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

70 DETERMINATIONS
We know (vide supra Rule 85) that
s = [ (n - 1) / 2 · d + a ] n ,
whence on solving for a , we get
a = s / n - d / 2 · (n - 1).
Sub-rule for finding the common difference of a series in A.P.
(interpreted geometrically by a series-figure) when the first
term, the number of terms, and the sum of the series are
known :
86(ii). The sum of the series (phala) as divided by the
number of terms of the series (pada), being (first) diminished
by the first term (mukha) and then divided by half of the
number of terms minus 1, gives the common difference of the
series (pracaya).¹
That is,
d = (s/n - a) / (½(n - 1)) ,
where a , d , n , and s have their usual meanings.
This formula is obvious from that of Rule 86(i).
Sub-rule for finding the number of terms of a series in A.P.
(interpreted geometrically by a series-figure), when the first
term, the common difference, and the sum of the series are
given :
87. Multiply the sum of the series (phala) by 8 times
the common difference (uttara) and (to that product) add
the square of the difference between twice the first term (ādi)
and the common difference (pracaya) : take the square root
of that. That (square root) diminished by twice the first
term (mukha) and increased by the common difference and
(then) divided by twice the common difference, gives the

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¹ Cf. GSS, ii. 74(ii) ; vi. 292(ii) ; MSi, xv. 49 ; L(ASS), p. 117, vs.
126. Also see GSS, ii. 73(i); 75; SiŚe, xiii. 23(ii).