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# arithmetic sequnce

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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
#4
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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   Dec 10, 2019