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

54 DETERMINATIONS
(phala); then having divided them severally by the respec-
tive numbers of spectators, add the preceding results to
the succeeding ones. The results thus obtained being multi-
plied by the numbers of spectators who go away (at the end
of the respective parts of the day), give the amounts to be paid
(to the dancing party by the spectators who go away at the
end of the respective parts of the day).¹
By the word ‘fruit’ (phala) is meant here the amount to be paid to
the dancing party for the whole day.
The above rule is based on the assumption that the spectators
have to pay the dancing party in proportion to the time of seeing the
dance. That is, if the spectators who see the dance for the whole day
pay 96 rūpas in all, then the spectators who see the dance for ¼ of a day
will have to pay 96/4, i.e., 24 rūpas in all.
For the illustration of the rule see under the next example.
Ex. 86-87. One man saw a dance for one quarter of the
day, another for two quarters of the day, another for
three quarters of the day, and (yet) another till the end of the
day. The dancing party has to be paid by them a sum of
ninety six rūpas in all. If payment is to be made in propor-
tion to the time of seeing the dance, how much of that (sum)
should be paid by each of them separately?
Solution. Writing down the parts of the day during which the
various spectators saw the dance, we have
1/4 2/4 3/4 4/4
Diminishing the succeeding part by the preceding one, we get
1/4 1/4 1/4 1/4
Multiplying each of these by the fruit (i.e., by 96), we get
24 24 24 24
Dividing these numbers by the number of spectators who saw the
dance during the respective parts of the day, we get
6 8 12 24
¹ See GSS, vi. 230.

MIXTURES. THE CISTERN PROBLEM 55
Adding the preceding numbers to the succeeding ones, we get
6 14 26 50
Multiplying by the number of spectators who go away at the end of
the respective parts of the day, we get
6 14 26 50
Hence the spectator who saw the dance for ¼ of the day will have
to pay 6 rûpas ; the spectator who saw the dance for ²⁄₄ of the day will have
to pay 14 rûpas ; the spectator who saw the dance for ¾ of the day will
have to pay 26 rûpas ; and the spectator who saw the dance for the whole
day who will have to pay 50 rûpas.
Ex. 88. A palanquin is to be carried to a distance of
3 krośas by 10 men for 100 (rûpas as wages). Of those (men),
2, 3, and 5 stop away after going over 1, 2, and 3 krośas respec-
tively. (Calculate the wages of each of them separately).¹
Ex. 89-90. Five Brâhmaṇas, enchanters of stotras
(“hymns in praise of gods”), were invited by a certain person
to take part in the worship of the five faces of the five-faced
god Śiva on (a remuneration of) 300 rûpas. And they, on the
completion of worship of one, two, three, four, and five faces
respectively, went away (from the place of worship) one by
one.² Say what are their remunerations (separately).
(vii) THE CISTERN PROBLEM
Rule for finding the time in which a cistern would be filled
up by a number of drains flowing into it simultaneously, when
the times in which the cistern is filled up by each drain
separately are given :
69. Having divided unity severally by the (given)
fractions (of time), take the sum of the quotients and by that
sum divide unity (again) : this will give the time in which
the cistern is filled up (when all the drains are simultaneously
opened to flow into it).³
¹ Cf. GSS, vi. 231-232.
² What is meant is this : when the worship of the first face was com-
pleted by all the five Brâhmaṇas, one Brâhmaṇa left the place of worship ;
when the worship of the second face was completed by the remaining four
Brâhmaṇas, one more Brâhmaṇa left the place of worship ; when the
worship of the third face was completed by the remaining three Brâhmaṇas,
one more Brâhmaṇa left the place of worship ; and so on.
³ Cf. MSi, xv. 43; L (ASS), p. 91, vs. 96 ; GK, I, p. 94, line 2-3.

