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d=400+ 240-40 600. Wherefore, each payment exceeds the preceding by $40, and the last is $600.

Proof. 400 440+480+ 520+560+ 600=3000. 2. B buys a piece of land for $3000, which he is to pay by yearly instalments, each exceeding the preceding one by $40, and the first payment is to be $400. In how many years will he clear his purchase, and what is the last payment?

3000.

α= 400, d 40, s = By the formula, l=d+√(2ds + 1d2 — ad + a2) = — 20+√(240000 16000 160000) = 620 20= 600, and n 6000 .6000

+400

2s

=

=

a + 1 400+ 600 1000

6. Wherefore, he will

clear his purchase in 6 years, and the last payment is $600. In the same way may the formula be applied in the solution of all such problems.

3. B pays, for a piece of land, $3000 by instalments, each of which exceeds the preceding one by $40, the last payment being $600. Required the number of payments, and the amount of the first?

Ans. Number of payments, 6; and first amount, $400. 4. B pays, for his land, a sum in 6 payments, each of which exceeds the preceding one by $40, the first payment being $400. Required the cost of the land, and the last payment? Ans. The cost, $3000; the last payment, $600. 5. What is the sum of the natural numbers 1, 2, 3, &c., to 1000000 inclusive? Ans. 500000500000. 6. Required to insert 9 arithmetical means between 1 and 16.

Ans. 1.2.4. 51. 7. 81. 10.111.13.141.16. 7. Required the sum of 1.3.5.7, &c. to a million of terms. Ans. 1000000000000. 8. What is the sum of 150 terms of a decreasing series, the first term being 397, and the common difference 4?

Ans. 14850.

9. A basket being placed on level ground, and in a right line with it, on one side, 100 stones, 1 yard apart from each other, and the first 1 yd. from the basket, how far will the

person travel who shall begin at the basket and gather the stones into it, making a separate tour for each stone?

Ans. 58, or very nearly 5 miles. 10. At what times during the twelve hours will the hour and minute hands of a common clock be together?

Ans. At 55 min. past one; 1019 min. past two; 16,1 min. past 3; 21 min. past 4; 273 min. past 5; 328 min. past 6; 38 min. past 7; 437 min. past 8; 49 min. past 9; 54 min. past 10; and at 12.

Geometrical Progression.

462. This name is given to a series of numbers in continual geometrical proportion. Thus, 2: 6:18: 54: 162: 486 is a geometrical progression. The sign is to and so is, is used to signify the repetition of the intermediate terms, thus: 2 is to 6 as 6 is to 18, &c.

A geometrical progression is often called simply a geometrical series, and is said to be increasing when the numbers increase, and decreasing when they decrease. But as a de

creasing series is merely an increasing one reversed, and has consequently the same properties, the same reasoning will apply to it, if we substitute divide for multiply, and contains for is contained. We shall therefore consider the series as in

creasing.

The number found in dividing the greater of any two proximate terms by the less is called the ratio of the progression, and as this ratio is a quotient, we shall call it q. The other elements of the progression-namely, the first and last terms, the number of terms, and sum of the series-we shall designate by the letters a, l, n, s, as in arithmetical progression.

As the second term is formed by multiplying the first term by the ratio, and each successive term, by multiplying the preceding one by the ratio, it is evident that any term of the progression is composed of the first term multiplied by the ratio raised to a power the index of which is the number of terms less one. Hence, the last term 7 =aXq. When a = 1, we have 7=q-1. If we would find the eleventh term of the above progression, we have 7 = 2 × 310 118098, the term required.

=

463. The terms intervening between the first and last are said to be so many geometrical means between those two terms; and any three consecutive terms being taken, the middle one is a geometrical mean between the other two; also, if we re

peat the progression as indicated by the sign, it is evident (403) that the product of the first and last of any three such terms is equal to the square of the middle one. Wherefore,

to find a geometrical mean between any two numbers, we take the square root of their product.

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Now, as l say, the ratio of the progression is found in dividing the lasi term by the first, and extracting that root of the quotient the index of which is the number of terms diminished by a unit, or the number of means increased by a unit; therefore, when there is one geometrical mean between two numbers, the ratio of the progression is the square root of their quotient; when there are two means, it is the cube root; when there are three, it is the fourth root of their quotient, and so on for any number of means. To produce the means, multiply the first term by the ratio, and the product again by the ratio, and

so on.

464. We have seen in Evolution (366) that though we cannot always have the exact root, we can approach to it as near as we please. Hence it is evident that between two given numbers, though we cannot always insert in integral numbers as many geometrical means as we please, we can always obtain a series of numbers which shall approach these means within any prescribed degree of exactness.

3

If we would have 9 geometrical means between 3 and 3072, it is plain from the formula that the ratio q = 1/307 =1024, which, by mere inspection of the table of powers, we find to be 2. Wherefore, in continually multiplying by the ratio 2, we have for the required series 3: 6:12:24 : 48: 96: 192 : 384: 768 : 1536: 3072.

Again, if we would insert 5 geometrical means between 2 and 3, we have q √3 = 1,0699 +, or 1,07—; and multiplying by this ratio, we have 2: 2,14 2,29: 2,45 : 2,62: 2,80: 3, for the required series; the mean terms being true to within a hundredth of a unit.

465. If we take any geometrical proportion a:b::c:d; alternando ac::b:d; componendo a +c:c::b+d:d; again, alternando a +c: b+d::c:d; that is, the sum of the antecedents of two equal ratios is to the sum of the consequents as one antecedent is to its consequent. Again, if cdef, it is plain that a + c :b+d::ef; but we

have shown that the sum of the antecedents is to that of the consequents as one antecedent to its consequent; and, therefore, a+c+e:b+d+f::e:f. Hence we infer that the sum of the antecedents of any number of equal ratios is to that of their consequents as any one antecedent is to its consequent.

Now in the progression 2:6:: 6:18 :: 18: 54, &c., which is evidently a number of equal ratios, we easily perceive that the sum of all the antecedents comprehends all the terms except the last, and may be represented by s―l; and that the sum of all the consequents comprehends all the terms except the first, and may be represented by sa. Now the first term being a and the ratio q, the second term is aq, and therefore s -1: s a::a: aq; and hence (403) (s Daq (sa)a; dividing by a, (s—1)q = 8 $ = qla, or s(q-1)

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That is, to obtain the sum of the series,

we multiply the last term by the ratio, subtract the first term from the product, and divide the difference by the ratio diminished by a unit.

From the two equations

l=a X qn−1
ql a
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8 =

and in a manner exactly analogous to that pursued in constructing the table of arithmetical progression, we construct the following table, by means of which, if we know any three of the five things l, a, q, n, s, the other two can always be found with great facility. Also, with equal facility are performed, by the same table, the most difficult calculations of banking and finance, such as Compound Interest, Annuities, &c. In the fourth section of the formula, the capital L placed before a quantity signifies the logarithm of that quantity. This section, therefore, the student may omit, till he has read the next article, in which we shall treat of those most interesting and important numbers.

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