Horatio Nelson Robinson
Published: 2013-09
Total Pages: 26
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1844 edition. Excerpt: ...Jx. Assume Jx=v, then x=va, and x'=v. The last equa. then becomes, v--13u2=--12u, or a2--I3v----12. Now--12 has two factors, 3 and--4. And 32--13=--4. That is m'--a=n referred to the theoretical equations. Hence e=3, and x=9. But?-!= =24. Or, =24. Jx 3 81--y2 =72. 9=y2, or 3=y. Therefore, 6, 8, 12, are the the numbers. 5. Find two numbers such that their sum shall be 12, and the difference of their fourth powers, 1776. Ans. 7 and 5. Put x--y= the greater, and x--y= the less. Section XVI. On Arithmetical Progression The formulas connected with arithmetical progression, are very simple, and drawn out merely from inspection; any great attempt at explanation, serves rather to confuse, than enlighten; and although this does not profess to be an elementary work, we shall gives all the explanation necessary. Let a represent the first term of any arithmetical series, and d the common difference. Then a, (a+d), (o+2ri), (a+3d), (a+id), &c., will be the series itself, if ascending. If decending, a, (a--d), (a--2d(, (a--3d), (a--id), &c., will represent the series. Wherever we stop, is the last term. The first term exists without, and independent of the common difference. Therefore, to obtain any term, we add or subtract the common difference one less times than the number of terms. Let L be the last term, and n the number of terms. Then the general formula for the last term will be L=a+n--l)d (1) Now let S represent the sum of any arithmetical series. Then S = a+(a+d)+a+2d)+(a--3d). Also, S = (a--3d)--(a--2d)--(a--d)--a by simply changing the order of the terms and adding, 2S = (2a+3d)-f (2a+3d)+(2a+3cZ)+(2a-f3d). That is 2S is equal to the first and last terms of any series repeated as many times as there are terms Now if L is the...