Arithmetic, Part 1 · Section 21
Progreſſion.
IS alſo of two ſorts, the firſt is of certain numbers in Arithmetical Proportion from 1, that is, ſuch as differ equally, as 1, 2, 3, 4, 5, 6, where the common difference is 1, (as is eaſily ſeen,) or 1, 3, 5, 7, 9, 11, where the common difference is 2, or any other, as 1, 8, 15, 22, 29, 36, where the common difference is 7, this is called Arithmetical Progreſſion.
2 Secondly, of certaine numbers in Geometrical proportion from 1, that is ſuch as increaſe by a common Multiplication, as 1, 2, 4, 8, 16, 32, where the common multiplier is 2, that is the firſt by 2, produceth the ſecond, and the ſecond multiplied by 2, produceth the third, and ſo on.
Or as 1, 3, 9, 27, 81, 243, where the common Multiplier is 3, this is called Geometrical Progreſſion.
Both the common difference (in the firſt) and the common Multiplication (in the later) ſhall for ſhortneſſe hereafter be called the common exceſſe.
Firſt, now of the firſt ſort, or Arithmetical Progreſſion, the principal uſe of this is,
1 If the number of places, and common exceſſe be given, to find the laſt number.
2 When the number of places, and the laſt number is given, to find the aggregate, or total ſumme of all the numbers.
3 When the laſt number, and the total ſumme is given, to find the number of places.
4 The Number of places, and total ſumme being given to find the laſt number.
5 The laſt number, and number of places given, to find the common exceſſe.
6 The laſt number, and common exceſſe being given; to find the number of places.
I will inſtance in no more, few of theſe ever hapning to be uſed.
| For the firſt of theſe, let there be given the number of places | 100 |
| The common exceſſe | 1 |
| To find the laſt number alſo; | 100 |
THE RƲLE.
Multiply the number of places leſſe by 1, by the common exceſſe, and to the product adde the firſt number; the ſumme is equal to the laſt number.
So here multiply 99 by 1, the product is 99, (for 1 neither multiplies nor divides) to this adde the firſt number 1, it gives 100 for the laſt number.
Or let the numbers be 1, 7, 13, 19, 25, 31, where the common exceſſe is 6. and the number of places alſo 6.
Now (if the number of places leſſe by 1, that is) 5 be multiplied by the common exceſſe, which is 6, the product is 30, to which adding the firſt number which is 1, the laſt number 31, is thereby compoſed. This is ſo caſie that it needs no proof.
2 For the ſecond, which is, The laſt number, and the number of places given, to find the total ſumme of all the numbers.
THE RƲLE.
Adde the firſt and laſt numbers together, and multiply the ſumme by half the number of places, the product is equal to the aggregate or ſumme of all the numbers added together.
So if to the firſt number 1 be added the laſt number 100, it gives 101, which multiplied by 50 (which is half the number of places) produceth 5050, which is equal to all the hundred numbers added together.
And hereby may that vulgar queſtion be anſwered, which is,
If a man take up 100 ſtones placed a yard one from another all in a right line, by one at once, and bring them back one by one to his firſt ſtanding, how many yards doth he goe backwards and forwards?
It is ſhewed before that he goes forward 5050 yards, and he muſt needs come back juſt as much; that is, in all 10100 yards. which is 5 miles and 3 quarters; wanting 20 yards.
Or ſecondly, ſuppoſe the numbers were 1, 9, 17, 25, 33, 41. Whereof the common exceſſe is 8, the firſt and laſt added gives 42, which multiplied by 3, (half the number of places) the product is 126, which is the ſumme of them all.
3 For the third thing, that is, by the laſt number and the total, to find the number of places.
THE RULE.
Adde the firſt and laſt numbers, and by the ſumme of them divide the total, … the quotient will be equal to half the number of places.
This is ſo plain it needs no cleering.
4 For the fourth, if the total, and number of places bee given, to find the laſt number.
THE RULE.
Divide the total by half the number of places, the quotient is a number, from which if 1 be taken, the reſt is the laſt number.
As let the number be 1, 3, 5, 7, 9, 11, 13, 15, or any other (in arithmetical proportion) whatſoever. The ſumme of theſe is to be 64. And the number of places is 8, the half of it 4. Now if 64 be divided by 4, … the quotient is 16, from which if 1 be taken, there remains 15 for the laſt number.
5 Now for the fiſt variety, If the laſt number and number of places be given, to find the common exceſſe.
THE RƲLE.
From the laſt number take 1, and the remaine ſhall be the Dividend; then from the number of places alſo take 1, and make this later remaine the Diviſor▪ then the quotient of this Diviſion ſhall be the common exceſſe.
Example, Let the numbers be 1, 4, 7, 10, 13, 16, from 16 take 1. remains 15, for the dividend, then from 6, (which is the number of places) take alſo 1, remains 5 for the Diviſor.
Now when 15 is divided by 5, the quotient is 3. And 3 is alſo the common exceſſe, or difference between 1 and 4, or 4 and 7, &c.
6 Laſtly, let the laſt number, and the common exceſſe be given, to find the number of places.
THE RULE.
From the laſt number take 1, and divide the remaine by the common exceſſe; then to the quotient adde 1, the ſumme is the number of places.
As let the numbers be 1, 5, 9, 13, 17, 21, 25, 29, from 29 take 1, remains 28, which divide by 4, (which is the exceſſe) the quotient is 7, to which adde 1, the ſumme is 8, which is the number of places, as the reader may eaſily count.