56 DETERMINATIONS
That is, if n drains severally fill up a cistern in
a₁/b₁, a₂/b₂, ... , aₙ/bₙ of a day,
then all the drains working together will fill up the cistern in
1 / (b₁/a₁ + b₂/a₂ + ... + bₙ/aₙ) of a day.
Ex. 91. In what time will the (four) drains, which
severally fill up a cistern in ½, ¼, ⅕, and ⅙ of a day, fill up
that cistern if they are opened simultaneously (to flow into
it) ? 1
(viii) WAGES PAID FROM THE COMMODITY
Rule for finding the wages of a porter who carries a commo-
dity over a part of the stipulated distance, when the wages
are to be paid out of the commodity itself :
70. (To obtain the wages for a part of the stipulated
distance) divide half the product of the commodity and the
stipulated distance as diminished by the square root extracted
from the square of half the product of the commodity and the
stipulated distance as diminished by the (continued) product
of the distance (already) gone over, the distance to be gone
over, the commodity, and the stipulated wages, by the
distance which is still to be gone over.²
Assuming that the (stipulated) wages for carrying a articles over the
(stipulated) distance d are w articles (out of those a articles), let the
wages for carrying those a articles over a distance d₁, where d₁ < d, be
x articles (out of those a articles). Then calculating at the same rate, the
wages for carrying the remaining a - x articles over the remaining distance
d - d₁ are equal to
(a - x) (d - d₁)x / (ad₁) articles.
But the remaining wages being w - x articles, we have
(a - x) (d - d₁)x / (ad₁) = w - x ,
1 Similar examples occur in Pṛthûdaka Svâmî's (860 A.D.) comm.
on BrSpSi, xii. 9; L (ASS), p. 91, vs. 97 ; GK, I, p. 94, lines 5-6.

• Cf. GSS, vi. 226. For another rule, see GK, I, p. 103, lines 10-15.

• MIXTURES. WAGES PAID FROM COMMODITY 57
  or, (d - d₁) x² - adx + awd₁ = 0 ,
  whence x = [ad/2 ± √((ad/2)² - awd₁ (d - d₁))] / (d - d₁) .
  Taking the negative sign before the radical, we have the rule stated
  above. The positive sign is rejected because it makes x > w.
  Ex. 92. If for carrying 24 jack-fruits over a distance of
  5 krośas (a porter) is to get 9 of those jack-fruits, what will
  he get if he carries them over a distance of 2 krośas (only) ?¹
  Rule for finding the distances over which two porters carry
  a commodity in turn, when the wages paid to the porters
  out of the commodity carried by them are given :

71. The first and the second loads (i.e., the loads carried
    by the first and the second porters), each multiplied by
    the wages for the other, being added together, the resul-
    ting sum is (to be taken as) the divisor (of the following
    results) : the results obtained by multiplying the products
    of the wages received (for each of the loads) and the
    (total) distance (gone over by the porters), each by the other
    load. This gives the distances (gone over by the first and the
    second porters separately).²
    Suppose that the first porter carries a articles over a distance x
    (unknown) and gets b articles out of those a articles as wages, and that
    the second porter carries the remaining a - b articles over the remaining
    distance d - x and gets c articles as wages. Then, we have
    c = b(a - b)(d - x) / ax ,
    whence x / ((a - b)b) = (d - x) / ac = d / (ac + (a - b)b)
    Hence the above rule.
    Ex. 93-94. Twenty-four jack-fruits were carried (over
    a certain distance) by one (person) for 4 out of those (24) jack-
    fruits as wages ; the remaining jack-fruits were carried (over
    the remaining distance) by another person for 5 of them (as
    wages). The load was thus carried by the two (person) over

¹ Cf. GSS, vi. 227. Also see GK, I, p. 103, lines 17-20.
² See GSS, vi. 228.

58 DETERMINATIONS
a distance of 5 krośas (in all). Say, O learned, how much of
that distance was gone over by each of them ?¹
(ix) COMBINATIONS OF SAVOURS
Rule for finding the number of combinations that can be
formed out of a given number of savours (or things) by
taking 1, 2, 3, ...., all at a time :
72. Writing down the numbers beginning with 1 and
increasing by 1 up to the (given) number of savours in the
inverse order, divide them by the numbers beginning with 1
and increasing by 1 in the regular order, and then multiply
successively by the preceding (quotient) the succeeding one.
(This will give the number of combinations of the savours
taken 1, 2, 3, ... , all at a time respectively).²
Let the given number of savours be six. Then the numbers
1, 2, 3,..., 6 are written down in the inverse order thus :
6 5 4 3 2 1
These are divided by the numbers 1, 2, 3, ... 6 respectively, so that
we get
6/1, 5/2, 4/3, 3/4, 2/5, 1/6.
Now multiplication is made successively by the preceding quotient of the
succeeding one. Thus we get
6/1, 6/1 × 5/2, 6/1 × 5/2 × 4/3, 6/1 × 5/2 × 4/3 × 3/4, 6/1 × 5/2 × 4/3 × 3/4 × 2/5,
6/1 × 5/2 × 4/3 × 3/4 × 2/5 × 1/6.
These, according to the rule, are respectively the numbers of combinations
of the six savours taken 1, 2, 3,..., 6 at a time.
Ex. 95. Friend, a cook prepares varieties of food with
the six savours, pungent, bitter, astringent, acid, saline, and
sweet. Say what is the (possible) number of varieties ?³

¹ A similar problem occurs in GSS, vi. 229.
² Cf. GSS, vi. 218 ; MSi, xv. 45(ii)-46 ; L (ASS), p. 106, vv. 112-113
(i); GK, II, pp. 318-319, vs. 58.
³ The same example occurs in GSS, vi. 219 ; GK, II, p. 319, Ex. 22.

