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# If in an AP $m$ times $\;m^{th}\;$ term equals $n$ times $\;n^{th}$ term , then $\;(m+n)^{th}$ term will be

$(a)\;1\qquad(b)\;0\qquad(c)\;m+n\qquad(d)\;(m+n)\;a_{m}$

Can you answer this question?

Answer : (b) 0
Explanation : $\;m*a_{m}=n*a_{n}$
$m*[a+(m-1)d]=n*[a+(n-1)d]$
$(m-n)\;(a-d)+(m^2-n^2)\;d=0$
$(m-n)\;[a+(m+n-1)d]=0$
$a_{m+n}=0.$
answered Jan 21, 2014 by