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56

GĀŅITASĀRASAŇGRAHA.

(number) which is (immediately) next to it (in value) and then halved.

The rule for arriving at the (required) denominators (in the case of certain intended fractions), when their numerators are (each) one or other than one, and when the (fraction constituting their) sum has one for its numerator:--

78. When the sum (of certain intended fractions) has one for its numerator, then (their required denominators are arrived at by taking) the denominator of the sum to be that of the first (quantity), and (by taking) t!his (denominator) combined with its own (related) numerator to be (the denominator) of the next (quantity) and so on, and then by multiplying (further each such denominator in order) by that which is (immediately) nex to it, the last (denominator) being (however multiplied) by its own (related) numerator.

Examples in illustration thereof

79. The sums (of certain intended fractions) having for their numerators 7, 9, 3 and 18 (respectively) are (firstly) 1, (secondly) ${\tfrac {1}{4}}$ and (thirdly) ${\tfrac {1}{6}}$. Say what the denominators (of those fractional quantities) are.

The rule for arriving at the denominators (of certain intended fractions) having one for their numerators, when tho sum (of those fractions) has one or (any quantity) other than one for its numerator:--

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78. Algebraically, if the sum is ${\tfrac {1}{n}}$ and a, b, c and d are the given numerators, the fractions summed up are as below:--

==${\tfrac {a}{n(n+a)}}+{\tfrac {b}{(n+a)(n+a+b)}}+{\tfrac {c}{(n+a+b)(n+a+b+c)}}+{\tfrac {d}{d(n+a+b+c)}}$

==${\tfrac {a(n+a+b)+ba}{n(n+a)(n+a+b)}}+{\tfrac {c+n+a+b}{(n+a+b)(n+a+b+c)}}$

==${\tfrac {(n+a)(a+b)}{n(n+a)(n+a+b)}}+{\tfrac {1}{n+a+b}}$

==${\tfrac {a+b+n}{n(n+a+b)}}=={\tfrac {1}{n}}$