document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1028154>Question 1028154</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One day, Jack planted a magic beanstalk. (Let's call this day,  \"Day 1\"). The next day (day 2), Jack notices that the beanstalk's height had increased by 1/2. On day 3, the beanstalk's height had increased by 1/3 of what it was on the previous day (day 2).  One day 4, the beanstalk's height had increased by 1/4 of what it was on the previous day (day 3).  Suppose the beanstalk continues to grow in this manner.  \r<BR>\n" );
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document.write( "On what day will the beanstalk be 1000 times the height it was on the day Jack planted it? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #643338 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=643338\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it appears that it is increasing by .5 each day.\r<BR>\n" );
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document.write( "assume day 1 is x<BR>\n" );
document.write( "day 2 is x + 1/2 * x = 1.5 * x<BR>\n" );
document.write( "day 3 is 1.5 * x + 1/3 * 1.5 * x = 2 * x<BR>\n" );
document.write( "day 4 is 2 * x + 1/4 * 2 * x = 2.5 * x<BR>\n" );
document.write( "day 5 is 2.5 * x + 1/5 * 2.5 * x = 3 * x<BR>\n" );
document.write( "day 6 is 3 * x + 1/6 * 3 * x = 3.5 * x<BR>\n" );
document.write( "day 7 is 3.5 * x + 1/7 * 3.5 * x = 4 * x<BR>\n" );
document.write( "day 8 is 4 * x + 1/8 * 4 * x = 4.5 * x<BR>\n" );
document.write( "etc.....\r<BR>\n" );
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document.write( "so .....\r<BR>\n" );
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document.write( "day 1 = x<BR>\n" );
document.write( "day 2 = 1.5 * x<BR>\n" );
document.write( "day 3 = 2 * x<BR>\n" );
document.write( "day 4 = 2.5 * x<BR>\n" );
document.write( "day 5 = 3 * x<BR>\n" );
document.write( "day 6 = 3.5 * x<BR>\n" );
document.write( "day 7 = 4 * x<BR>\n" );
document.write( "etc....\r<BR>\n" );
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document.write( "the common difference appears to be .5 * x\r<BR>\n" );
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document.write( "let's assume that day 1 was 5 feet.<BR>\n" );
document.write( "day 2 would be 5 + 1/2 * 5 = 7.5 feet.<BR>\n" );
document.write( "day 3 would be 7.5 + 1/3 * 7.5 = 10 feet.<BR>\n" );
document.write( "day 4 would be 10 + 1/4 * 10 = 12.5 feet.<BR>\n" );
document.write( "day 5 would be 12.5 + 1/5 * 12.5 = 15 feet.<BR>\n" );
document.write( "day 6 would be 15 + 1/6 * 15 = 17.5 feet.<BR>\n" );
document.write( "day 7 would be 17.5 + 1/7 * 17.5 = 20 feet.<BR>\n" );
document.write( "etc.....\r<BR>\n" );
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document.write( "the bean stalk is growing at 2.5 feet per day, assuming it's original height was 5 feet.\r<BR>\n" );
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document.write( "this make it an arithmetic progression because the common difference is 2.5 feet per day.\r<BR>\n" );
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document.write( "you want to know on what day it will be 1000 * it's original height.\r<BR>\n" );
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document.write( "the formula for the nth term of an arithmetic progression is An = A1 + (n-1)*d\r<BR>\n" );
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document.write( "An is the nth term in the progression.<BR>\n" );
document.write( "A1 is the first term.<BR>\n" );
document.write( "n is the number of terms<BR>\n" );
document.write( "d is the common difference.\r<BR>\n" );
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document.write( "in our example above:<BR>\n" );
document.write( "A1 = 5<BR>\n" );
document.write( "An is equal to 1000 * 5 = 5000<BR>\n" );
document.write( "n is what we want to find.<BR>\n" );
document.write( "d is the common difference of 2.5\r<BR>\n" );
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document.write( "the formula becomes 5000 = 5 + (n-1) * 2.5\r<BR>\n" );
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document.write( "simplify to get 5000 = 5 + 2.5 * n - 2.5\r<BR>\n" );
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document.write( "combine like terms to get 5000 = 2.5 + 2.5 * n\r<BR>\n" );
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document.write( "subtract 2.5 from both sides of the equation to get 5000 - 2.5 = 2.5 * n\r<BR>\n" );
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document.write( "combine like terms to get 4997.5 = 2.5 * n\r<BR>\n" );
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document.write( "divide both sides of the equation by 2.5 to get 4997.5 / 2.5 = n\r<BR>\n" );
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document.write( "solve for n to get n = 4997.5 / 2.5 = 1999\r<BR>\n" );
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document.write( "it appears that the beanstalk will be 1000 times it's original height on day 1999.\r<BR>\n" );
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document.write( "let's see if we can get the same answer using x rather than 5.\r<BR>\n" );
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document.write( "A1 = x<BR>\n" );
document.write( "An = 1000 * x<BR>\n" );
document.write( "d = .5 * x\r<BR>\n" );
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document.write( "formula of An = A1 + (n-1) * d becomes (1000 * x) = x + (n-1) * (.5 * x)\r<BR>\n" );
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document.write( "simplify to get (1000 * x) = x + (.5 * x * n) - (1 * .5 * x)\r<BR>\n" );
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document.write( "simplify further to get (1000 * x) = x + (.5 * x * n) - (.5 * x)\r<BR>\n" );
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document.write( "combine like terms to get (1000 * x = (.5 * x) + (.5 * x * n)\r<BR>\n" );
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document.write( "subtract (.5 * x) from both sides of the equation to get (1000 * x) - (.5 * x) = (.5 * x * n)\r<BR>\n" );
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document.write( "simplify to get (999.5 * x) = (.5 * x * n)\r<BR>\n" );
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document.write( "divide both sides of the equation by (.5 * x) to get:\r<BR>\n" );
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document.write( "(999.5 * x) / (.5 * x) = n\r<BR>\n" );
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document.write( "the x in the numerator and denominator on the left side of the equation cancel out and you are left with:\r<BR>\n" );
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document.write( "999.5 / .5 = n\r<BR>\n" );
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document.write( "solve for n to get n = 999.5 / .5 = 1999.\r<BR>\n" );
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document.write( "you get that you get 1000 * the original value on day 1999.\r<BR>\n" );
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document.write( "let's see if that's true.\r<BR>\n" );
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document.write( "An = A1 + (n-1) * d\r<BR>\n" );
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document.write( "A1 = x<BR>\n" );
document.write( "n = 1999<BR>\n" );
document.write( "d = .5 * x\r<BR>\n" );
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document.write( "formula becomes An = x + 1998 * .5 * x\r<BR>\n" );
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document.write( "this results in An = x + 999 * x\r<BR>\n" );
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document.write( "combine like terms to get An = 1000 * x.\r<BR>\n" );
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document.write( "the solution is confirmed as good.\r<BR>\n" );
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document.write( "the key was to determine that it was an arithmetic progression.\r<BR>\n" );
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document.write( "this was determined by evaluating the difference between each successive term and finding that it was constant.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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