2. The extremes of a geometrical progression are I and 65536, and the ratio 4; what is the sum of the series ? Ans. 87381. 3. The extremes of a geometrical series are 1024 and 59049, and the ratio is ; what is the sum of the series? Ans. 175099. PROBLEM II. Given the first term and the ratio, to find any other term assigned. RULE.* 1. Write down a few of the leading terms of the series, and place their indices over them, beginning with a cypher. 2. Add together the most convenient indices, to make an index less by 1 than the number expressing the place of the term sought. 3. Multiply the terms of the geometrical series together, belonging to those indices, and make the product a dividend. 4. Raise * DEMONSTRATION. In example 1, where the first term is equal to the ratio, the reason of the rule is evident; for as every term is some power of the ratio, and the indices point out the number of factors, it is plain from the nature of multiplication, that the product of any two terms will be another term corresponding with the index, which is the sum of the indices standing over those respective terms. And in the second example, where the series does not begin with the ratio, it appears, that every term after the two first contains some power of the ratio, multiplied into the first term, and therefore the rule, in this case, is equally evident. Y The 4. Raise the first term to a power, whose index is I less than the number of terms multiplied, and make the result a divisor. 5. Divide The following table contains all the possible cases of geometric 5. Divide the dividend by the divisor, and the quotient will be the term sought. NOTE. When the first term of the series is equal to the ratio, the indices must begin with an unit, and the indices added must make the entire index of the term re quired; and the product of the different terms, found as before, will give the term required. EXAMPLES. 1. The first term of a geometrical series is 2, the number of terms 13, and the ratio 2; required the last term. 1, 2, 3, 4, 5, indices. 2, 4, 8, 16, 32, leading terms. Then 4+4+3+2 = index to the 13th term. In this example the indices must begin with 1, and such of them be chosen, as will make up the entire index to the term required. 2. Required the 12th term of a geometrical series, whose first term is 3, and ratio 2. 0, 1, 2, 3, 4, 5, 6, indices. 3, 6, 12, 24, 48, 96, 192, leading terms. Then 6+5 = index to the 12th term. The number of terms multiplied is 2, and 2-1=1 is the power to which the term 3 is to be raised; but the first power of 3 is 3, and therefore 18432+3=6144 the 12th term required. 3. The first term of a geometrical series is 1, the ratio 2, and the number of terms 23; required the last term. QUESTIONS Ans. 4194304. TO BE SOLVED BY THE TWO PRECEDING PROBLEMS. 1. A person, being asked to dispose of a fine horse, said he would sell him on condition of having one farthing for the first nail in his shoes, 2 farthings for the second, one penny for the third, and so on, doubling the price of every nail to 32, the number of nails in his four shoes: what would the horse be sold for at that rate ? Ans. 44739241. 5s. 3d. 2. A young man, skilled in numbers, agreed with a farmer to work for him eleven years without any other reward than the produce of one wheat corn for the first year, and that produce to be sowed the second year, and so on from year to year till the end of the time, allowing the increase to be in a tenfold proportion : what quantity of wheat is due for such service, and to what does it amount * at a dollar per bushel ? Ans. 226056 bushels, allowing 7680 wheat corns to be a pint; and the amount is 226056 dollars.. 3. What |