पृष्ठम्:Ganita Sara Sangraha - Sanskrit.djvu/२२४

GANITASĀRASANGRAHA. difference. (In splitting up the number of terms from the miśra- dhana), the (required) number of terms (is obtained) in accordance with the rule for obtaining the number of terms, provided that the first term is taken to be increased by one (so as to cause a corresponding increase in all the terms). 82. The misradhana is diminished by the first term and the number of terms, both (of these) being optionally chosen; (then) that quantity, which is obtained (from this difference) by applying the rule for (splitting up) the utta: a-misradhana, happens to be the common difference (required here). This is the mathod of work in (splitting up) the all-combined (m/radhana) 26 Examples in illustration thereof. 83. Forty exceeded by 2, 3, 5 and 10, represents (in order) the adi-maradhanu and the other (maradhanas) Tell me what (respectively) happens in these cases to be the first term, the common difference, the number of terms and all (these three). series in arithmetical progression. There are accordingly four different kinds of miś, adhana mentioned here, and they are respectively ddi-mis adhana, uttara- misradhana, gaccha-miśradhana and sarva-misradhana. For adidhana and uttara- dhana see note under stanzas 63 and 64 in this chapter n (n-1) 2 n + 1 Sa- Algebraically, stanza 80 works out thus a=- Sa is the ddi-misradhana, s.e., S + a. And stanza 81 gives b=; n (n-1) So na where So is the uttara-mis adhana, + 1 2 1.e., 8+b and further points out that the valve of n may be found out, when the value of Sa, which, being the gaccha-misradhana, is equal to S+, is given, from the fact that, when S= a + (a + b) + (a + 2b) +.. .up to n terms, Sn= (a + 1) + (a+1+b) (a+1+2b) +..... up to the same n terms. Since, in stanza 82, the choice of a and n are left to our option, the problem of finding out a, n, and b from the given value of Sano, which, being the sarva-miśradhana, is equal to S+a+n+b, resolves itself easily to the finding out of b from any given value of So in the manner above explained. b where 83. The problem expressed in plainer terms is :-(1) Find out a when Sa = 42, b=3 and n = 5. (2) Find out b, when Sb= 43, a= 2 and n = 5. (3) Find out when S+= 15, a= 2 and b 3. And (4) find out a, b, and n when S+ a + b + = 50.