What is the answer to solvefor d,S= n/2 [2a+(n-1)d] ?

The answer to solvefor d,S= n/2 [2a+(n-1)d] is d=(2S-2na)/(n(n-1));\quad n\ne 0,n\ne 1

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

solve for

d, S = \frac{n}{2} [2 a + (n - 1) d]

d = \frac{2 S - 2 na}{n ( n - 1 )}; n \neq = 0, n \neq = 1

Solution steps

S = \frac{n}{2} [2 a + (n - 1) d]

Switch sides

\frac{n}{2} [2 a + (n - 1) d] = S

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

\frac{n ( 2 a + d ( n - 1 ) )}{2} = S

Multiply both sides by

2

n (d (n - 1) + 2 a) = 2 S

Divide both sides by

n; n \neq = 0

d (n - 1) + 2 a = \frac{2 S}{n}; n \neq = 0

Move

2 a

to the right side

d (n - 1) = \frac{2 S}{n} - 2 a; n \neq = 0

Divide both sides by

n - 1; n \neq = 0, n \neq = 1

d = \frac{2 S - 2 na}{n ( n - 1 )}; n \neq = 0, n \neq = 1

4 x - 5 = 3

- 9 n + 19 n - 17 = 13

- 10 + 3 c + 1 = 3 + 5 c

s, v^{2} = u^{2} + 2 a s

x, 2 x - 2 y = 0

What is the answer to solvefor d,S= n/2 [2a+(n-1)d] ?
The answer to solvefor d,S= n/2 [2a+(n-1)d] is d=(2S-2na)/(n(n-1));\quad n\ne 0,n\ne 1