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September 4th, 2017, 02:28 AM   #1
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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?

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September 4th, 2017, 03:18 AM   #2
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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.
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September 4th, 2017, 04:42 AM   #3
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It's stated that the A.P. has first term -4, not 4.
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September 7th, 2017, 08:09 AM   #4
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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.
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September 7th, 2017, 10:01 AM   #5
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Quote:
Originally Posted by kirtanadevaraj View Post
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?
arithmetic: a = 1st term, d = common difference, n = number of terms
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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