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

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.

PROBLEMS ON ARITHMETIC SERIES 81
common difference (and the padas for the number of terms,
and find the sums of the series. These will give the vṛddhis
for those padas).
Now taking the sum of all the padas (for the pada), find
the sum as in the case of the first pada; then diminish that
(sum) by twice the vṛddhis corresponding to the padas of the
lesser group (i.e. the padas corresponding to the person who
wins lesser number of casts); then multiply (the remainder)
by the (given) common difference; and then add that (pro-

• duct) to the product of the (given) first term (ādi) and the
  difference of the padas (of the greater and lesser groups) :
  this gives the amount by which the person with greater
  number of padas (i.e., casts of dice in his favour) is victorious.
  If that quantity be negative, then it gives the amount by
  which the person with lesser number of padas is victorious.
  Suppose that two persons A and B gamble with dice, and that they
  alternately win p₁, p₂, p₃ and p₄ casts. If the stake-moneys of the casts
  be in the arithmetic progression
  a, a + d, a + 2d, ... ,
  then the amount won by A
  = [a + (a + d) + (a + 2d) +... to p₁ terms]

• [{a + (p₁ + p₂)d} + {a + (p₁ + p₂ + 1)d} +...to p₃ terms].
  = (p₁ / 2) [2a + (p₁ − 1) d] + (p₃ / 2) [2(a + p₁ + p₂ d) + (p₃ − 1)d] ,
  and the amount won by B
  = [{a + p₁d} + {a + (p₁ + 1)d} + {a + (p₁ + 2)d} + .... to p₂ terms]
• [{a + (p₁ + p₂ + p₃)d} + {a + (p₁ + p₂ + p₃ + 1)d} + .... to p₄ terms]
  = (p₂ / 2) [2(a + p₁d) + (p₂ − 1) d] + (p₄ / 2) [2{a + (p₁ + p₂ + p₃)d} + (p₄ − 1)d].
  Suppose that p₁ + p₃ > p₂ + p₄, then the person with greater number
  of casts in his favour, viz. A, is victorious by the amount
  = (p₁ / 2) [2a + (p₁ − 1) d] + (p₃ / 2) [2(a + p₁ + p₂ d) + (p₃ − 1)d]
  − (p₂ / 2) [2(a + p₁d) + (p₂ − 1)d] − (p₄ / 2) [2(a + p₁ + p₂ + p₃ d) + (p₄ − 1)d]
  = a (p₁ + p₃ − p₂ − p₄) + d [(1/2)(p₁ + p₂ + p₃ + p₄ − 1) (p₁ + p₂ + p₃ + p₄)
  − 2 { (p₂ / 2)(2p₁ + p₂ − 1) + (p₄ / 2)(2(p₁ + p₂ + p₃) + p₄ − 1) } ] .
  Hence the rule.
  Indological Truths

82 DETERMINATIONS
Ex. 113. In a gamble two (persons) alternately won
30 , 10 , 100 and 8 casts of dice (with stake-moneys) be-
ginning with 9 and increasing (successively) by 6. Say who
is the winner.
Ex. 114. If the casts of dice (alternately won by the two
persons) be 7, 3, 9 and 12, and the first term and com-
mon difference (of the series formed by the stake-moneys)
as stated before, then say after calculation who wins, if you
know (the method).
Ex. 115. In a gamble two (persons) alternately won
from each other 4, 3, 2 and 2 casts of dice (with stake-moneys)
beginning with 1 and increasing (successively) by 6. Say
who is the winner.
(v) SERIES OF SQUARES, CUBES, AND SUCCESSIVE SUMS OF
NATURAL NUMBERS
r = n r = n
Rule for finding the sum of ∑ r + n² + n³ and of ∑ r² :
r = 1 r = 1
102. The number of terms plus one, as multiplied by
twice the number of terms plus one, being (further) multiplied
by half the number of terms, gives the sum of (i) the sum of a
series of natural numbers (from 1 up to the given number of
terms), (ii) the square of the number of terms, and (iii) the
cube of the number of terms.
That divided by 3 gives the sum of a series of squares of
natural numbers.
That is
r = n
(1)¹ ∑ r + n² + n³ = [(2n + 1) (n + 1) n] / 2 ,
r = 1
r = n
(2)² ∑ r² = [(2n + 1) (n + 1) n] / (2 × 3) .
r = 1

¹ For an alternative formula, see GSS, vi. 296; GK, I, p. 116, lines 3-4.
² Cf. Ā, ii. 22; BrSpSi, xii. 20 (i); SiŚe, xiii. 22 (i); L (ASS), I, p. 113, vs. 19 (i); GK, I, p. 117, lines 1-2.

