document.write( "Question 316907: So what I really need is the formula to use here. I can figure out the answer but would like the formula for different numbers. \r
\n" ); document.write( "\n" ); document.write( "I have $150,000,000 and planning on purchasing an item that costs $200,000. I can only buy one at a time and every time I purchase this item the price increases by $1,800. How many times can I purchase this item?
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Algebra.Com's Answer #226894 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
You have an arithmetic progression with a starting value of 200000 and a difference of 1800.
\n" ); document.write( "\"An=200000%2B1800%2A%28n-1%29\", with n=1,2,3,...
\n" ); document.write( "The sum of an arithmetic progression is given as,
\n" ); document.write( "\"Sn=%281%2F2%29%282A1%2Bd%28n-1%29%29n\"
\n" ); document.write( "\"%281%2F2%29%282%28200000%29%2B1800%28n-1%29%29n=150000000\"
\n" ); document.write( "\"400000n%2B1800n%28n-1%29=300000000\"\r
\n" ); document.write( "\n" ); document.write( "\"400000n%2B1800n%5E2-1800n=300000000\"
\n" ); document.write( "\"1800n%5E2%2B398200-300000000=0\"
\n" ); document.write( "\"18n%5E2%2B3982n-3000000=0\"
\n" ); document.write( "\"9n%5E2%2B1991n-1500000=0\"
\n" ); document.write( "Using the quadratic equation,
\n" ); document.write( "\"N=312.6\"
\n" ); document.write( "You will be able to buy the 312th item but won't have enough money left for the next item.
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