Question

Show that the sequence, defined by its nth term $\frac{3+n}{4},$ forms an AP. Also, find the common difference of it.

Answer 1

To show

$\frac{3+D}{4}$ term on $AP$

Supposed let $t_1=\frac{3+Ss}{4}$

and $n_1=1$ and $d=4$

$\therefore t_1=\frac{3+1}{4}=1$

Similarly $t_2=\frac{3+(n_1+4)}{4}$

$=\frac{3+5}{4}$

$=\frac{8}{4}$

$=2$

$t_3=\frac{3+(n_2+4)}{4}$

$=\frac{3+9}{4}$

$=\frac{13}{4}$

$=3$

$\therefore t_1=1,t_2=2,t_3=3$ shows that $\frac{3+n}{4}$ is an $AP$ and the common difference is one.