p * [ 1/(1+i) + 1/(1+i)^{2} + 1/(1+i)^{3} + ... + 1/(1+i)^{360} ] =

p * [ (1 - 1/(1+i)^{360}) * ( 1/(1+i) + 1/(1+i)^{2} + 1/(1+i)^{3} + ... ) ].

Now recall that that the sum for n=1 to infinity of x^{n} = x/1-x, where x is less than 1, so
we can further simplify:

p * [ (1 - 1/(1+i)^{360}) * (1/(1+i))/(1-(1/(1+i))) =

p * [ (1 - 1/(1+i)^{360}) / i = $100,000.

Solving for p (remember that i=.075/12):

p = $100,000 * i / (1 - 1/(1+i)^{360}) =~ $699.21

For your own information here are what the monthly payments would be under various other interest rates:

0% $277.78 1% $321.64 2% $369.62 3% $421.60 4% $477.42 5% $536.82 6% $599.55 7% $665.30 8% $733.76 9% $804.62 10% $877.57 11% $952.32 12% $1028.61 13% $1106.20 14% $1184.87 15% $1264.44 16% $1344.76 17% $1425.68 18% $1507.09 19% $1588.89 20% $1671.02 25% $2084.58 30% $2500.34 35% $2916.76 40% $3333.36 45% $3750.01 50% $4166.67 55% $4583.33 60% $5000.00 65% $5416.67 70% $5833.33 75% $6250.00 80% $6666.67 85% $7083.33 90% $7500.00 95% $7916.67Michael Shackleford, A.S.A.