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 :

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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.