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an arithmetic sequence of positive integers has a common difference that is three times the first term. If the sum of the first five terms is equal to the absolute difference between the first term and its square, what is the first term in the sequence

 Dec 10, 2019
 #1
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Let the first term = F
Common difference=3F
F^2 - F = F + 3F*4
F^2 = 2F + 12F
F^2 = 14F, solve for F
F = 14 - First term

 Dec 10, 2019
 #2
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Sorry!. I missed one F:

 

Let the first term = F
Common difference=3F
F^2 - F = 2F + 3F*4
F^2 = 3F + 12F
F^2 = 15F, solve for F
F = 15 - First term

 Dec 10, 2019
 #3
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thanks

Guest Dec 10, 2019
 #4
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Really sorry about this. I must be asleep!. This is the correct answer.

 

Let the first term = F
Common difference=3F
F^2 - F =5/2* [ 2F + 3F*4]
F^2 - F =5F + 30F
F^2 = 6F + 30F
F^2 = 36F
F = 36 -First term

Guest Dec 10, 2019
 #5
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Let the first term  =   a

 

Sum of first  5 terms  =

 

a  + (a + 3a)  + (a + 2*3a)  +  (a + 3*3a)  +  ( a + 4*3a)  =    35a

 

So

 

l  a - a^2   l   =   35a

 

Then either......

 

a - a^2  =  35a               or         a^2  - a  = 35a

 

a^2 + 34a  = 0                            a^2 - 36a  =  0

 

a(a + 34)  =0                              a ( a - 36)   =  0

 

a =  0        a = -34        or a  = 36

 

Since a is positive......then  a  =  36

 

 

cool cool cool

 Dec 10, 2019

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