division is then divided by the sum of the remaining provisional numerators, i.e., 189, giving the quotient 1, which, combined with the numerator of the first fraction, namely 2, becomes the numerator in relation to the second denominator. The remainder in this second division, viz, 50, is divided by the provisional numerator 90 of the last fraction, and the quotient 1, when combined with the numerator of the previous fraction, namely 3, gives rise to the numerator in relation the last enumerator. Hence the fractions, of which ${\displaystyle {\tfrac {877}{990}}}$ the sum, are ${\displaystyle {\tfrac {2}{9}},{\tfrac {3}{10}}}$ and ${\displaystyle {\tfrac {4}{11}}}$.

It is noticeable here that the numerators successively found out thus become the required numerators in relation to the given denominators in the order in which they are given.

Algebraically also, given the denominators a, b & c, in respect of 3 fractions whose sum is ${\displaystyle {\tfrac {bcx+(x+1)ac+(x+2)ab}{abc}}}$ , the numerators x,x + 1 and x + 2 are easily found out by the method as given above.

75. In working out an example according to the method stated herein, it will be found that when there are n fractions, there are, after leaving out the first term and the last fraction n - 2 terms in geometrical progression with ${\displaystyle {\tfrac {1}{3}}}$ as the first term and ${\displaystyle {\tfrac {1}{3}}}$ as the common ration. The sum of theese n - 2 terms is ${\displaystyle {\tfrac {{\tfrac {1}{3}}{\big \{}1-\left({\tfrac {1}{3}}\right)^{n-2}{\big \}}}{1-{\tfrac {1}{3}}}}}$, which when reduced becomes ${\displaystyle {\tfrac {1}{2}}-{\tfrac {1}{2}}.{\tfrac {1}{3^{n-2}}}}$, which is the same