Question

$s=\frac{n}{2}[2a+(n-1\left)d\right]$
Hence, find ${}^{\prime }{d}^{\prime }$, if $s=144,a=1$ and $n=12$

Answer 1

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

Or $2S=n(2a+nd-d)$
Or $2S=2an+n^{2}d-nd$
Or $n^{2}d-nd=2S-2an$
Or $dn(n-1)=2S-2an$
Or $d=\frac{2S-2an}{n(n-1)}$
Place value $S=144,a=1,n=12$
Then $d=\frac{2\times 144-2\times 1\times 12}{12(12-1)}=\frac{288-24}{12\times 11}=\frac{264}{132}=2$