Question

C. Make n subject of the formula

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

please provide your working out ​

Answer 1

`n=(4as -2ds)/((1-2ds))`

Explanation:

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

Simplifying the fraction by multiplying d into the (n-1) term,

`s=(n)/(2 [2a + (n - 1)d] ) = (n)/(2[2a + dn - d] )`

Simplifying the fraction by multiplying 2 throughout,

`s= (n)/(4a + 2dn-2d)`

Multiply
`(4a + 2dn-2d)` on both sides

`(4a + 2dn-2d)s= (n)/(4a + 2dn-2d) (4a + 2dn-2d)`

Cancel the
`(4a + 2dn-2d)` on the right hand side

`(4a + 2dn-2d)s= n`

Multiply s to the terms,

`4as + 2dns-2ds= n`

Move
`2dns` to the right hand side by subtracting
`2dns` on both sides
`4as + 2dns-2ds-2dns= n-2dns`

`4as -2ds= n-2dns`

On the right hand side of the equation, take out
`n`

`4as -2ds= n(1-2ds)`

Divide Left hand side by
`(1-2ds)`,

`n=(4as -2ds)/((1-2ds))`