# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/२७४

एतत् पृष्ठम् परिष्कृतम् अस्ति
78
GAŅIITASĀRASAŃGRAHA.

at last 12 ascetics were seen (to remain without being included among those mentioned before). O(you) excellent ascetic, of what numerical value was (this) collection of ascetics ?

46. Five and one-fourth times the square root (of a herd) of elephants are sporting on a mountain slope; ${\displaystyle {\tfrac {5}{9}}}$ of the remainder sport on the top of the mountain; five times the square root of the remainder (after deducting this) sport in a forest of lotuses; and there are 6 elephants then (left) on the bank of a river. How many are (all) the elephants here ?

Here ends the Śēșamūla variety (of miscellaneous problems on fractions).

The rule relating to the Śēșamūla variety involving two known (quantities constituting the) remainders:--

The (coefficient of the) square root (of the unknown collective quantity), and the (final) quantity known (to remain), should (both) be divided by the product of the fractional (proportional) quantities, as subtracted from one (in each case); then the first known quantity should be added to the (other) known quantity (treated as above). Thereafter the operation relating to the Śēșamūla variety ( of miscellaneous problems on fractions is to be adopted).

47. Algebraically, this rule enables us to arrive at the expressions ${\displaystyle {\tfrac {c}{(1-b_{1})(1-b_{2})\times \&c}}}$ and ${\displaystyle {\tfrac {a_{2}}{(1-b_{1})(1-b_{2})\times \&c}}+a_{1}}$, which are required to be substituted for c and a respectively in the formula for Śēșamūla, which is ${\displaystyle x-bx={\Big \{}{\tfrac {c}{2}}+{\sqrt {{\tfrac {c}{2}}^{2}+a}}{\Big \}}^{2}}$. In applying the value of b becomes zero, as the mūla square root involved in the dviragra-śēşamūla is that of the total collective quantity and not of a fractional part of that quantity. Substituting a desired, we get ${\displaystyle x={\Big \{}{\tfrac {c}{2(1-b_{1})(1-b_{2})\times \&c}}+{\sqrt {\left({\tfrac {c}{s(1-b_{1})(1-b_{2})\times \&c}}\right)^{2}+{\tfrac {a_{2}}{(1-b_{1})(1-b_{2})\times \&c}}+a_{1}}}{\Big \}}^{2}}$. This result may easily be obtained from the equation ${\displaystyle x-a_{1}-b_{1}(x-a_{1})-b_{2}{\Big \{}x-a_{1}-b_{1}(x-a_{1}){\Big \}}-.....-c{\sqrt {x-a_{2}}}=0}$ where b1,b2, &c are, the various fractional parts of the successive remainders; and 11 and a2 are the first known quantity and the final known quantity respectively.