Question

Change the subject of formula for $d$.

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

• A. $\frac{s+2an}{n(n-1)}$
• B. $\frac{2s-an}{n(n-1)}$
• C. $\frac{2s-2an}{n(n-1)}$
• D. $\frac{2s+2an}{n(n-1)}$

Answer 1

The correct option is C $\frac{2s-2an}{n(n-1)}$

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

$2S=n[2a+(n-1\left)d\right]$

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

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

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

$\therefore d=\frac{2S-2an}{n(n-1)}$