EXAMPLES IN ARITHMETICAL PROGRESSION. 1. Given the first term 3, the common difference 2, and the sum of the series 399, to find the number of terms and the last term. By the 4th Case in the table we have, √3×2--212+399×2×8-3×2—2—19, the number of terms. 2×2 And, 3x2--+399x2x8-2-39, the last term. 2 2. Given the first term 3, the number of terms 19, and the sum of the terms 399, to find the common difference. By the first rule of Case 6th, we have, 2X399-3X19 19--1×19 =2, the common difference. 3. Given the common difference 2, the last term 39, and the sum of the series 399, to find the first term and the number of terms. By Case 7, we have, 2+√39×2+22-399×2×8 =3, the common difference. 19 the number of And, 39x2+2+39x2+21-399x2×3. terms. 2×2 4. A merchant owed to several persons $1080; to the greatest creditor he paid $142, to the greatest but one $132, and so on, in Arithmetical Progression; What was the number of creditors, and what did the least creditor receive? Ans. The number of creditors was 15, and the least creditor received $2. 5. Given the last term 39, the number of terms 19, and the sum of the series 399, to find the common difference. By the 2d rule in Case 8, we have, 2X13X39-399 19-1X19 =2, the common difference. 6. Sixteen persons gave in charity to a poor man in such a manner as to form an arithmetical series; the last gave 65 cents, and the whole sum was $5 60c.; What did cach give less than the other, from the last down to the first. Ans. 4 cents. GEOMETRICAL PROPORTION. THEOREM I. IF four quantities, 2. 6. 4. 12, be in Geometrical Proportion, the product of the two means, 6X4 will be equal to that of the two extremes, 2×12, whether they are continued, or discontinued,* and, if three quantities, 2. 4. 8, the square of the mean is equal to the product of the two extremes. THEOREM 2. If four quantities, 2. 6. 4. 12, are such, that the product of two of them, 2×12 is equal to the product of the other two, 6X4, then are those quantities proportional.† *It was stated under Proportion in General, that the geometrical ratio of two quantities is expressed by the quotient, arising from dividing the antecedent by the consequent; thus, the geometrical ratio of 6 to 2 is 3,=, and of 2 to 6, is = or, and of 3 to 8 is: and that in a proportion there must be two, or more, couplets which have equal ratios. Hence, four numbers will be geometrical proportionals, when the ratios, obtained in this manner, are equal. Thus 2, 4, 8, 16, are geometrically proportional, because 28 each to; and thus also, 9, 3, 12, 4, because each to 3. From these principles, it is easy to prove in a given example, the theorems in Geometrical Proportion. The factors should be kept separate by the sign of multiplication. Let 2,4,3,6, be the geometrical proportionals; then 23. Multiply both fractions by the product of the second and fourth terms, and the fractions will obvi2X4×6 3X4×6 ously still be equal, and we haveThen cancel the equal 4 6 terms in the fractions, and 2×6=3x4, that is, the product of the extremes, 2×6, is equal to the product of the means, 4X3. The same may be shown in any other case, and, hence the general rule be inferred. Again; Three numbers are geometrical proportionals, when the ratio of the first and second terms is equal to the ratio of the second and third. Thus, 2, 4, 8, are three geometrical proportionals, for 2=4each to 3. Proceed as 2×4×8 before, and we have4 4X4X8 and 2×8=4X4, or 42, that is, the product of the extremes, 2x8, is equal to the square of the mean, 4×4 or 42. + Let the four quantities be 2, 6, 4, and 12, so that 2x12=6×4. Divide these equal products by the quantity 6×12, and the quotients will obviously 2×12 6×4 Cancel the equal terms in these two fractions, and be equal, or 2 6X12 6X 12 T2 we have ; whence 2:6::4: 12, by the definition of geometrical proportionals. In the same way it may be shown, that if 2×8=-4X4 or 42, then 2:4:: 4:3, and 2, 4, and 8, are three geometrical proportionals. Ff THEOREM 3. If four quantities, 2. 6. 4. 12, are proportional, the product of the means, divided by either extreme, will give the other extreme.