Question

If $m$times the $m^{th}$term of an A.P. is equal to $n$times the $n^{th}$term, then ${(m+n)}^{th}$term of the A.P is _______ .

Answer 1

According to $n^{th}$ term of an arithmetic progression, we have

$t_n=$$n^{th}$term of AP $=a+\left(n-1\right)d............\left(1\right)$

$t_m=$$m^{th}$term of AP $=a+\left(m-1\right)d............\left(2\right)$

From the given information, we can write

$m\times t_m=n\times t_n$

From $\left(1\right)and\left(2\right)$ we can write,

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

Now, simplify the equation, we get

$$\begin{gathered} am+m^{2}d-md=an+n^{2}d-nd \\ a(m-n)+(m+n)(m-n)d-(m-n)d=0 \end{gathered}$$

Now, take out common terms

$(m-n)[a+(m+n-1\left)d\right]=0$

$(a+(m+n-1\left)d\right)=0.......\left(3\right)$

From equation $\left(3\right),$we can say

$(m-n)[a+(m+n-1\left)d\right]=t_{m+n}......\left(4\right)$

Now comparing equations $\left(3\right)and\left(4\right)$, we can conclude that

$t_{m+n}=0$

Hence, ${(m+n)}^{th}$term of A.P. is $0.$