SERIES OF SQUARES, CUBES AND SUCCESSIVE SUMS 83
Ex. 116 (i). Say what is the sum of (i) the sum of
the first five natural numbers, (ii) the square of 5, and (iii) the
cube of 5.
Ex. 116 (ii). Also (say), if you know, the sum of the
squares of 5 terms of the series whose first term and common
difference are each unity.
Rule for finding the sum of ∑(r = 1)^(r = n) r³ and of ∑(m = 1)^(m = n) ∑(r = 1)^(r = m) r :
103. One-half of what is obtained by adding the
number of terms to the square of the number of terms, when
multiplied by itself, gives the sum of the cubes of natural
numbers (from 1 up to the given number to terms) ; and when
multiplied by the number of terms plus 2 and divided by 3,
gives the sum of the successive sums of those natural numbers.
That is,
(1)¹ ∑(r = 1)^(r = n) r³ = {(n² + n) / 2}² ;
(2)² ∑(m = 1)^(m = n) ∑(r = 1)^(r = m) r = [((n² + n) / 2)(n + 2)] / 3 .
Ex. 117. Friend, quickly say what is the sum of the
cubes of 10 terms of the series whose first term and common
difference are each unity ; and also the sum of the successive
sums of those terms.
Rule for finding the sum of ∑(m = 1)^(m = n) ∑(r = 1)^(r = m) r + ∑(r = 1)^(r = n) r² + ∑(r = 1)^(r = n) r³ :
104. The number of terms, as multiplied by the square
of (the sum of ) the number of terms plus one, when (further)
multiplied by the number of terms plus two and divided by
four, gives the sum of (i) the sum of successive sums of natural
numbers (from 1 up to the given number of terms), (ii) the
sum of squares of those natural numbers, and (iii) the sum of
cubes of those natural numbers.
¹ Cf. Ā, ii. 22(ii) ; BrSpSi, xii. 20(ii) ; GSS, vi. 301 ; SiŚe, xiii. 22(ii);
L(ASS), I, p. 113, vs. 119(ii); GK, I, p. 117, lines 3-4.
² Cf. Ā, ii. 21; BrSpSi, xii. 19; SiŚe, xiii. 21; L(ASS), I, p. 112, vs. 117.
Indological Truths

84 DETERMINTAIONS
That is
m = n r = m r = n r = n n(n+1)² (n+2)
∑ ∑ r + ∑ r² + ∑ r³ = —————————— .
m = 1 r = 1 r = 1 r = 1 4
Ex. 118. Friend, if you know then say after calculation
the sum of (i) the sum of successive sums of the first 6 natural
numbers, (ii) the sum of the squares of the first 6 natural
numbers, and (iii) the sum of the cubes of the first 6 natural
numbers.
(vi) SERIES OF SQUARES AND CUBES, ETC. OF THE TERMS
OF AN ARITHMETIC SERIES
Rule for finding the sum of the squares of the terms of the
r = n
series ∑ {a+(r-1)d} :
r = 1
105. The sum of the ( arithmetic ) series with
twice the common difference, when multiplied by the first
term and (then) increased by the sum of the squares of the
natural numbers ranging from 1 to one-less the number of
terms, as multiplied by the square of the common difference,
gives the sum of the squares of the terms of the (given arith-
metic) series.¹
That is,
r = n
∑ {a+(r-1)d}²
r = 1
= [a+(a+2d) +(a+4d) +...to n terms] × a

• [1² + 2² + 3² +...+ (n-1)²] × d².
  Ex. 119. Tell me the sum of the squares of (the first)
  six terms of the (arithmetic) series whose first term is two and
  common difference three.
  Rule for finding the sum of the sums of series of natural num-
  bers each beginning with 1, whose number of terms are the
  terms of a given arithmetic series :

106. Having obtained the sum of the squares of the
     terms of the (arithmetic) series with (given) first term,

¹ The same rule occurs in GK, I, p. 119, lines 7-8, and p. 120,
lines 1-2. For further rules, see GSS, vi. 298, 299.

