You could do it algebraically if you have all day, but I suspect that somebody that knows how to do matrix algebra on a computer could probably answer it in minutes.

This leaves you with 7 linear equations, in 7 unknowns.

Solve for the unknowns A1 through A7. You can do this manually by a process of reduction (adding and subtracting combinations of these equations in order to eliminate unknowns), or by a matrix inversion and multiplication process. Any good text on linear algebra would show the procedure.

In this case, reduction is trivial. For example, to calculate A2, you would just subtract the second equation from the first, leaving you with

A2 = 1/200 - 1/250 Do the same trick for the subsequent equations (subtract each from the first) to calculate A3 through A7

A3 = 1/200 - 1/300 A4 = 1/200 - 1/600

and do forth.

Plug all of these values into the first equation, simplify, and you have the value of A1.

Then, invert each value A1 through A7 to calculate R1 through R7, and you're done.

--
Dave Platt AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior

7 resistors to make 200 ohms = 1400 ohms each
6 resistors to make 250 ohms = 1500 ohms each etc. etc. no computations just a little mental arithmetic. Or did I miss somthing??

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