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 April 6th, 2014, 11:52 PM #1 Newbie   Joined: Mar 2014 From: Australia Posts: 3 Thanks: 0 Geometric Series Hi everyone, And thank you for the assistance last time asked for help. I have a question posed to me by my tutor. Unfortunately, the tutor is away for 2 weeks so I have no one to ask for direction. The question is "For the Geometric progression with T3 = 3 and T10 = 6561, determine the first term, common ratio, and the sum of the first 20 terms." I am using the following formula........... an = a1 + (n - 1)d 3 = a1 + (3 - 1)d 6561 = a1 + (10 - 1)d a1 = 3 - (3 - 1)d a1 = 6561 - (10 - 1)d 3 - (3 - 1)d = 6561 - (10 - 1)d 6561 - 3 = (10 - 1)d - (3 - 1)d 6558 = 10d - d - 3d + d 6558 = 7d d = 6558/7 d = 936.86 For T3 = 3 3 = a1 + (3 - 1)936.86 3 = a1 + 2 x 936.86 a1 = 3 - 1873.72 a1 = -1870.72 I am not asking for answers......... just want to know if I am on solid foundations. Thanks Phil
 April 7th, 2014, 03:03 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond You say you are working with a geometric series but you are using the formula for an arithmetic series! nth term of a geometric series (with a as the first term and r as the common ratio): $a\cdot r^{n-1}$ So we have $a\cdot r^2=3$ and $a\cdot r^9=6561$. Solve for a in terms of r in the first and substitute that into the second to find r. Now use the formula for the sum of a geometric series with $n=20$: $$S=\frac{a(1-r^n)}{1-r}$$ Last edited by greg1313; April 7th, 2014 at 03:16 AM.

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