If m times the mth term of an AP is equal to n times its nth term, find the value of the
(m + n)th term.
well, after expanding and factorizing I stuck with dividing the whole thing by m-n/...
Hello,
Well, I think you're on the right track ^^
$\displaystyle a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr$
$\displaystyle a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr$
where $\displaystyle a_0$ is the first term of the sequence and r the constant progression.
The text says that $\displaystyle m \cdot a_m=n \cdot a_n$
$\displaystyle m(a_0+mr)=n(a_0+nr)$
$\displaystyle m \cdot a_0+m^2r-n \cdot a_0-n^2r=0$
$\displaystyle r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0$
Factoring by m-n :
$\displaystyle (m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0$
Conclude ^^
It's the same. See, if you start at 0 instead of 1, you will have n instead of n-1.
Ok, if you want to do this way...
$\displaystyle m[a_1+(m-1)r]=n[a_1+(n-1)r]$
$\displaystyle ma_1+m^2r-mr-na_1-n^2r+nr$
$\displaystyle a_1(m-n)+r[m^2-m-n^2+n]=0$
$\displaystyle a_1(m-n)+r[m^2-n^2-(m-n)]=0$
$\displaystyle a_1{\color{red}(m-n)}+r[{\color{red}(m-n)}(m+n)-{\color{red}(m-n)}]=0$
Factor by m-n :
$\displaystyle {\color{red}(m-n)}[a_1+(m+n-1)r]=0$