# Equation

Algebra

### Equation

Solve for d.

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

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

Unsure how to proceed.
Guest

### Re: Equation

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

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

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

$$\frac{S}{n} = 2a + (2n - 2)d$$

$$\frac{S}{n} - 2a = (2n - 2)d$$

$$\frac{S - 2an}{n} = (2n - 2)d$$

$$\frac{S - 2an}{n(2n - 2)} = d$$

$$d = \frac{S - 2an}{n(2n - 2)}$$

Is it clear?
Guest

### Re: Equation

The first line $$S_{n }$$=$$\frac{2a_{1 }+(n-1)d}{2}$$.n

On the second line there is a Big Error.

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

### Re: Equation

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

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

S−2ann =(2n−2)d

S−2ann(2n−2) =d

I don't understand these steps also.
Guest

Guest

### Re: Equation

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

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

$$\frac{2S_{n }}{n}$$=2$$a_{1 }$$+(n-1)d

$$\frac{2S_{n }}{n}$$-2$$a_{1 }$$=(n-1)d

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

d=$$\frac{2(S_{n }-na_{1 }}{n(n-1)}$$
Guest

### Re: Equation

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

### Re: Equation

$$a_{1 }$$- first member
Guest

Ok, thanks.
Guest