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

MIXTURES. SPECIAL PROBLEMS 61
two as multiplied by 2 and divided by 5 went away
towards the north, and three (cows) are left. (What is the
numerical strength of the herd ?)¹
Here the fractions (of the herd of cows) that went away in the
various directions are
½, ¼, and ⅖(½ - ¼)
i.e., ½, ¼, and ⅒ ;
and the visible quantity (i e., the number of cows left) is 3. Hence the
numerical strength of the herd
= 3 / [1 - (½ + ¼ + ⅒)] , i.e., 20 cows.
Rule for solving problems involving remainders due to sub-
traction of square root, etc. (mūlādiśeṣoddeśaka) : ²
75. When the visible quantity stands near a square
root, multiply the visible quantity by 4, then increase that by
the square of the pada (i.e., the co-efficient of the square root
of the unknown), then extract the square-root of that and
increase that by the pada, and then take the square of half of
that ; (and when the visible quantity stands near a fraction)
divide the visible quantity by one minus the fraction.³
The first part of this rule relates to the solution of the quadratic
equation of the type
x - p√x = d,
where p is the pada, d the visible quantity (dṛśya), and √x the positive
square root of x. The solution is correctly stated as
x = [ {√(4d + p²) + p} / 2 ]²

¹ For similar examples see GSS, iv. 23-27; GT, p. 46, lines 20-25;
L (ASS), p. 53, vs. 55.
² Mahāvīra and Śrīpati have called mūlādiśeṣa-jāti by the name
śeṣamūla-jāti.
³ Cf. GSS, iv. 40; GT, p. 48, lines 24-25.

62 DETERMINATIONS
The second part of the rule relates to the solution of the simple
equation of the type
x − (a/b) x = d′,
where a/b is the fraction and d′ the visible quantity. The solution is
correctly stated as
x = d′ / (1 − a/b).
Ex. 99. A number is diminished by its square root,
what remains is diminished by its one-sixth, what remains
after that is diminished by its square root, what remains after
that is diminished by its one-fifth, and what remains after
that is diminished by twice the square root of itself; the
residue now left is 8. (Find the number).¹
This problem reduces to the solution of the following equation :
x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)] − ⅕ {x − √x − ⅙ (x − √x)
− √[x − √x − ⅙ (x − √x)]} − 2 [x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)]
− ⅕ {x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)]}]^½ = 8,
which can be written as
x − √x = y, (1)
where y − ⅙ y = z, (2)
z − √z = u, (3)
u − ⅕ u = v, (4)
v − 2 √v = 8. (5)
Solving these equations in the inverse order, we get x = 36.
Hindu Method of Solution. Writing down the square roots and frac-
tions, which have been successively subtracted from the original number,
in order and then the visible quantity, we get
Square root, 1/6 part of the remainder, square root of the remainder, 1/5 part of the
remainder, 2 times the square root of the remainder, visible quantity 8.
──────────────────────────────────────────────────────────
¹ For other examples see, GSS, iv. 41, 42-45, 46; GT, p. 49, lines
7-10; p. 50, lines 2-5.

MIXTURES. SPECIAL PROBLEMS 63
Since the visible quantity stands near a square root, therefore per-
forming the operations prescribed in the first part of the rule, we get
[ (√(4 × 8 + 2²) + 2) / 2 ]², i.e., 16.
This is the number which when diminished by 2 times the square root
of itself yields 8 as remainder.
Now treat this (16) as the visible quantity. Since it stands near a
fraction, therefore performing the operation prescribed in the second part
of the rule, we get
16 / (1 - 1/5), i.e., 20.
This is the number which when diminished by 1/5 of itself yields 16
as remainder.
Now treat this (20) as the visible quantity. Since it stands near a
square root, therefore performing the operations prescribed in the first
part of the rule, we get
[ (√(4 × 20 + 1) + 1) / 2 ]², i.e., 25.
This is the number which when diminished by its square root yields 20 as
remainder.
Now treat this (25) as the visible quantity. Since this stands near a
fraction, therefore performing the operation prescribed in the second part
of the rule, we get
25 / (1 - ⅙), i.e., 30.
This is the number which when diminished by its ⅙ yields 25 as
remainder.
Now treat this (30) as the visible quantity. Since this stands near a
square root, therefore performing the operations prescribed in the first
part of the rule, we get
[ (√(4 × 30 + 1) + 1) / 2 ]², i.e., 36.
This is the number which being diminished by its square root yields 30
as remainder.
Since all quantities in the statement of the problem are now
exhausted, no further operation is needed. The required number is thus
obtained to be 36.

