term, the common ratio and the number of terms thereof being 3, 5 and 15 (respectively) ?
The rule for finding out the common ratio and the first term in relation to the (given) sum of a series in geometrical progression :--
101. That (quantity) by which the sum of the series divided by the first term and (then) lessened by one is divisible throughout (when this process of division after the subtraction of one is carried on in relation to all the successive quotients) time after time--(that quantity) is the common ratio. The sum, multiplied by tho common ratio lessened by one, and (then) divided by that self-multiplied product of the common ratio in which (product) that (common ratio) occurs as frequently as the number of terms (in the series), after this (same self-multiplied product of the common ratio) is diminished by one, gives rise to the first term.
102. When the first term is 3, the number of terms is 6, and the sum is 4095 (in relation to a series in geometrical progression), what is the value of the common ratio ? The common ratio is 6, the number of terms is 5, and the sum is 3110 (in relation to another series in geometrical progression). What is the first term here ?
101. The first part of the rule will become clear from the following
example:-
The sum of the series is 4095, the first term 3, and the number of terms 6. Here, dividing 4095 by 3 we get 1365. Now, . Choosing by trial 4, ; ; ; ; ; ; ; . Hence 4 is the common ratio. The principle on which this method is based will be clear from the following:--
- ; and , which is obviously divisible by r
- The second part expressed algebraically is .
