Difficult arithmatic progression question

**ice_syncer** · July 4th 2008, 02:57 AM

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/...

**Moo** · July 4th 2008, 03:21 AM

Hello,

Originally Posted by ice_syncer

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/...

Well, I think you're on the right track ^^

$a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr$
$a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr$

where $a_0$ is the first term of the sequence and r the constant progression.

The text says that $m \cdot a_m=n \cdot a_n$

$m(a_0+mr)=n(a_0+nr)$

$m \cdot a_0+m^2r-n \cdot a_0-n^2r=0$

$r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0$

Factoring by m-n :

$(m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0$

Conclude ^^

**ice_syncer** · July 4th 2008, 03:26 AM

Originally Posted by Moo

Hello,

Well, I think you're on the right track ^^

$a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr$
$a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr$

where $a_0$ is the first term of the sequence and r the constant progression.

The text says that $m \cdot a_m=n \cdot a_n$

$m(a_0+mr)=n(a_0+nr)$

$m \cdot a_0+m^2r-n \cdot a_0-n^2r=0$

$r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0$

Factoring by m-n :
$(m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0$

Conclude ^^

Hey uh.... I was taught that the nth term is
a1 + (n-1)d = nth term
where a1 is the first term of AP

**Moo** · July 4th 2008, 03:31 AM

Originally Posted by ice_syncer

Hey uh.... I was taught that the nth term is
a1 + (n-1)d = nth term
where a1 is the first term of AP

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...

$m[a_1+(m-1)r]=n[a_1+(n-1)r]$

$ma_1+m^2r-mr-na_1-n^2r+nr$

$a_1(m-n)+r[m^2-m-n^2+n]=0$

$a_1(m-n)+r[m^2-n^2-(m-n)]=0$

$a_1{\color{red}(m-n)}+r[{\color{red}(m-n)}(m+n)-{\color{red}(m-n)}]=0$

Factor by m-n :

${\color{red}(m-n)}[a_1+(m+n-1)r]=0$

**ice_syncer** · July 4th 2008, 03:31 AM

Originally Posted by Moo

Hello,

Well, I think you're on the right track ^^

$a_m ~:~ \text{m-th term } \rightarrow a_m=a_0+mr$
$a_n ~:~ \text{n-th term } \rightarrow a_n=a_0+nr$

where $a_0$ is the first term of the sequence and r the constant progression.

The text says that $m \cdot a_m=n \cdot a_n$

$m(a_0+mr)=n(a_0+nr)$

$m \cdot a_0+m^2r-n \cdot a_0-n^2r=0$

$r\underbrace{(m^2-n^2)}_{(m-n)(m+n)}+a_0(m-n)=0$

Factoring by m-n :

$(m-n)[\underbrace{r(m+n)+a_0}_{\text{(m+n)th term !!!}}]=0$

Conclude ^^

ok I got it the last factorization step was what helped me, thanks alot!

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