64 DETERMINATIONS
Rule for solving problems in which the visible quantity is the
remainder due to the subtraction of a fraction of the unknown
and also a multiple of the square root of the unknown
(bhâgamûlâgroddeśa) :¹
76. After having divided the pada (i.e., the co-efficient
of the square root of the unknown) and the visible quantity
(or ultimate remainder, agra) by one minus the fraction, add
the square of half the first quotient to the second quotient,
then take the square root of that, then increase that by half
the first quotient, and then multiply that by itself.²
This rule is meant to solve problems reducing to a quadratic
equation of the type
x - (a/b)x - p√x = d ,
where a/b is the 'fraction', p the 'pada', d the visible quantity, and √x
the positive square root of x.
The solution is correctly stated as
x = [ p / {2(1 - a/b)} + √{ (p / {2(1 - a/b)})² + d / (1 - a/b) } ]²
Ex. 100. One-third of a troop of monkeys together
with one-third of itself has gone to the tank ; the square root
(of the whole troop) is afflicted with thirst ; and the remain-
ing two monkeys are sitting under the mango tree. (What is
the number of monkeys in the troop ?)³
Rule for solving problems involving two visible quantities,
the square root of the original number, and remainders due
to subtraction of fractional multiples of remainders (ubhayâgra-
mûlaśeṣoddeśa):⁴
77. Take the continued product of units severally
diminished, as before, by the fractional multiples of remain-
――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――
¹ Mahâvîra calls bhâgamûlâgra-jâti by the name mûla-jâti, Śrîpati by
the name mûlâgra-bhâga-jâti, and Nârâyaṇa by the name ṛṇâṁśavimûla-jâti.
² Cf. GSS, iv. 33; GT, p. 50 lines 27-28 (contd. on p. 51, lines 1-2).
Also see GK, I, p. 21, lines 2-6.
³ For other examples see GSS, iv. 34, 35, 36, 37, 38, 39; GT, p. 51,
lines, 11-14; p. 52, lines 16-19; p. 53, lines 17-20; GK, I, p. 23, lines 15-16,
18-19, and p. 24, lines 1-2.
⁴ Mahâvîra calls ubhayâgramûlaśeṣa-jâti by the name dviragraśeṣamûla-
jâti, and Śrîpati by the name ubhayâgradṛśya-jâti.

MIXTURES. SPECIAL PROBLEMS 65
ders as the divisors of the pada as well as the (last) visible
quantity ; to the second quotient also add the first visible
quantity. After that apply the previous rule (starting with
‘add the square of half the first quotient’).¹
This rule applies to problems which reduce to the solution of the
quadratic equation of the type
(1 - a/b) (1 - c/d) (1 - e/f) (x - d₁) - p√x = d₂ ,
where a/b, c/d, e/f are the fractional multiples of remainders, p is the pada,
d₁, d₂ the first and last visible quantities, and √x denotes the positive
square root of x, as before.
According to the above rule, this equation should be first trans-
formed into the form
x - {p / [(1 - a/b)(1 - c/d)(1 - e/f)]} √x = {d₂ / [(1 - a/b)(1 - c/d)(1 - e/f)]} + d₁ ,
and then solved by the previous rule.
Ex. 101. After giving away one (out of a certain
number), then one-sixth of what remains, then one-fourth of
what remains after that, then one-third of what remains after
that, and then the square root of the original number, the
residue left is 5. (What is that number ?)²
Rule of inversion :
78. (Proceeding from the visible quantity backwards,
make) addition subtraction, subtraction addition, multi-
plication division, division multiplication, square square-
root, and square-root square : this is stated to be the method
of inversion.³
Ex. 102. Say what (is that number which) being multi-
plied by 5/2, then divided by 3, then squared, then increased
by 9, then reduced to its square root, and then diminished by
1, becomes 4.⁴
¹ Cf. GSS, iv. 47 ;GT, p. 54, lines 21-24.
² For other examples see GSS, iv. 48, 49, 50; GT, p. 55, lines 7-10;
p. 56, lines 8-11 ; p. 5 , lines 19-22.
³ Cf. Ā, ii. 28; BrSpSi, xviii. 14(17); GSS, vi. 286; MSi, xv. 23;
GT, p. 65, vs. 83; SiŚe, xiii. 13; L (ASS), p. 42, vs. 48; GK, I, p. 46,
lines 13-16.
⁴ For similar examples see GSS, vi 287; GT, p. 66, vs. 84; p. 67,
vs. 85; L (ASS), p. 43, vs. 50.

