पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

SERIES IN ARITHMETIC PROGRESSION 73
terms, and (then) added to the ‘undestroyed result’ as mul-
tiplied by the fractional part of the number of terms (vikala),
gives the sum of the series (gaṇita).
That is to say, if n+p/q be the number of terms, then
s = (n / 2) (dn + a + a − d) + (p / q) (dn + a) ,
where the symbols have their usual meanings.
It will be noted that
(n / 2) (dn + a + a − d)
denotes the sum of the n terms of the series, and
(p / q) (dn + a)
is equal to the p/qᵗʰ part of the (n+1)ᵗʰ term of the series.
Ex. 106. One man gets 3 (rūpas), and the other men get
2 rūpas more in succession ; say, what do (the first) 4½ men get.
By saying 4½ men is meant, according to the commentator, that the
fifth man does only half the work due from him.
Ex. 107. If a labourer gets 1½ in the first month and ⅓
more in succession in the following months, what will he
get in (the first) 3½ months ?
Sub-rule for finding the first term of a series in A.P. (inter-
preted symbolically), when the common difference, the number
of terms (which is partly integral and partly fractional), and
the sum of the series are known :
90. The integral part of the number of terms (nirvikala-
pada) minus 1, halved and increased by the fractional part of
the number of terms (vikala), should be multiplied by the com-
mon difference (caya) and also by the integral part of the
number of terms (vikala-vihīna-pada). The sum of the series
(dhana) minus that, when divided by the (given) number of
terms, gives the first term of the series (prabhava).
That is, if n+p/q be the number of terms, then
a = [ s − {½(n − 1) + p/q}dn ] / (n + p/q) .
We know (vide supra Rule 89) that
s = (n / 2) (dn + a + a − d) + (p / q) (dn + a)
= a(n + p/q) + {½(n − 1) + p/q}dn.
∴ a = [ s − {½(n − 1) + p/q}dn ] / (n + p/q) .