If m times the mth term of an AP is equal to n times its nth term, then show that (m + n)th term of an AP is zero.

Explanation

Let a be the first term and d be the common difference of the given A.O. Then m times mth term = n times nth term. " "…(Given)
`thereforemt_(m)=nt_(n)`
Now, `t_(m)=a+(m-1)dand t_(n)=a+n-1)d`
`thereforem[a+(m-1)d]=n[a+(n-1)d]`
`thereforem[a+(m-1)d]-n[a+(n-1)d]=0`
`therefoream+m(m-1)d-na-n(n-1)d=0`
`therefoream+na+m(m-1)d-n(n-1)d=0`
`thereforea(m-n)+(m^(2)-m)d-(n^(2)-n)d=0`
`thereforea(m-n)+d[m^(2)-m-n^(2)+n]`
`thereforea(m-n)+d[m^(2)-n^(2)-m+n]=0`
`thereforea(m-n)+d[(m-n)(m-n)-(m-n)]=0`
`thereforea+d(m+n-1)=0 ...[becausem-n""ne0]...(1)`
Now, `(m+n)th` term of the A.P. `= a+(m+n-1)d` ...(2)
`therefore` from (1) and (2) ,
`(m+n)th term of the A.P. is zero.