66 DETERMINATIONS
(2) Determinations pertaining to series (śreḍhî-vyavahâra)
(i) SERIES IN ARITHMETIC PROGRESSION
(GEOMETRICAL INTERPRETATION)
Form of a series-figure (śreḍhî-kṣetra) :
79. As in the case of an earthen drinking glass
(śarâva) the width at the base is smaller and at the top
greater, so also is the case with a series-figure (śreḍhî-kṣetra).
The altitude (lambaka) of that (series-figure) is equal
to the number of terms (gaccha) of the (corresponding)
series.
The series-figure contemplated here is a plane figure resembling a
trapezium with equal flank sides
If a series be
a+(a+d)+(a+2d)+... to n terms,
then, according to the second part of the verse, the altitude of the corres-
ponding series-figure = n units, say n cubits.
80(i). The (partial) areas (phala) of the series-figure
for the successive cubits (kara) of the altitude form a series
which begins with the given âdi ('first term of the series')
and increases successively by the given caya ('common diffe-
rence of the series').
That is, the area of the series-figure for the first cubit of the altitude
= a, i.e., the first term of the series; the area of the series-figure for the
second cubit of the altitude = a+d, i.e., the second term of the series; the
area of the series-figure for the third cubit of the altitude = a+2d; and
so on.
Construction of a series-figure :
80(ii). I shall now describe the method for finding
the lengths of the base (i.e., lower side, bhû) and the face
(i.e., upper side, mukha) of the series-figure (corresponding
to the first term of the series).
81. The number of terms (pada), i.e., one, is the alti-
tude of the (corresponding) series-figure ; the first term of
Indological Truths

SERIES IN ARITHMETIC PROGRESSION 67
the series (mukha) as diminished by half the common diffe-
rence of the series is the base (dharā) ; and that (base) in-
creased by the common difference of the series is the face
(vaktra). All these should be shown by means of threads.
That is,
base = a − d/2,
and face = (a − d/2) + d, i.e., a + d/2.
82. (Two) threads should then be stretched out, one
on either side, joining the extremities of those base and face :
these are the flank sides (bāhu) of the series-figure.
When the base is negative, these threads should be
stretched out crosswise.
Thus the series-figure will be of one of the following two forms :
(i) (ii)
[चित्र १: समलम्ब चतुर्भुज (Trapezoid) — ऊपर 'face', नीचे 'base', बाएँ व दाएँ 'flank side', मध्य में लम्ब 'altitude']
[चित्र २: परस्पर सम्मुख दो त्रिभुज (Crossed/Hourglass) — ऊपर 'face', नीचे 'base', मध्य में लम्ब 'altitude']
Forms (ii) corresponds to the negative base.
83. (When the base is negative the series-figure reduces
to two triangles situated one over the other.) In the upper
triangle, the altitude is equal to the face as divided by
face minus base ; and that subtracted from one gives the
altitude in the lower triangle.
That is,
(i) altitude of the upper triangle = face / (face − base), i.e., (2a + d) / 2d,
(ii) altitude of the lower triangle = 1 − face / (face − base), i.e., (d − 2a) / 2d.
Indological Truths