SERIES OF SQUARES, CUBES AND SUCCESSIVE SUMS 85
common difference and number of terms, as before, add to
it the sum of the same (arithmetic) series, and (then) reduce
that (resulting sum) to half : the result (thus obtained) is the
sum of sums of the natural series (beginning with 1 and ending
with the terms of the given arithmetic series).¹
That is to say,
(1+2+3+...to a terms) + (1+2+3+... to a+d terms)+
( +2+3+... to a+2d terms)+... to n terms.
r=n {a+ (r - 1)d} {a+(r - 1)d+1}
≡ ∑ ─────────────────────────────
r=1 2
r=n r=n
= ½ [ ∑ (a + r - 1d)² + ∑ (a + r - 1d) ] .
r=1 r=1
Ex. 120. O the best of mathematicians, say the sum of
the sums of series of natural numbers (each beginning with 1),
whose number of terms are the first six terms of the arithmetic
series with 3 as the first term and 5 as the common difference.
Rule for finding the sum of cubes of the terms of an arithme-
tic series :
107. To the square of the sum of the (given arithmetic)
series, as multiplied by the common difference, add the
product of the first term and the sum of the series, as multi-
plied by the first term minus the common difference : the
result is the sum of cubes of the terms of the (arithmetic)
series with given first term and common difference.²
That is,
a³ +(a+d)³ +(a+2d)³ +...+{a+(n - 1)d }³
r=n
= ∑ {a+(r - 1)d}³
r=1
= S² × d + aS × (a - d),
r=n
where S = ∑ {a+(r - 1)d} .
r=1
Ex. 121. Say after adding together the cubes of the
four terms, which begin with 5 and increase successively by 2.
────────────────────────────────────────────────────────────
¹ For other rules, see GSS, vi. 305-305½ and GK, I, p. 117, lines 11-16.
² Cf. GSS, vi. 303; GK, I, p. 121, lines 2-3, and p. 122, lines 1-2.

86 DETERMINATIONS
(3) Determinations pertaining to plane figures
(Kṣetra-vyavahāra)
(i) INTRODUCTION
Fundamental property of rectilinear figures :
108. (In rectilinear figures) the sum of all sides except
one, is neither equal to nor less than the side excepted,
because a curved path is neither less than nor equal to the
straight path.
What the author means here to say is that a rectilinear figure, such
as a quadrilateral or a triangle, is possible only when each of its sides is
less than the sum of the remaining sides.
Nārāyaṇa has made this statement more clearly :
“When in a rectilinear figure, the sum of the other sides is less than
or equal to the greatest side, it is impossible.”¹
Āryabhaṭa II states the same condition as follows :
“When (in a rectilinear figure) each side is less than half the sum
of all the sides, it is possible ; when (on the other hand) half the sum of
all the sides is less than (or equal to) any one of the sides, it is im-
possible.”²
Condition for the existence of ‘the perpendiculars dropped
from the vertices to the base’ and ‘the segments into which
the base is divided by them’ in the case of a quadrilateral :
109. Only in those (quadrilateral) figures in which the
square of base minus face exceeds the product of the difference
and sum of the flank sides, do ‘the perpendiculars dropped
from the vertices to the base’ (lamba) and ‘the segments into
which the base is divided by them’ (abadhā) exist.
Nārāyaṇa has criticised this rule.³ His criticism, however, is not
well-grounded as it is based on an impossible quadrilateral.

¹ GK, II, p. 48, lines 8-11. So also says Bhāskara II. See L (ASS),
pp. 151-152, vs. 163.
² MSi, xv. 64.
³ GK, II, p. 49.