MIXTURES. SPECIAL PROBLEMS 59
Rule for writing down the combinations of 2 or more savours
in a serial order :
73. To get the 2-savour combinations add the pre-
ceding savour to all the succeeding one’s in order ; and to get
the combinations of 3 or more savours add the preceding
savour to the (succeeding) combinations of 2 or more savours
which do not contain the preceding savour.
Let the six savours be denoted by
a b c d e f .
Then the 2-savour combinations are :
ab, ac, ad, ae, af ; bc, bd, be, bf ;
cd, ce, cf ; de, df ; ef.
The 3-savour combinations are :
abc, abd, abe, abf, acd, ace, acf, ade, adf, aef ;
bcd, bce, bcf, bde, bdf, bef ; cde, cdf, cef ; def.
The 4-savour combinations are :
abcd, abce, abcf, abde, abdf, abef, acde, acdf ;
acef, adef ; bcde, bcdf, bcef, bdef, cdef.
The 5-savour combinations are :
abcde, abcdf, abcef, abdef, acdef, bcdef.
The 6-savour combination is :
abcdef.
(x) CERTAIN SPECIAL TYPES OF PROBLEMS
Rule for solving problems on pillars (stambhoddeśaka¹) or
problems involving remainders (śeṣoddeśaka) :
74(i). To solve problems on pillars or problems involv-
ing remainders, divide the visible quantity (dṛśya) by one minus
(the sum of) the fractional parts (of the whole).²
¹ Mahāvīra calls stambha-jāti by the name bhāga-jāti, and Śrīpati
by the name dṛśya-jāti. See GSS, iv. 4(i) and GT, p. 41, vs. 55.
² Cf. GSS, iv. 4(i); GT, p. 41, vs. 55. Also see GK,I, p. 17, lines 14-19.
Mahāvīra gives a separate rule for the śeṣa-jāti. See GSS, iv. 4(ii). The
same rule is given by Āryabhaṭa II and Śrīpati. See MSi, xv. 20; GT,
p. 44, line 11.

50 DETERMINATIONS
Ex. 96. One-fourth, one-third, and one-sixth of a pillar
are respectively buried under the water, mud, and sand of a
river, and three cubits (of the pillar) are visible. Give out
the measure (of the length) of that (pillar).¹
Here the given fractions (of the pillar) are ¼, ⅓, and ⅙ ; and the
visible quantity (i.e., the visible part of the pillar) is equal to ‘ 3 cubits.’
Hence the length of the pillar
3
= ————————————————— , i.e., 12 cubits.
1 - (¼ + ⅓ + ⅙)
Ex. 97. After giving away one-half of a quantity, then
⅔ of what remains, then ¾ of what remains thereafter, and
then ⅘ of what remains thereafter, the residue left is 3.
(What is that quantity ?)²
Here the fractions (of the whole quantity) given away are
½, ⅔ of (1 - ½), ¾ of (1 - ½) (1 - ⅔) , and
⅘ of (1 - ½) (1 - ⅔) (1 - ¾) ,
i.e., ½, ⅓, ⅛, and ¹/₃₀ ;
and the visible quantity (i.e., the residue left) is 3 .
Hence the required quantity
3
= ——————————————————————— , i.e., 360 .
1 - (½ + ⅓ + ⅛ + ¹/₃₀)
Rule for solving problems involving differences (viśeṣoddeśaka):
74(ii). On subtracting the smaller quantity from the
greater quantity what remains is called the difference
(viśeṣa).³ (This having been done) the procedure (to be adop-
ted is the same as the other one (i.e., the previous one).⁴
Ex. 98. Of a herd of cows, one-half went away towards
the east and one-fourth towards the west, the difference of the

¹ Cf. GSS, iv. 5; GT, p. 41, vs. 56. For other examples see GSS, iv.
6-22 ;GT, p. 42, vs. 57 ; L (ASS), pp 47-48, vs. 53; GK, I, p. 20, lines 4-7,
10-13, 16-19.
² For other examples, see GSS, iv. 29-30, 31, 32; GT, p. 44, lines
20-23 ; p. 45, lines 16-19; L (ASS), pp. 49-51, vs. 54.
³ Mahāvīra does not differentiate between bhāga-jāti and viśeṣa-jāti
(or viśleṣa-jāti)). So he sets problems on the latter under the bhāga-jāti.
See GSS, iv. Ex. 23-27.
⁴ Cf. GT, p. 46, lines 10-13.

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) .