68 DETERMINATIONS
Rule for finding the face of the series-figure corresponding to
the given series :
84. Having constructed the series-figure (for altitude
unity) in this manner, one should determine the face for the
desired altitude (i.e., for the desired number of terms of the
series) (by the following rule) :
The face (for altitude unity) minus the base (for altitude
unity), multiplied by the desired altitude, and then increased
by the base (for altitude unity), gives the face (for the desired
altitude).
This rule, on simplification, reduces to the following formula :
face for altitude n = a + (n - ½) d.
Rule for finding (i) the sum of a series in A. P. (interpreted
geometrically by a series-figure), and (ii) the area of the
corresponding series-figure :
85. The common difference as multiplied by one-half
of the number of terms minus one, being increased by the first
term, and then multiplied by the number of terms, gives the
sum of the series.¹
And the area of the (corresponding) series-figure is equal
to the product of one-half of the sum of the base and the face,
and the altitude.
That is, the sum of the series
a + (a + d) + (a + 2d) + ... to n terms
is equal to { (n - 1)/2 d + a } n ; (1)
and the area of the corresponding series-figure is equal to
(base + face) / 2 × altitude, (2)

¹ Cf. Ā, ii. 19 ; GSS, ii. 61, vi. 29.) ; GK, I, 105, lines 12-13. The
rule of the GK is literally the same as above.
Brahmagupta (BrSpSi, xii. 17) states the result in the modern form :
Sₙ ≡ a + (a + d) + (a + 2d) + .. to n terms = (n/2)[2a + (n - 1)d].
The same form is given in GSS, ii. 62; MSi, xv, 47 ; Siśe, xiii. 20 ; L (ASS)
p. 114, vs. 121 ; GK, I, p. 105, vs. 1.

SERIES IN ARITHMETIC PROGRESSION 69
where, according to vv. 80(ii) to 84,
base = a − d/2,
face = a + (n − ½)d,
and altitude = n.
It may be easily seen that (1) and (2) are the same.
Ex. 103 (i). What is the sum of 5 terms of the series
whose first term is 2 and common difference 3 ? And what
of one-half of a term ?
Ex. 103 (ii). (Also) say (the sum) of one-fifth of a term
of the series whose common difference is 5 and first term 2.
Ex. 104-105. In a leathern oil-bottle (kutapa) full of
oil there occurs a minute hole, and the oil leaks through it.
The bottle has to be carried to a distance of 3 yojanas. If the
wages for the first yojana be 10 paṇas and those for the subse-
quent yojanas successively less by 2 paṇas, what are the wages
for a krośa ?
Sub-rule for finding the first term of a series in A.P. (inter-
preted geometrically by a series-figure), when the common
difference, number of terms, and the sum of the series are
known :
86(i). The sum of the series (gaṇita) as divided by the
number of terms of the series (pada), being diminished by
half the common difference (caya) as multiplied by the
number of terms (gaccha) minus 1, gives the first term of the
series (ādi).¹
That is,
a = s/n − (d/2)(n − 1),
where a, d, n, and s respectively denote the first term, common
difference, number of terms, and the sum of the series.

¹ Cf. GSS, ii. 74(i) ; vi. 292(i) ; MSi, xv. 48 ; SiŚe, xiii. 23(i); L(ASS),
p. 116, vs. 124 ; GK, I, p. 106, vs. 2. Also see GSS, ii. 73(ii), 76

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

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).

SERIES IN ARITHMETIC PROGRESSION 71
number of terms of the series (gaccha).¹
That is,
n = [√(8ds + (2a - d)²) - 2a + d] / 2d ,
where a, d, n and s have their usual meanings.
We know (vide supra Rule 85) that
s = [((n - 1) / 2) d + a] n ,
∴ dn² + (2a - d)n - 2s = 0.
On solving this quadratic for n, we get²
n = [√(8ds + (2a - d)²) - 2a + d] / 2d ,
taking the positive sign of the radical, because n is positive.
Hence the above rule.
Sub-rule for finding the first term of a series in A.P. (inter-
preted geometrically by a series-figure), when the sum
of the series, the number of terms, and the sum of the first
term and common difference are known :
88. Having subtracted the sum of the series (phala)
from the mixed amount (miśra-dhana) (i.e., the sum of the first
term and common difference) as multiplied by one-half of (the
difference of) the number of terms squared minus the number
of terms, divide the residue by one-half of (the difference of)