* THEOREM 4. The products of the corresponding terms of two Geometrical Proportions are also proportional. That is, if 26 :: 4 : 12, and 2:4: 5: 10, then will 2×2 : 6x4:: 4X5: 12 x 10.f THEOREM 5. If four quantities, 2, 6, 4, 12, are directly proportional, (1. Directly, 2:6 :: 4: 12 2. Inversely, Because the product of the meaus, in each case, is equal to that of the extremes, and therefore the quantities are proportional by Theorem 1. THEOREM 6. If three numbers, 2, 4, 8, be in continued proportion, the square of the first will be to that of the second, as the first number to the third; that is, 2×2: 4×4 :: 2 : 8. * Let the four proportionals be 2, 4, 5, and 10; then 2×10=4×5, by Theo2X10 4X5 rem 1. Divide both expressions by 2, and 2 2 vide, as before, by 10, and 2X10 4×5 4x5 or 2 10 4X5 ; or 10- ; or, di2 that is, the product of the means divided by one extreme, gives the other extreme. Hence, if the two means and one extreme be given, the other extreme, or geometrical proportional may be found.' + Let there be given, 2:6 :: 4 : 12, whence 24 by Theorem 1: and also, 3:56 10, whence. Multiply the corresponding parts of these equal tionals, 2x36x5 :: 6x4: 10x 12, and is the theorem. Hence, if four quantities are proportional, their squares, cubes, &c. will likewise be proportional. Thus, let the terms be 2: 6:4: 12, then 2×2:6×6 :: 4×4 : 12X12, or 22:62 :: 42: 122, and hence also, 23:63: 43: 123 and 25: 65:45:125. For since 2: 4 :: 4 : 8, thence will 2x8=4×4, by Theorem 1; and therefore 2×2×8=2×4×4, by equal multiplication; consequently,2×2:4×4 :: 28, by Theorem 2. In like manner it may be proved that, of four quantities continually proportional the cube of the first is to that of the second, as the first quantity to the fourth, THEOREM 7. In any continued Geometrical Proportion, 1, 3, 9, 27, 81, &c. the product of the two extremes, and that of every other two terms equally distant from them are equal.* THEOREM 8. The sum of any number of quantities, in continued Geometrical Proportion, is equal to the difference of the product of the second and last terms, and the square of the first, divided by the difference of the first and second terms.† GEOMETRICAL PROGRESSION. A GEOMETRICAL Progression is, when a rank or series of numbers increases, or decreases, by the continual multiplication, or division, of some equal number, which is called the ratio. * For, the ratio of the first term to the second being the same as that of the last but one to the last, these four terms are in proportion; and theretore by Theorem 1, the product of the extremes is equal to that of their two adjacent terms; and after the same manner, it will appear that the product of the third and last but two is equal to that of their two adjacent terms, the second and last but one, and so of the rest; whence the truth of the proposition is manifest. 2+6+18+54+162+486 † Take any series of continued geometrical proportionals, as 2, 6, 18, 54, 162, 486, and its sum is, 2+6+18+54+162+486,-728, by addition. Multiply the whole, by that number by which any term of the series is multiplied or divided to form the succeeding term, which is in this example, 3, and you have, 6+18+54+162+486+1458=2184. Subtract the first series, 728. As all the terms in the upper series are cancelled by those in the lower, except the last in the former and the first in the latter, those two terms become-2+1458, or, 1458-2 2184-728. Now, the upper series exceeds the lower three times, and, of course, from thrice the series there has been taken once the series, and the remainder must be twice, or 3-1 times, the series and equal to twice or 3—1 times, 1458-2 the sum of the series, that is, 3-1 728, the sum of the series. As must also be equal to the sum of the series, and is the rule. For 6 is the second term of the given series; 486, the greatest term; 4, the square of the second term; and the divisor, 6-2, is the difference of the first and second terms. In this demonstration it is shown that 3×486-2 the sum of the series; that is, that the greatest term multiplied by the number by which the series increases, and the product diminished by the least term, and this divided by a number one less than that by which the series increases, the quotient is the sum of the series. Let the series be 1, 4, 16, 64, 256, 1024, whose multi4× 1024-1 plier is 4. Then 1365, the sum of the series. |