and so on; the law of continuation of the successive terms being evident. Hence, collecting the results, our continued proportion or B geometrical progression becomes A, A × A x (53); A' A x Ах (3)* ̄*, (2), as required. (2) shows that the second term of the progression is equal B to the first term A multiplied by the ratio and that the A' third term equals the first term A multiplied by the second B power or square of the ratio and that the fourth term A' equals the first term multiplied by the third power or cube of the ratio; and, generally, that the term whose number is denoted by n is equal to the first term ▲ multiplied by the power of the ratio B (3), whose exponent is the number n B — 1. It is also evident, from (2), if the ratio is greater than A unity, that the series will be increasing, and if the ratio is less than unity, the series will manifestly decrease; also, if B A equals unity, the terms of the series are evidently equal to each other. It may not be improper here to observe that the first term. A of the geometrical progression (2) is frequently said to be to the third term A x in the duplicate ratio of the first term to the second, and that the first term is to the fourth term in the triplicate ratio of the first to the second, and that the first term is to the fifth in the quadruplicate ratio of the first to the second, and so on; and it is clear that the meaning is the same as to say (as we have done) that the third term equals the first term multiplied by the square of the ratio of the second to the first term, and that the fourth term equals the first term multiplied by the cube of the ratio of the second to the first term, and so on. (35.) It is easy to see, from (2), that the ratio of the term whose number is n to the first term, is expressed by (B)" ; so that if we extract the (n - 1)" root of this ratio, we shall B get which is the ratio of the second term to the first. A' Hence, if we know only the first and n term, we take the (n − 1)th root of the ratio of the nth term to the first term for the ratio of the second term to the first term. Then we multiply the first term by the ratio, and the product will equal the second term, and the second term multiplied by the ratio gives the third term, and the third term multiplied by the ratio gives the fourth term, and so on; so that all the terms of (2) can be found from the first and nth terms. Instead of multiplying the first term by the ratio for the second term, etc., we may divide the n" term by the ratio, and the quotient will clearly equal the term whose number is n − 1, and dividing this term by the ratio, the quotient will equal the term whose number is n 2, and so on. Instead of finding the ratio of the nth term to the first term, we may find the ratio of the first term to the nth term, and then extracting the (n - 1) root of the ratio thus found, we get the ratio of the first to the second term. Hence, if we divide the first term by the ratio of the first to the second term, the quotient will equal the second term, and if we divide the second term by the same ratio, the quotient will equal the third term, and so on, for the successive terms. Also, if we multiply the nth term by the ratio of the first term to the second, the product will equal the term whose number is n 1, and if we multiply this term by the same ratio, the product will equal the term whose number is n 2, and so on, for the preceding terms until we arrive at the first term. (36.) From what has been done, it is easy to see how to insert any number of geometrical means between any two given numbers or quantities of the same sort. For, let a denote the first and b the second of the given numbers or quantities, and suppose that the number m equals the number of means that are to be inserted between a and b, then when the means are inserted, the whole number of terms in the geometrical progression thus found will be expressed by m +2, whose first term will be a and last term b, whose number will be expressed by m + 2 when we count the terms from a to b. Consequently, for n - 1, in what has been previously shown, we must use m + 1, and we shall have for the ratio of the last term of the progression to its b a first term, whose (m + 1)a root will be expressed by (2) I , which expresses the ratio of the second term of the progression to the first term. Hence the terms of the progression will b (3), which can evidently be found by dividing b, the last term, by the ratio 1 b\m + 1 for the last term but one, and a dividing the last term but one by the same ratio, the quotient will be the last term but two, and so on. It is also easy to see that (3) can be found by dividing a by 1 (a), (which expresses the ratio of the first to the second term), for the second term, and dividing the second term by the same ratio for the third term, and so on; also, if we mul tiply the last term by the ratio ( 1 am+1 the product will equal the last term but one, and this term multiplied by the same ratio will equal the last term but two, and so on, which is in accordance with what has been previously shown. An example or two will serve to show the use of (3), and to make what has been said more evident. For the first example we shall suppose that we are required to insert two geometrical means between 4 and 108. Here we have a = 4, b + 1 1 = 108, m = 2, m + 1 3, and (-2) = (27)1 = 3. Hence (3) becomes 4, a 12, 36, 108; and 12 and 36 are the two means which have been inserted between 4 and 108, as required. For the second example we shall suppose that it is required to insert three geometrical means between 4375 and 7. In this example, we have a = 4375, b = 7, m = 3, m + 1 (3) becomes 4375, 875, 175, 35, 7; and 875, 175, 35 are the three means which have been inserted between 4375 and 7, as required. For the third and last example, we shall suppose that six means are to be inserted between c and cm". = Here a=c, b = cm", m = 6, m +1=7, and (m'7)✈ = m'. Hence, c, cm', cm3, cm'3, cm', cm3, cm", cm", is the required progression. It is evident, from what has been done, if we insert the same number of geometrical means between every two successive terms of any geometrical progression, that the series thus formed will be a geometrical progression. B (37.) If (for simplicity) we put r for the ratio, of the A second term of the series (2) to the first term, then the series (2) will become A, Ar, Ar2, Ar3, Ar1, . . . . Ap2-1, (4); and if we represent the sum of the terms of the series by S, we shall have S = A + Ar + Ar2 + Aμ3 + Apt + .... +Ap2-1, (5). A205 If we multiply both sides of (5) by r, we shall have Sr = Ar + Ar2 + Apa2 + Apu2 + ApNo5 + .... + A?", and if we subtract (5) from this, we get Sr-S= A?" — A, or S(r− 1) = AA, which gives S = A(”—1), (6). If we put l= r - 1 A-1, (a), then we get Ar"= lr, which reduces (6) to S = Ir - A (b). From (b) we derive the following rule for finding the sum of a series of numbers or quantities of the same sort that are in geometrical progression. RULE. Multiply the last term by the ratio, and from the product subtract the first term; then divide the remainder by the ratio less one, and the quotient will be the sum required. If r is sensibly less than unity, and n indefinitely great, then I will become indefinitely small, and may therefore be rejected, which will reduce (b) to S = A (c). (c) shows that if r is sensibly less than one, the sum of the series continued to infinity is expressed by the first term divided by one, less the ratio, Since A, l, r, n and S are connected by the equations (a) and (b), it is evident when any three of them are known that the remaining two can be found by the equations (a) and (b). We will now show the use of (a), (b), and (e) by a few examples. EXAMPLES. 1. The first term of a geometrical progression is 6 and the ratio 2; it is required to find the ninth term and the sum of the nine terms. Ans. The 9 term A-16 x 28=15367, and by (b) lr A 1536 × 2 6 or the rule we have S = p 1 2 1 = 3066. 2. The first term of an infinite number of terms in geomet 1 rical progression is 3 and the ratio is; required, the value 3. The first term of a geometrical progression is 2 and the ratio is 5; to find the seventh term and the sum of the seven terms. Ans. The seventh term is 31250, and S = 39062 is the sum of the seven terms. 4. The first term of a geometrical progression is 8000 and 1 the ratio is; to find the fifth term and the sum of the five terms. 4 = Ans. The fifth term is 12 and S 9996 is the sum of the five terms. |