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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.