¹ Cf. BrSpSi, xii. 18 ; GSS, ii. 70. Also see GSS, ii. 69.
² Âryabhaṭa I (Â, ii. 20) puts the result in the following form :
n = ½ [ {√(8ds + (2a - d)²) - 2a} / d + 1 ].
Âryabhaṭa II (MSı, xv. 50), Bhâskara II (L, p. 118, vs. 128) and
Nârâyaṇa (GK, I, p. 107, lines 4-7) put it in the form :
n = [√(2ds + (a - d/2)²) - a + d/2] / d .
Śrîpati (SiŚe, xiii. 24) puts it in the form :
n = √[ s / (d/2) + ((a - d/2) / d)² ] - (a - d/2) / d .
Indological Truths

72 DETERMINATIONS
the number of terms minus 1, as diminished by 1 and multi-
plied by the number of terms. Thus is obtained the first term
of the series (âdi).
That is
a = [ {(n² - n) / 2}(a + d) - s ] / [ {(n - 1) / 2 - 1} n ] ,
where the symbols have their usual meanings.
We know (vide supra Rule 85) that
s = { (n - 1)/2 · d + a } n
= [ (n - 1)/2 · (a + d̅ - a) + a ] n
= {(n² - n) / 2}(a + d) - { (n - 1)/2 - 1 } na
∴ a = [ {(n² - n) / 2}(a + d) - s ] / [ { (n - 1)/2 - 1 } n ] .
(ii) SERIES IN ARITHMETIC PROGRESSION
( SYMBOLICAL INTERPRETATION )
According to the symbolical interpretation, the series
a + (a + d) + (a + 2d) + ... to (n + p/q) terms
means the sum of n terms together with the p/qᵗʰ part of the (n + 1)ᵗʰ
term.
Rule for finding the sum of a series in A.P. (interpreted
symbolically), when the number of terms is partly integral
and partly fractional, the first term, common difference,
and the number of terms being known :
89. The common difference (caya) as multiplied by the
integral part of the number of terms (nirvikalapada) should be
increased by the first term (âdi), and the result (obtained)
should be kept undestroyed (at one place). The same result
(written in another place) being increased by the first term
(mukha), (then) diminished by the common difference, (then)
multiplied by one-half of the integral part of the number of

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) .

74 DETERMINATIONS
Sub-rule for finding the common difference of a series in A.P.
(interpreted symbolically), when the first term, the number
of terms which is partly integral and partly fractional, and
the sum of the series are known :
91. The sum of the series (dhana) as diminished by the
product of the first term and the number of terms, should be
divided by the sum of the series in which the first term and
common difference are each unity and the number of terms
is equal to the given number of terms minus 1. The result
which is thus obtained is the common difference.
That is, if n+p/q be the number of terms, then
s - a(n+p/q)
d = ------------ ,
S
where S is the sum of (n+p/q - 1) terms of the series 1+2+3+ ..., i. e.,
(n - 1)n p
-------- + ― n.
2 q
(Vide supra, Rule 89).
Sub-rule for finding the number of terms of a series in A.P.
which is partly integral and partly fractional, when the first
term, the common difference, and the sum of the series are
known, the series being interpreted symbolically :
92-93. To the sum of the series (dhana) as multiplied
by twice the common difference (caya), add the square of (the
difference of) the first term minus half the common difference:
of that obtain the nearest (integral) square root. That (square
root), diminished by the square root of the previous square
(i.e., the square of the difference of the first term minus half
the common difference), (then) divided by the common
difference, and (then) rid of its fractional part (vikala), is
the so called ‘undestroyed quantity.’
That (‘undestroyed quantity’) lessened by 1, being
multiplied by half the common difference, (then) increased
by the first term (mukha), and (then) multiplied by the ‘un-
destroyed quantity’, should be subtracted from the sum of the
series (gaṇita). That being divided by the first term as

