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

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.

PLANE FIGURES 87
Enumeration of primary plane figures :
110–111. The rectangular quadrilateral, the equilateral
quadrilateral, the equi-bilateral quadrilateral, the equi-trila-
teral quadrilateral, the inequilateral quadrilateral, the equi-
lateral triangle, the scalene triangle, the isosceles triangle, the
circle, and the segment of a circle—these are the ten (primary)
plane figures ; the areas of these (figures) should be determined
by applying their own rules. And by considering (the shape
in terms of ) these (figures) should be obtained the areas of
other figures, such as (those of the shape of ) an elephant's
tusk, a felloe, etc.
Nârâyaṇa, too, like Śrīdhara takes the primary plane figures to be 10
in number. But his list of primary plane figures differs from that of
Śrīdhara in so far as "the segment of a circle" has been replaced by him
by "the conch-figure."
Mahâvîra gives the following list of 16 primary plane figures : .
(1-3) Three varieties of triangles, equilateral, isosceles, and scalene ;
(4-8) Five varieties of quadrilaterals, equilateral, equidichastic (dvi-dvi-
sama), equibilateral, equitrilateral, and inequilateral ;
(9-16) Eight varieties of curvilinear figures, a circle, a semi-circle, an
ellipse, a conchiform area, a concave circle, a convex circle, an
outlying annulus, and an in-lying annulus.
Mahâvîra’s classification of quadrilaterals deserves special attention,
as it differs from that of Śrīdhara. Śrīdhara’s commentator has referred
to this classification, and has tried to justify Śrīdhara’s classification.
On a formula for the area of a triangle or a quadrilateral :
112–114. (It is said that) the product of half the sums
of the sides and counter sides (i.e., the product of half the
sum of the base and face and half the sum of the flank sides)
of a triangle or a quadrilateral, gives the gross value of the
area.¹ But this result is true only for those figures in which the
difference between the altitude and the flank sides is small.
In the case of other figures the above result is far removed from
¹ Reference is evidently to BrSpSi, xii. 21(i) where this rule is given.
This rules occurs also in GSS, vii. 7(i), MSi, xv. 66. and GK, II, p. 3, vs. 8.

88 DETERMINATIONS
the truth ; as for example, in the case of the triangle having 13
for the two (flank) sides and 24 for the base, the gross area is
156, whereas the correct area is 60.
I shall therefore state the methods for obtaining the
accurate results only.
(ii) AREA OF THE QUADRILATERAL WITH EQUAL
ALTITUDES AND OF THE TRIANGLE
Rule for finding the area of a quadrilateral in which the per-
pendiculars dropped from the vertices to the base are equal,
and that of a triangle :
115. In the case of a quadrilateral, in which the per-
pendiculars dropped from the vertices to the base are equal,
and a triangle, half the sum of the base and the face, multi-
plied by the altitude, gives the area.¹
That is,
(i) area of a quadrilateral with parallel base and face
base + face
= ——————————— × altitude;
2
(ii) area of a triangle = ½ base × altitude.
Ex. 122. In an equilateral quadrilateral, the face, the
base, and the altitude are all equal to the flank sides, each
being 1½ cubits in length. Say, friend, what is the area of
that (quadrilateral).
Ex. 123. Give out the area of that rectangular quadri-
lateral in which the base and face are each 5½ cubits, and the
flank sides and altitude each 3 cubits.
Ex. 124. In a triangle the (flank) sides are 4 - ¼ and
3¼ cubits, the base is 3½ cubits, and the altitude is 3 cubits.
What is the area of that ?
Ex. 125. In an equilateral triangle the base is 8½ cubits,
and the altitude is 7 cubits and 8⅔ aṅgulas. What is the area
thereof ?
―――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――
¹ The same rule is given in MSi, xv. 78; SiŚe, xiii. 30; L(ASS), p. 162,
line 3; GK, II, p. 42, lines 10-11; and page 50, lines 13-14.

