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

78 DETERMINATIONS
95(ii). The multiplication of the number of bangles by
half the sum of prices of the first and last bangles, gives the
price (of all the bangles).
This rule is evidently based on the following formula for the sum of
an arithmetic series :
Sum of an arithmetic series = (first term + last term) / 2 × (no. of terms).
Ex. 110. The first bangle is obtained for 8 paṇas, and
the last bangle for 13 paṇas. If the total number of the
bangles be 24, say what is the price of all of them.¹
The first bangle means the smallest bangle which lies at the wrist
ahead of all the other bangles, and the last bangle means the biggest
bangle which lies behind all the other bangles.
Rule for finding the time in which two persons, one travelling
with a constant speed and the other accelerating his initial
speed by a given quantity per day, would traverse the same
distance :
96. Having subtracted the initial speed from the
constant speed, divide the remainder as multiplied by 2 by
the acceleration in velocity. The quotient plus 1 gives the
time elapsed (in term of days) when the distances traversed
are equal.²
Suppose that one person travels with a constant speed u per day,
while the other starts with speed v per day on the first day and then
accelerates his speed by f per day per day. Also suppose that the two
persons traverse the same distance in n days. Then
un = (n / 2) [ 2v + (n - 1)f ].
Hence
n = [2(u - v) / f] + 1.
Ex. 111. One man goes with initial speed 3 (yojanas)
per day and acceleration 1 (yojana) per day per day, and
another man goes with the (constant) speed of 10 yojanas per
day. In what time will they cover the same distance ?³

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¹ Similar examples are found to occur in Bhâskara I's comm. on Â,
ii. 19.
² Cf. BM, III, B1, 8 recto; GSS, vi. 319 (i); GK, I, p. 112, lines 4-5.
³ Cf. BM, III, B3, 7 verso; B1, 8 recto; GSS, vi. 320.