Equation

Algebra

Equation

Postby Guest » Tue Sep 25, 2018 12:58 am

Solve for d.

S = n/2[2a + (n - 1)d]

2S = n [4a + (2n - 2)2d]

Unsure how to proceed.
Guest
 

Re: Equation

Postby Guest » Tue Sep 25, 2018 1:37 am

[tex]2S = n [4a + (2n - 2)2d][/tex]

[tex]2S = 2n [2a + (2n - 2)d][/tex]

[tex]S = n [2a + (2n - 2)d][/tex]

[tex]\frac{S}{n} = 2a + (2n - 2)d[/tex]

[tex]\frac{S}{n} - 2a = (2n - 2)d[/tex]

[tex]\frac{S - 2an}{n} = (2n - 2)d[/tex]

[tex]\frac{S - 2an}{n(2n - 2)} = d[/tex]

[tex]d = \frac{S - 2an}{n(2n - 2)}[/tex]

Is it clear?
Guest
 

Re: Equation

Postby Guest » Tue Sep 25, 2018 2:34 am

The first line [tex]S_{n }[/tex]=[tex]\frac{2a_{1 }+(n-1)d}{2}[/tex].n

On the second line there is a Big Error.

I think that : d=[tex]\frac{2(S_{n }-a_{1 }.n)}{n(n-1)}[/tex]
Guest
 

Re: Equation

Postby Guest » Tue Sep 25, 2018 12:04 pm

2S=2n[2a+(2n−2)d]

S=n[2a+(2n−2)d]

Sn =2a+(2n−2)d

Sn −2a=(2n−2)d


I don't understand these steps.
Guest
 

Re: Equation

Postby Guest » Wed Sep 26, 2018 2:22 am

2S=2n[2a+(2n−2)d] divide the both sides by 2

S=n[2a+(2n−2)d] divide the both sides by n

S/n =2a+(2n−2)d subtract the both sides by 2a

S/n −2a=(2n−2)d
Guest
 

Re: Equation

Postby Guest » Wed Sep 26, 2018 9:34 am

S−2ann =(2n−2)d

S−2ann(2n−2) =d

I don't understand these steps also.
Guest
 

Re: Equation

Postby Guest » Fri Sep 28, 2018 12:57 pm

Please reply. Thanks.
Guest
 

Re: Equation

Postby Guest » Fri Sep 28, 2018 11:52 pm

Here's my decision : [tex]S_{n }[/tex]=n.[tex]\frac{2a_{1 }+(n-1)d}{2}[/tex] ;d=?

2[tex]S_{n }[/tex]=n[2[tex]a_{1 }[/tex]+(n-1)d] /:n[tex]\ne[/tex]0

[tex]\frac{2S_{n }}{n}[/tex]=2[tex]a_{1 }[/tex]+(n-1)d

[tex]\frac{2S_{n }}{n}[/tex]-2[tex]a_{1 }[/tex]=(n-1)d

(n-1)d=[tex]\frac{2S_{n }-2n.a_{1 }}{n}[/tex] /:(n-1)[tex]\ne[/tex]0

d=[tex]\frac{2(S_{n }-na_{1 }}{n(n-1)}[/tex]
Guest
 

Re: Equation

Postby Guest » Sat Sep 29, 2018 12:12 am

I don't' understand what the 1 next to a means.
Guest
 

Re: Equation

Postby Guest » Sat Sep 29, 2018 1:15 am

[tex]a_{1 }[/tex]- first member
Guest
 

Re: Equation

Postby Guest » Sat Sep 29, 2018 1:36 am

Ok, thanks.
Guest
 


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