PLANE FIGURES 89
Ex. 126. If you know the method of finding the area
of plane figures, say the area of the isosceles triangle, whose
(flank) sides are (each) 5 cubits, altitude is 3 cubits, and base
8 cubits.
Ex. 127. In a quadrilateral, the face is 1⅓ cubits, the
base is 9⅓ cubits, the (flank) sides are (each) 5 cubits, and
the altitude is 3 cubits. What is the area of that ?
Ex. 128. Say what will be the area of the (quadrila-
teral) figure, whose base is 39 (cubits), (flank) sides and face
are (each) 25 (cubits), and altitude is 24 (cubits).
Ex. 129–30. In an inequilateral quadrilateral with equal
altitudes, the base is 10 cubits, the face is 4⅛ (cubits), the flank
sides are 9–⅓ and 6+½ (cubits), and the altitude is 6½ cubits
minus 1/60 of an aṅgula. What is the area thereof ?
Instruction regarding plane figures of the shape of an elep-
hant’s tusk, a felloe, a crescent moon, and a thunderbolt :
116. A figure of the shape of an elephant’s tusk (may
be considered) as a triangle, of a felloe as a quadrilateral,
of a crescent moon as two triangles, and of a thunderbolt as
two quadrilaterals.¹
Ex. 131. What is the area of (the figure of the shape of)
an elephant’s tusk whose base is 2 cubits and altitude 3 cubits ;
and also of the figure of the shape of a felloe whose base and
face are (each) 3 cubits and altitude is 10 cubits ?
Ex. 132. The central length of (a figure of the shape
of ) a crescent moon is 8 cubits, and the central width 3
cubits. Treating it as made up of a pair of triangles, quickly
say what its area is ?
Ex. 133. In (a figure of the shape of ) a thunderbolt,
the central length is 10 cubits, the faces are each 5 cubits,
and the central width is 2 cubits. What is its area, if it be
regarded as made up of two quadrilaterals ?
¹ Similar statements are made in MSi, xv. 101 and GK, II, p. 10,
lines 2-4.

90 DETERMINATIONS
(iii) AREA OF THE QUADRILATERAL WITH UNEQUAL
ALTITUDES
Rule for finding the area of a quadrilateral with unequal
altitudes, when its sides are given :
117. Set down half the sum of the (four) sides (of the
quadrilateral) in four places, (then) diminish them (respec-
tively) by the (four) sides (of the quadrilateral), (then) multi-
ply (the resulting numbers) and take the square root (of the
product) : this gives the area of quadrilaterals having (two or
more) equal sides but unequal altitudes and also of quadri-
laterals having unequal sides and unequal altitudes.¹
That is to say, if a, b, c, d be the sides of a quadrilateral with un-
equal altitudes, and s half the sum of those sides, then the area of the
quadrilateral
= √((s - a) (s - b) (s - c) (s - d)) .
Our author, like Brahmagupta and other early Hindu mathema-
ticians, has committed here an error in declaring the above formula as
applicable to all quadrilaterals (with unequal altitudes), when in fact it is
applicable to cyclic quadrilaterals only. The formula which is applicable
to all quadrilaterals, in general, of which it is only a particular case, is
as follows :
area of a quadrilateral = √{(s - a) (s - b) (s - c) (s - d) - abcd cos²α}
where a, b, c, d are the sides of the quadrilateral, s is half the sum of those
sides, and α is half the sum of a pair of opposite angles of the quadri-
lateral.
The earliest Hindu mathematician who pointed out the inapplicabi-
lity of the above formula to a quadrilateral having only its sides given was
Âryabhaṭa II (c. 950 A.D.). He writes :

¹ According to Brahmagupta and Mahâvîra, this rule gives the
accurate area for a quadrilateral in general. See BrSpSi, xii. 21(ii) ; and
GSS, vii. 50 (ii). The same rule occurs also in SiŚe, xiii. 28. Âryabhaṭa II
has pointed out that this rule gives the accurate area in the case of a
triangle only and not in the case of a quadrilateral. See MSi, xv. 69. So
also write Bhâskara II and Nârâyaṇa. See L (ASS), p. 156, vs. 169 ;
and GK, II, p. 39, lines 13-14 and p. 40, lines 1-2.

