September 4th, 2017, 02:28 AM  #1 
Newbie Joined: Sep 2017 From: Manila Posts: 1 Thanks: 0  Sequences and Series
Hey! I need help with this question.. An arithmetic sequence has the first term 4 and common difference 1. A geometric sequence has the first term 8 and common difference 0.5. After how many terms would the sum of the arithmetic sequence exceed the sum of the geometric sequence? Thanks! 
September 4th, 2017, 03:18 AM  #2 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 876 Thanks: 60 Math Focus: सामान्य गणित 
Like Arithmetic sequence has common difference, Geometric sequence has common ratio. A.P. = 4, 5, 6, 7, ........... G.P. = 8, 4, 2, 1, ........... Sum upto 3rd term, A.P. = 4 + 5 + 6 = 15 G.P. = 8 + 4 + 2 = 14 So, the sum of A.P. exceeds the sum of G.P. after third term. 
September 4th, 2017, 04:42 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,786 Thanks: 1029 Math Focus: Elementary mathematics and beyond 
It's stated that the A.P. has first term 4, not 4.

September 7th, 2017, 08:09 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,097 Thanks: 849 
So for the a.p., $\displaystyle a_n= 4+ n$, (the first term is $\displaystyle a_0$) and for the g.p. $\displaystyle b_n= 8(1/2)^n$ (again the first term is $\displaystyle b_0$. You want $\displaystyle 4+ n> 8(1/2)^n$. Clearly, the right side is positive so n must be at least 5. $\displaystyle 8(1/2)^3< 1$ so taking $n= 5$, $\displaystyle a_n= a_5= 1$ and $\displaystyle 8(1/2)^n= 8/8^5= 1/8^4= 0.000244140625$ so $\displaystyle a_5> b_5$.
Last edited by skipjack; September 7th, 2017 at 10:33 AM. 
September 7th, 2017, 10:01 AM  #5  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,106 Thanks: 797  Quote:
geometric: b = 1st term, r = common ratio, n = number of terms Formula for sum of 1st n terms: arithmetic: (2a + d(n  1)) / 2 [1] geometric: b(r^n  1) / (r  1) [2] [1] = [2] : solve for n Try at Wolfram: you'll be told to get lost!!  

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