Question

C. Make n subject of the formulaS= n/2 [2a + (n - 1)d]please provide your working out ​

Answer 1

Answer:

$n=\frac{4as -2ds}{(1-2ds)}$

Step-by-step explanation:

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

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

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

Simplifying the fraction by multiplying 2 throughout,

$s= \frac{n}{4a + 2dn-2d}$

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

$(4a + 2dn-2d)s= \frac{n}{4a + 2dn-2d} (4a + 2dn-2d)$

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 $(1-2ds)$,

$n=\frac{4as -2ds}{(1-2ds)}$