PLANE FIGURES 91
“The mathematician who wishes to tell the area or the altitudes
of a quadrilateral without knowing a diagonal, is either a fool or a
blunderer.”¹
Rule for finding an approximate value of the square root of
a non-square number :
118. Of the non-square number as multiplied by some
large square number, extract the square root, neglecting the
remainder ; and divide that by the square root of the multi-
plier.²
That is to say, multiply the non-square number by some large
square number (chosen at pleasure) ; then find out the square root of
that product, neglecting the remainder ; and then divide that square
root by the square root of (the square number, which was taken as)
the multiplier.
We illustrate this rule by the following two examples :
Ex. 1. Show that √10 = 3⅐ approximately.
Choose 49 for the square number. Then multiplying 10 by 49 , we
get 490. Extracting the square root of 490, we get 22 as the square root
and 6 as the remainder. Neglecting the remainder, and dividing 22 by
the square root of 49, i.e., by 7 , we get 3⅐ .
Ex. 2. Find the square root of 3 , correct to 2 decimal places.
Choose 1000000 for the square number. Then multipling 3 by
1000000 , we get 3000000 . Extracting its square root we get 1732 as
the square root and 175 as the remainder. Neglecting the remainder, and
dividing 1732 by the square root of 1000000, i.e., by 1000, we get 1·732.
Hence the square root of 3 correct to 2 decimal places is 1·73 .
¹ कर्णज्ञानेन विना चतुरश्रे लम्बकं फलं यद्वा ।
वक्तुं वाञ्छति गणको योऽसौ मूर्खः पिशाचो वा ॥ (MS, xv. 70)
Also see L (ASS), pp. 159–160.
² For similar rules, see MSi, xv. 55, or Siśe, xiii. 36 ; L (ASS),
p. 132, vs. 140 ; GK, II, p. 33, lines 4-7.

92 DETERMINATIONS
It is evident that the larger the square number is chosen, the better
will be the approximation. Hence the above rule.
In the Bakhshālī Manuscript the following formula is used for finding
an approximate value of the square root of a non-square number :
√(a² + b) = a + b/(2a) − ((b/(2a))²) / (2(a + b/(2a))) appr