GEOMETRIC SERIES 75
increased by the product of the ‘undestroyed quantity’ and
the common difference and then added to the ‘undestroyed
quantity’, gives the number of terms of the series.
That is, if n denote the integral part of
√[2ds + (a - d/2)²] - (a - d/2)
───────────────────────────────── ,
d
then the number of terms of the series is equal to
s - {(n - 1)d/2 + a}n
n + ─────────────────────────── .
nd + a
The rationale of this rule is as follows :
Let N be the (unknown) number of terms of the series. Then (vide
Rule 85)
s = ( (N - 1)/2 · d + a ) N
or dN² + 2 (a - d/2) N - 2s = 0
giving
N = { √[2ds + (a - d/2)²] - (a - d/2) } / d .
Now let N = n + p/q. Then
n = integral part of { √[2ds + (a - d/2)²] - (a - d/2) } / d ,
and p/q is obtained by the formula (vide Rule 89)
s = (dn + a + a - d) n/2 + (dn + a) p/q
i.e., p/q = [s - {(n - 1)d/2 + a}n] / (nd + a) .
(iii) SERIES IN GEOMETRIC PROGRESSION
Rule for obtaining the amount which a given sum increasing
in a geometric progression would become after a given
period :
94. When the number of terms of the series (i.e., the
number denoting the period) is odd, subtract 1 from it
and write ‘multiply (by the common ratio)’; and when the
number of terms of the series is even, halve it and write
‘square’. (Apply the same rule to the resulting number, and
continue the process till the number reduces to zero.)

76 DETERMINATIONS
Having thus written down ‘multiply’ and ‘square’ in a
sequence (write 1 thereafter). Then starting from 1
backwards, perform the operations of multiplication and
squaring, and finally multiply the resulting quantity by the
first term of the series (i.e., the given sum).¹
Given the geometric series to n terms
a, ar, ar², ..., arⁿ⁻¹,
the object of this rule is to obtain the value of the (n+1)ᵗʰ term,
i. e., arⁿ .
Ex. 108. Some businessman, taking 3 rûpas with him,
went out to make profit. If his capital becomes double after
every month, what will it become after 3 years ?
Here the given sum = 3 rûpas, increase-ratio = 2, and the period = 3
years, i. e., 36 months.
Therefore we have to consider the geometric series, whose first
term = 3, common ratio = 2, and the number of terms = 36.
Proceeding according to the rule, we have
No. of terms of the series | Operation | Write
36 (even) | 36/2 (=18) | square
18 (even) | 18/2 (= 9) | square
9 (odd ) | 9-1 (= 8) | multiply
8 (even) | 8/2 (= 4) | square
4 (even) | 4/2 (= 2) | square
2 (even) | 2/2 (= 1) | square
1 (odd ) | 1-1 (= 0) | multiply
Denoting ‘square’ by s and ‘multiply’ by m, and writing them in a
sequence and 1 in the end, we get
s s m s s s m 1

¹ Cf. GSS, ii. 94, vi. 311½; MSi, xv. 52-53 (i); SiŚe, xiii. 25; L (ASS),
p. 119, vs. 131; GK, I, p. 127, lines 8-11.

PROBLEMS ON ARITHMETIC SERIES 77
Proceeding from 1 backwards, and performing the operations of
'multiplication (by the common ratio 2)' and 'squaring', we suc-
cessively get
s s m s s s 2
s s m s s 2²
s s m s 2⁴
s s m 2⁸
s s 2⁹
s 2¹⁸
2³⁶
Finally multiplying 2³⁶ by the first term of the series, we get 3×2³⁶,
i. e., 206158430208.
Rule for obtaining the sum of a series in G.P. when the first
term, common ratio, and number of terms are known :
95(i). The result obtained according to the previous
rule, being diminished by the first term of the series, and
(then) divided by the common ratio minus 1, gives the sum
of the series.¹
That is,
a+ar+ar²+... to n terms= (arⁿ - a)/(r - 1), r > 1.
Ex. 109. One man gets 3, and the other men in
succession get in the ratio of 2. Quickly say, how much
money will (the first) five men get.
(iv) MISCELLANEOUS PROBLEMS ON SERIES IN
ARITHMETIC PROGRESSION
Rule for finding the sum of the prices of a number of bangles,
which are in an arithmetic progression, when the prices of
the first and last bangles are known :

¹ Cf. GSS, ii. 94, vi. 311½; MSi, xv. 53(ii); L (ASS), pp. 119-120,
vs. 130; GK, I, p. 127, lines 8-11.

