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

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