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 April 14th, 2012, 12:35 AM #1 Senior Member   Joined: Apr 2012 Posts: 112 Thanks: 0 Number Theory Hello, thank you all for all your help in my previous questions, I really appreciated that! Unfortunately, I'm back with some more. 1a.) Show that if Un is gometric and Sn is the sum of the first n terms then the common ratio is given by r = (Sn-U1)/(Sn-Un) I understand how the common ratio was derived, except the U1 part. Doesn't that mean the first term? How do you know that Sn is the sum of n numbers except the first term? I thought it was the sum of n numbers except one particular term, and my difficulty was to find which term it was. b.) A GP has a first term of -0.08 and a last term of 0.6075, and a sum of 0.3325. Find the common ratio. 0.06075=(-0.0r^(n-1) -7.59=r^(n-1) 0.03325=((-0.0(1-r^n))/(1-r) 0.03325-0.03325r=-0.08+0.08r^n 0.4125=0.3325r+0.08r^n A man is appointed to a position at a salary of $30000 per annum and each year, his salary is to be increased by 5%. FInd his salary at the end of 20 years of service and his average annual salary for twenty years. I know how to approach this problem but my answer was incorrect. First year -$30 000 2nd Year - 30 000(1.05) 3rd year - 30 000(1.05)^2, Following this pattern, where ar^(n-1), I thought to get 30 000(1.05)^19 for 20 years but the answer is 30 000(1.05)^20. The growth of a certain tree during any year is 90% of its growth during the previous year. It is now 10m high and one year ago it was 9m high. What will be its growth this year? Next year? Find the height of the tree in 15 years time. I got an equation of 10(1.09) its growth this year. Is it right? Is it right to say that, the height in 15 years time will be 10(1.09)^14?
 April 14th, 2012, 02:03 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Number Theory 1a) We have: $S_n=U_1\frac{1-r^n}{1-r}$ $(1-r)S_n=U_1$$1-r^n$$$ $S_n-rS_n=U_1-U_{n}r$ $S_n-U_1=r$$S_n-U_n$$$ $r=\frac{S_n-U_1}{S_n-U_n}$ 1b) Using the result of part a), we have: $r=\frac{0.3325+0.08}{0.3325-0.6075}=-1.5$ 2) At the end of the nth year the man's salary is 30000(1.05)^n, so at the end of the 20th year, his salary is 30000(1.05)^(20). 3) During the previous year, the tree's growth was 1 m. This year it will grow 90% of that or 0.9 m. Next year it will be (0.9)² m = 0.81 m. The height of the tree in n years is (in meters): $h=10+\sum_{k=1}^{n}(0.9)^k=19-9(0.9)^n$ Thus, in 15 years, the height is: $19-9(0.9)^{15}\approx17.15\text{ m}$
 April 14th, 2012, 09:25 AM #3 Senior Member   Joined: Apr 2012 Posts: 112 Thanks: 0 Re: Number Theory How come when U1was multiplied into,(1-r^n) become U1-Un^r? It is the transformation from r^n to Un^r I do not understand. Thank you so much!
 April 14th, 2012, 10:49 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Number Theory It stems from: $U_1\cdot r^n=$$U_1\cdot r^{n-1}$$r=U_n\cdot r$
 April 14th, 2012, 12:26 PM #5 Senior Member   Joined: Apr 2012 Posts: 112 Thanks: 0 Re: Number Theory Thank you! By the way, "At the end of the nth year the man's salary is 30000(1.05)^n, so at the end of the 20th year, his salary is 30000(1.05)^(20)", I have always thought Un=ar^(n-1)? Thus at the end of nth year, his salary will be 30000(1.05)^(19).
 April 14th, 2012, 01:01 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Number Theory He is given a raise at the end of each year and there are 20 years, so 20 raises.
 April 14th, 2012, 08:56 PM #7 Member   Joined: Apr 2012 Posts: 92 Thanks: 0 Re: Number Theory For question 2, could you elaborate on how you got 19-9(0.9)^n from the summation? Thank you much.
 April 14th, 2012, 09:09 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Number Theory I used the computer, but it is easy enough to derive: $h=10+\sum_{k=1}^{n}(0.9)^k$ We need to adjust the lower bound for the index of summation to zero, so we subtract the equivalent term, so that the net addition is zero: $h=10-1+\sum_{k=0}^{n}(0.9)^k$ Use the formula for the sum of a geometric series: $h=9+\frac{1-(0.9)^{n+1}}{1-0.9}$ The rest is algebra: $h=9+10$$1-(0.9)^{n+1}$$$ $h=9+10-10(0.9)^{n+1}$ $h=19-10(0.9)(0.9)^{n}$ $h=19-9(0.9)^n$

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