Sect. I.

PROBLEM I.

Having two Numbers given, to find a third Geometrically proportional unto them, and to three a fourth, and to find a fifth, &c.

Find one of the Numbers given upon the Line B, and set it against the other given Number on the Line A, then find the same Number upon B (which was last countered upon A) and against it you have this third proportional upon A; and against this third upon B, is the fourth upon A; in like manner, against the fourth upon B, you have the fifth on A, &c.

    Example. Let it be required to find a third proportional to these two Numbers 2 and 4, which may bear the same proportion to 4, that 4 bears to 2.

    Draw out the sliding Rod till 2 upon B stand against 4 upon A, then against 4 upon B is 8, (the third proportional) upon A; and against this third (viz. 8) upon B, is 16 upon A; which is the fourth proportional: Likewise against 16 upon B, is 32 the fifth upon A, and against 32 upon B, is 64 the sixth proportional: But now proceeding forward, I find that 64 upon B, will reach beyond the end of the Line A, I therefore seek 64 towards the left hand upon B, and against it I find 128 the seventh proportional; and so proceeding further, you may find the eighth to be 256, the ninth 512, &c. Contrariwise, if it were required to find a third proportional to the same numbers 2 and 4, which may bear the same proportion to 2, that 2 bears to 4.

    Set 4 in the second Radius upon A, to 2 upon B, then against 2 upon A (towards the left hand) is 1, the third proportional, and against 1 upon A is .5 the fourth upon B; also against this fourth (viz.) upon A, is .25, the fifth proportional on B, &c.

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This page was last updated on 29 March 2000.