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

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.

PROBLEMS ON ARITHMETIC SERIES 79
Rule for finding the number of days elapsed when two
travellers, starting at a specified interval of time, meet each
other for the second time after their first meeting :
97-98. In relation to the first traveller, assume an
arbitrary number, greater than the caya (i.e., acceleration)
for the second traveller, for the caya (i.e., acceleration) ; and
another arbitrary number (iṣṭa) for the mukha (i.e., initial
speed). In relation to the second traveller, assume another
arbitrary number to denote the pada (i.e., the number of days
elapsed at the first meeting) ; and from the corresponding
pada (i.e., the number of days elapsed at the first meeting) for
the first traveller, calculate the âdi (i.e., the initial speed) for
the second traveller.
Now divide the phala (i.e., the distance travelled by each
traveller at their first meeting) severally by the padas (for the
two travellers) and take the difference of the two; diminish
that (difference) by half the difference of caya × pada for
the two travellers ; and then divide that by half the diffe-
rence between the cayas (for the two travellers) : the quotient·
gives the days elapsed at the second meeting (of the two
travellers) (since the first meeting).¹
That is to say, the number of days (D) elapsed at the second meeting
of the two travellers since their first meeting is given by
( s / (n - d) - s / n ) - ½ {n f₁ - (n - d) f₂}
D = ---------------------------------------------------- ,
½ (f₁ - f₂)
where f₁ and f₂ are the accelerations of the two travellers, n is the number
of days elapsed at the first meeting since the start of the first traveller,
(n - d) is the number of days elapsed at the first meeting since the start of
the second traveller, and s is the distance travelled by the two travellers at
their first meeting.

¹ Cf. GK, I, p 110, lines 9-14, and p. 111, lines 1-2

80 DETERMINATIONS
Let the initial speeds, etc., of the two travellers be as follows :
Initial speed Acceleration No. of days elapsed since start
at the first at the second
meeting meeting
I v₁ f₁ n n+D
II v₂ f₂ n-d n-d+D
Then s, the distance travelled by the travellers at the first meeting,
and s₁, the distance travelled by them at the second meeting, are given by
s = n/2 [2v₁ + (n - 1) f₁] = (n - d)/2 [2v₂ + (n - d - 1) f₂], (1)
s₁ = (n + D)/2 [2v₁ + (n + D - 1) f₁] = (n - d + D)/2 [2v₂ + (n - d + D - 1) f₂]. (2)
Using (1), (2) can be written as
(n + D)/2 [2s/n + Df₁] = (n - d + D)/2 [2s/(n - d) + Df₂],
whence
D = ((s/(n - d) - s/n) - ½{n f₁ - (n - d) f₂}) / (½ (f₁ - f₂)) .
Ex. 112. After one man had travelled for 6 days with
some (unknown) initial speed (ādi) and acceleration (uttara),
another man went by the same track with (an unknown
initial speed and) acceleration 2 per day per day. Say how
will they meet each other two times (on the way).¹
Rule for finding the amount by which one gambler defeats
his opponent in a play with dice, when the moneys staked at
the successive casts of dice are in an arithmetic progression :
99-101. Diminish the first pada (i.e., the number of
casts of dice won in the beginning by either of the two
persons) by one ; (taking the remainder as the number
of terms) find the sum of the series whose first term (ādi)
and common difference (caya) are each one : this is the vṛddhi
(for the first pada). In regard to the other padas (i.e., the
number of casts of dice won subsequently), take the sum of
the preceding padas for the first term (prabhava), one for the

¹ Cf. GK, I, p. 111, lines 4-7.