Question

If $m$ time the $n^{th}$ terms of an A.P. is equal to $m$ times the $n^{th}$ term, find $m+n^{th}term$?

Answer 1

According to question,

$m(a+(n-1\left)d\right)=n(a+(m-1\left)d\right)$

$\to am+mnd-md=an+mnd-nd$

$\to a(m-n)=d(m-n)$

$a=d$

$\therefore {(m+n)}^n$ term $=a+\left[\right(m+n)-1]a$

$=(m+n)a$