ANSWERS TO EXAMPLES 93

1. 55 ; 210 ; 465 ; 820 ; 1275 ; 1830 ; 2485 ; 3240 ; 4095 ;
2. 10 ; 20 ; 30 ; 40 ; 50 ; 60 ; 70 ; 80 ; 90 ; 100.
3. 4995 ; 4840 ; 4585 ; 4230 ; 3775 ; 3220 ; 2565 ; 1810 ; 955 ; 0.
4. 27216 ; 33152 ; 483900 . 4. 1, 4, 9, 16, 25, 36, 49, 64, 81, 625, 1296,
   3969, 186624, and 60871204. 5. 1 ; 8 ; 27 ; 64 ; 125 ; 216 ; 343 ; 512 ;
   729 ; 3375 ; 16777216 ; and 8365427. 6. 1 1/12 ; 11 1/4 . 7. 1 7/8 ; 3/8 ; 3/4 .
5. 1/4 ; 1/6 . 9. 13 1/2 . 10. 3 3/4 ; 150 2/5 . 11. 2 1/2 ; 17 5/14 . 12. 6 1/4 ;
   232 9/16 ; 1/4 ; 1/9 . 13. 421 7/8 ; 5132 61/64 ; 5/64 ; 2/7 . 14. 1 9/20 ; 5 1621/25200 .
6. 3103/25200 . 16. 30 . 17. 5 11/16 . 18. 15 1/12 . 19. 5 15/16 . 20. 12 11/12 .
7. 1 13/16 . 22. 5 347/6400 purâṇas. 23. 4 23/48 . 24. 10 11/12 . 25. 7 7/10
   paṇas ; or 4 purâṇas, 13 paṇas, 2 kâkiṇîs, and 16 varâṭakas. 26. 10 1/4 palas ;
   or 10 palas, 1 karṣa, 3 mâṣas, and 4 8/9 guñjâs. 27. 87 21/31 . 28. 2 droṇas,
   1 âḍhaka, and 2 177/256 prasthas. 29. 3 158/159 rûpas. 30. 8 months and 26 2/3 days.
8. 33600 years. 32. 9 34/49 days. 33. 19 4/11 days. 34. 26 2/5 necklaces.
9. 250 suvarṇas. 36. 244 suvarṇas, 5 mâṣas, and 4 7/11 guñjâs.
10. 450 blankets. 38. 122 suvarṇas, 8 mâṣas, and 4 24/49 guñjâs.
11. Interest = 36. 40. 20 135/136 . 41. 2362 1/2 . 42. 1 111/832 . 43. 38 1/4 paṇas ;
    or 2 purâṇas, 6 paṇas, and 1 kâkiṇî. 44. 60 . 45. 33 3/4 . 46. 49 7/8 .
12. 3 droṇas, 0 âḍhaka, 3 prasthas, and 1 5/7 kuḍavas. 48. 2 palas of long
    pepper. 49. 25 wood-apples. 50. 64. 51. 200. 52. Capital = 60 ;
    Interest = 36. 53. Capital = 33 1/3 ; Interest = 3. 54. Capital = 500 ;
    Interest = 300 ; share of surety = 60 ; share of calculator = 30 ; and share
    of scribe = 15. 55-56. The debtor is relieved of his debt in 2 months
    and 21 1371/1789 days. In this time the rich man gets 109 1371/1789 rûpas in all, the
    profit being 9 1371/1789 rûpas. 57-58. Principal = 1000 ; time = 8 months
    and 3 days ; and rate per cent. per month = 4. 59. Principal = 100 ;
    time = 8 months and 17 days ; and rate per cent. per month = 4 1/2 .
    60(i). 20 months. 60(ii). 7 months, 4 days, 17 ghaṭikâs, and 8 4/7 caṣakas.
13. 11 5/31 varṇas. 62. 9 15/55 varṇas. 63. 13 11/17 varṇas. 64. 16 mâṣas.
14. 11 3/4 varṇas. 66. 5 mâṣas.
    67-68. varṇa gold of varṇa 16 gold of varṇa 10
    16 2 mâṣas 0 mâṣas
    15 3/4 23/12 ,, 1/12 ,,
    15 1/2 22/12 ,, 2/12 ,,
    15 1/4 21/12 ,, 3/12 ,,
    15 20/12 ,, 4/12 ,,
    Indological Truths

94 ANSWERS
14¾ 19/12 ,, 5/12 ,,
14½ 18/12 ,, 6/12 ,,
14¼ 17/12 ,, 7/12 ,,
14 16/12 ,, 8/12 ,,
13¾ 15/12 ,, 9/12 ,,
13½ 14/12 ,, 10/12 ,,
13¼ 13/12 ,, 11/12 ,,
13 12/12 ,, 12/12 ,,
12¾ 11/12 ,, 13/12 ,,
12½ 10/12 ,, 14/12 ,,
12¼ 9/12 ,, 15/12 ,,
12 8/12 ,, 16/12 ,,
11¾ 7/12 ,, 17/12 ,,
11½ 6/12 ,, 18/12 ,,
11¼ 5/12 ,, 19/12 ,,
11 4/12 ,, 20/12 ,,
10¾ 3/12 ,, 21/12 ,,
10½ 2/12 ,, 22/12 ,,
10¼ 1/12 ,, 23/12 ,,
10 0 2 ,,
69. One is of weight 2 māṣas and varṇa 12 ; the other is of weight 3 māṣas
and varṇa 8. 70. One is of weight 5 māṣas and varṇa 14 ; the other is of
weight 7 māṣas and varṇa 10. 71. 30 prasthas ; 45 prasthas ; 75 prasthas ;
and 60 prasthas respectively. 72. 900 prasthas, 600 prasthas, and 200 prasthas
respectively. 73-74. 49/32 kuḍavas of mudga (seeds of phaseolus mungo) and
49/64 kuḍavas of rice, the respective prices being 63/32 paṇas and 49/32
paṇas. 75. 14/37 palas of asafoetida, 14/37 palas of long pepper, and 14/37
palas of dry ginger, the respective prices being 28/37 paṇa, 7/37 paṇa, and
2/37 paṇa.
76. There will be infinite solutions. The commentator obtains two
solutions in the first case, viz.,
rate of purchase : 17 artieles for 1 } , 20 articles for 1 } ,
rate of sale : 6 artieles for 1 } 7 articles for 1 }
and one solution in the second case, viz.,
rate of purchase : 240 articles for 1,
rate of sale : 7 articles for 1.
77. One solution as given by the commentator is :
rate of purchase : 40 articles for 1,
rate of sale : 22 articles for 1.

