Question

If $m$ times the $m^{th}$ term of an AP is equal to $n$ times its $n^{th}$ term, show that the $(m+n{)}^{th}$ term of the AP is zero.

Answer 1

$m\times a_m=n\times a_n$

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

$am+m^{2}d-md=an+n^{2}d-nd$

$am-an=-m^{2}d+n^{2}d-nd+md$

$a(m-n)=d(n^2-m^2+m-n)$

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

$a(m-n)=d(m-n)\{-1(n+m)+1\}$

$a=d(-n-m+1)$...(1)

To prove: $(m+n)th$ term is zero

$(m+n)thterm=a+(x-1)d$

here, $x=$ number of terms

$=a+(m+n-1)d$...(2)

Substituting $a=d(-n-m+1)$ in (2)

$=d(-n-m+1)+(m+n-1)d$

$=-dn-dm+d+md+nd-d$

$=0$