Question

If $m$ times the $m^{th}$ term of an A.P. is equal to $n$ times its $n^{th}$ term. Find the value of $(m+n{)}^{th}$ term of the A.P.

• A. $2$
• B. $1$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

We know $a_n=a+(n-1)d$
$a_{m+n}=a+(m+n)d-----------\left(i\right)$
Let the first term and common difference of the A.P be $aandd$
Then
$m^{th}term=a+(m-1)d$
$n^{th}term=a+(n-1)d$
By the given condition
$m\times a_m=n\times a_n$
$m[a+(m-1)d]=n[a+(n-1)d]$
$\Rightarrow ma+m(m-1)d=na+n(n-1)d$
$\Rightarrow ma+(m^2-m)d-na-(n^2-n)d=0$
$\Rightarrow ma-na+(m^2-m)d-(n^2-n)d=0$
$\Rightarrow a(m-n)+d(m^2-n^2-m+n)=0$
$\Rightarrow a(m-n)+d[(m+n)(m-n)-(m-n)]=0$
Now divide both sides by $m-n$
$\Rightarrow a\left(1\right)+d[(m+n)\left(1\right)-\left(1\right)]=0$
$\Rightarrow a+d(m+n-1)=0...\left(ii\right)$
From equation number 1 and 2
$a_{m+n}=a+(m+n-1)d$
And we have shown
$a+d(m+n-1)=0$
So $a_{m+n}=0$