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[[five column table]] | |Nomb|Primos|Seconds|Thirds| |[[line]] | |[[strikethrough]]Nomb|Primos|Thirds][[/strikethrough]] |R1| 1| | | | |R2|1|4|1|4| |R3|1}7|3|2| |R4|2| | | | |R5|2|2|3|6| |R6|2|4|4|9| |R7|2|6|4|5| |R8|2|8|2|8| |R9|3| | | | |R10|3|1|6|2| |R11|3|3|1|6| |R12|3|4|6|2| |R13|3|6|0|5| |R14|3|7|4|3| |R15|3|8|7|2| |R16|4| | | | |R17|4|1|2|3| |R18|4|2|4|2| |R19|4|3|5|8| |R20|4|4|7|2| |R21|4|5|8|2| |R22|4|6|9|1| |R23|2|7|9|2| |R24|4|8|9|9| [[/five column table]] Now if we take ye summ of ye 12 first numbers, it is something more than 29, and 12 times ye 12th number viz 3462 gives a product something greater than 415 and by consequence that summ wch is ye parabola is greater than 2/3 of ye product wch is ye rectangle: but if we take ye 24 numbers we shall find something less than 79 for ye parabola, and ye product of ye half 4899 by 24 [[strikethrough]] numbers [[/strikethrough]] is something more than 117 where of ye 2/3 is 78 and so ye summ of ye 24 numbers doth not differ from 2/3 of ye product but aboute an unite and it approacheth nearer than when we take ye 12 first numbers, and if we continue to increase ye table by a greater number of divisions, ye difference of that summ and ye product will diminish always, so that we may judge it would at last come to be precisely 2/3 We see also that if we take ye 6 numbers from ye middle of twelve, if they do [[?surpass]] ye summ of 3 first and ye 3 last together, and that ye summ of ye 6 first and ye 6 last of 24, shall be less than ye summ of 12 of ye middle wch ought necessarily to happen and may be demonstrated in this manner: The extremes of ye Squares of numbers wch are in an arithmeticall
Transcription Notes:
mandc: makes absolutely no sense to me. Need mathematician help.