Question

If $m$ times the $m^{th}$term is equal to $n$times the $n^{th}$term of an A.P. Prove that ${(m+n)}^{th}$term of A.P. is zero

Answer 1

Solve for ${(m+n)}^{th}$ term.

An arithmetic progression is a series which has consecutive terms having a common difference between the terms as a constant value. It is used to generalize a set of patterns, that we observe in our day to day life. The general form of arithmetic progression is given by $a,a+d,a+2d,a+3d,.......$

The formula to find the $n^{th}$term is:

$a_n=a+(n-1)\times d$

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

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

From the given conditions, i.e. $m$ times the $m^{th}$term is equal to $n$ times the $n^{th}$term of an A.P.

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

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

$\Rightarrow a(m-n)+(m+n)(m-n)d-(m-n)d=0$

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

Rejecting the non-trivial case of $m=n,$we assume that $m$and $n$ are different.

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

The L.H.S of this equation denotes the ${(m+n)}^{th}$term of the AP, which is zero.

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