Question

Determine the $AP$ whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12$

Answer 1

By using nth term $t_n=a+(n-1)d$ we have

As given $a+2d=16$.......$2^{nd}$ term

And $a+6d-(a+4d)=12$

By using abovee equations,

$\Rightarrow d=6$

$\therefore a+2\left(6\right)=16$, then $a=4$

Then A.P is $a,a+d,a+2d,a+3d...............$

So, $4,4+6,4+2\left(6\right),4+3\left(6\right),4+4\left(6\right)..............$

$4,10,16,22,28,\mathrm{34............}$