ANSWERS 95
78-79. The following are the 16 valid integral solutions :
(1) (2) (3) (4)
Number Price Number Price Number Price Number Price
Pigeons 15 9 | 55 33 5 3 | 30 18
Cranes 28 20 | 21 15 56 40 | 21 15
Swans 45 35 | 9 7 27 21 | 36 28
Peacocks 12 36 | 15 45 12 36 | 13 39
(5) (6) (7) (8)
Number Price Number Price Number Price Number Price
Pigeons 10 6 20 12 15 9 | 35 21
Cranes 42 30 14 10 63 45 | 42 30
Swans 36 28 54 42 9 7 | 9 7
Peacocks 12 36 12 36 13 39 | 14 42
(9) (10) (11) (12)
Number Price Number Price Number Price Number Price
Pigeons 20 12 | 40 24 60 36 | 25 15
Cranes 49 35 | 28 20 7 5 | 35 25
Swans 18 14 | 18 14 18 14 | 27 21
Peacocks 13 39 | 14 42 15 45 | 13 39
(13) (14) (15) (16)
Number Price Number Price Number Price Number Price
Pigeons 45 27 35 21 5 3 | 10 6
Cranes 14 10 7 5 21 15 | 7 5
Swans 27 21 45 35 63 49 | 72 56
Peacocks 14 42 | 13 39 | 11 33 | 11 33
Of these sixteen solutions, the first four have been given by the com-
mentator.
80. The commentator gives the following solutions :
(1) (2) (3) (4) (5)
Fruits Prices Fruits Prices Fruits Prices Fruits Prices Fruits Prices
Pomegranates 16 32 | 17 34 | 19 38 | 18 36 | 15 30 |
Mangoes 60 36 | 45 27 | 15 9 | 30 18 | 75 45 |
Wood-apples 24 12 | 38 19 | 66 33 | 52 26 | 10 5 |
81-82. After 8 days, 21 1/11 ghaṭikās. 83. The meeting takes place
after 20 days when the travellers are at a distance of 40 yojanas from the
starting place. 84-85. 2 paṇas and 3 kākiṇīs. 86-87. 6, 14, 26, and
50 respectively. 88. The two persons stopping away after going
over 1 krośa receive 6 2/3 ; the three persons stopping away after going over
2 krośas get 22 1/2 ; and the five persons who go over 3 krośas receive 70 5/6 .
89-90. 12, 27, 47, 77, and 137 respectively. 91. 1 1/7 of a day.

96 ANSWERS
92. 4 jack-fruits. 93-94. 2 krośas and 3 krośas respectively.
95. 63. 96. 12 cubits. 97. 360. 98. 20 cows. 99. 36.
100. 9. 101. 25. 102. 24/5. 103(i). 40 ; 8/3. 103(ii). 0.
104-105. 2 paṇas and 2¾ kākiṇīs. 106. 29½. 107. 6¾. 108.
206158430208. 109. 93. 110. 252 paṇas. 111. 15 days. 112.
If the first traveller travel with speed 1 and acceleration 6 and meet
the second traveller for the first time after 10 days, then the second
meeting of the two travellers will occur 8 days after the first meeting.
113. The first man wins, the amount won being 48360. 114. The
second man wins, the amount won being 101⁷. 115. The second
man wins, the amount won being 25. 116(i). 165. 116(ii). 55.
117. 3025 ; 220. 118. 588. 119 699. 120. 986. 121.
2528. 122. 2¼ square cubits. 123. 15½ square cubits. 124.
5¼ square cubits. 125. 31 ⁴¹/₁₄₄ square cubits. 126. 12 square
cubits. 127. 16 square cubits. 128. 768 square cubits.
129-130. 46 ¹²⁷/₁₄₅₆ square cubits. 131. 3 square cubits ; 30 square
cubits. 132. 12 square cubits. 133. 35 square cubits.