As in arithmetical progressions, all the questions which can be proposed for solution in geometric progressions reduce to 10, the solutions of which are deduced from 1 from which it appears generally that as the number increases, the difference of the logarithms of x and 1+x diminishes. Also, since is greater than the whole series, - being diminished by more than it is increased, we have If x consist of five places, its least value is 10000. Therefore the greatest value of (1+x)-l is less than -0.00005. 1 20000 Hence we may infer that the logarithms of every two consecutive whole numbers consisting of five places must agree in the first four decimal places at least. If a consist of five places, its least value is 10000, and, therefore, the greatest value of A-A' is less than 1 1 20000 X 10002 200040000' which, when reduced to a decimal, has no significant figure within the first eight places. Hence, in tables which extend only to seven places, we may assume that A-A=0, or A=A'. Thus we infer that, under the circumstances which have been supposed, the logarithms of numbers in arithmetical progression will themselves be in arithmetical progression. Let now n and n+1 be two consecutive whole numbers, and n+ an intermediate fraction. These may be looked upon as three terms of an arithmetical progression, whose first term is n, whose common difference is, whose (p+1)th term is n+, and whose (9+1)th q 1 term is n+1. By what has been already shown, the logarithms of the several terms of this series will also be in arithmetical progression. Let & be their common difference. The (p+1)th term of this series will be In+p8, which will be the logarithm of the (p+1)th term of the former series; 234. A series of quantities is called a harmonical progression when, if any three consecutive terms be taken, the first is to the third as the difference of the first and second to the difference of the second and third. Thus, if a, b, c, d.... be a series of quantities in harmonical progression, we shall have a:c::a-b:b-c; b:d::b-c:c-d, &c. 235. The reciprocals of a series of terms in harmonical progression are in arithmetical progression. Let a, b, c, d, e, f.... be a series in harmonical progression. Hence the differences of the logarithms are as the differences of the numbers. or a:c::a-b:b-c; b:d::b-c:c-d; c:e::c-d:d-e, &c. ac ce ed cde' To insert m harmonic means between a and b. Since the reciprocals of quantities in harmonical progression are in arithmetical progression, let us insert m arithmetic means between and Generally, in arithmetical progression, l=a+(n−1)8 1 1 a ō 236. THE solution of all questions connected with interest and annuities may be greatly facilitated by the employment of the algebraical formulæ. In treating of this subject we may employ the following notation: Let p dollars denote the principal. r the interest of $1 for one year. i the interest of p dollars for t years. s the amount of p dollars for t years at the rate of interest denoted by r. t the number of years that p is put out at interest. SIMPLE INTEREST. PROBLEM I.-To find the interest of a sum p for t years at the rate r. Since the interest of one dollar for one year is r, the interest of p dollars for one year must be p times as much, or pr; and for t years t times as much as for one year; consequently, PROBLEM II.-To find the amount of a sum p laid out for t years at simple interest at the rate r. The amount must evidently be equal to the principal, together with the interest upon that principal for the given time. Required the interest of $873.75 for 21 years at 43 per cent. per annum. i=ptr. The amount of the above sum at the end of the given time will be s=d+dtr $873.75+$103.757. PRESENT VALUE AND DISCOUNT AT SIMPLE INTEREST. The present value of any sum s due t years hence is the principal which in the time t will amount to s. The discount upon any sum due t years hence is the difference between that sum and its present value. PROBLEM III.-To find the present value of s dollars due t years hence, simple interest being calculated at the rate r. By formula (2) we find the amount of a sum p at the end of t years to be s=p+ptr. Consequently, p will represent the present value of the sum s due t years hence, and we shall have p=1+tr (3) for the expression required. * r is the interest of $1 for one year. To find the value of r when interest is calculated at the rate of $41 or $4.75 per cent. per annum, we have the following proportion: PROBLEM IV. To find the discount on s dollars due t years hence, at the rate r, simple interest. Since the discount on s is the difference between s and its present value, we shall have Required the discount on $100, due 3 months hence, interest being calculated at the rate of 5 per cent. per annum. PROBLEM V. To find the amount which must be paid at the end of t years, for the enjoyment of an annuity a, simple interest being allowed at the rate r. At the end of the first year the annuity a will be due; at the end of the second year a second payment a will become due, together with ar the interest for one year upon the first payment; at the end of the third year a third payment a becomes due, together with 2ar the interest for one year upon the former two payments, and so on; the sum of all these will be the amount required. Thus : At the end of the first year, the sum due is a. &c. &c. &c. At the end of the tth year, the sum due is a+(t-1)ar. Hence, adding these all together for the whole amount, s=ta+ar(1+2+3+..... (t−1)). Or, taking the expression for the sum of the arithmetical series, 1+2+3 +(-1) s=ta+ra. t(t-1) (5) |