Arithmetic progression problem

[Government$]

Homework Statement

Let [itex]a_{m+n}=A[/itex] and [itex]a_{m-n}=B[/itex] be members of arithmetic progression then [itex]a_{m}[/itex] and [itex]a_{n}[/itex] are? (m>n).

The Attempt at a Solution

I fugured that [itex]a_{m}=\frac{A+B}{2}[/itex] but i have no idea what [itex]a_{n}[/itex] is.
In my textbook solution is [itex]a_{n}=\frac{(2n-m)A + mB}{2}[/itex]
How did they arrived to that solution?

 

[tiny-tim]

Hi Government$!

In my textbook solution is [itex]a_{n}=\frac{(2n-m)A + mB}{2}[/itex]
How did they arrived to that solution?

a_n = a_m + (n-m)∆

 

[Government$]

Hi,

using the fact that [itex]A-B=2dn[/itex] and plugging that into your equation and then finding
[itex]a_{n-m}[/itex] from [itex]a_{n}=\frac{A+a_{n-m}}{2}[/itex] i got solution [itex]a_{n}=\frac{(2n-m)A + mB}{2n}[/itex]. So i have one extra [itex]n[/itex] that i can't get rid of.

 

[symbolipoint]

using the fact that A−B=2dn

How did you calculate that?

 

[Curious3141]

Hi,

using the fact that [itex]A-B=2dn[/itex] and plugging that into your equation and then finding
[itex]a_{n-m}[/itex] from [itex]a_{n}=\frac{A+a_{n-m}}{2}[/itex] i got solution [itex]a_{n}=\frac{(2n-m)A + mB}{2n}[/itex]. So i have one extra [itex]n[/itex] that i can't get rid of.

I get the same answer you do. I think the textbook has a typo.

 

[Curious3141]

How did you calculate that?

Because there are 2n terms between term (m-n) and term (m+n), which gives a difference of 2nd, where d is the common difference of the arithmetic progression. So A - B = 2nd.

 

[symbolipoint]

Because there are 2n terms between term (m-n) and term (m+n), which gives a difference of nd, where d is the common difference of the arithmetic progression. So A - B = 2nd.

I am beginning to understand. The difference would need to be even, since there are TWO differences involved among the m and the n terms.

 

[SammyS]

Hi,

using the fact that [itex]A-B=2dn[/itex] and plugging that into your equation and then finding
[itex]a_{n-m}[/itex] from [itex]a_{n}=\frac{A+a_{n-m}}{2}[/itex] i got solution [itex]a_{n}=\frac{(2n-m)A + mB}{2n}[/itex]. So i have one extra [itex]n[/itex] that i can't get rid of.

[itex]\displaystyle a_{n}=\frac{(2n-m)A + mB}{2n}\ \ [/itex] looks right to me.

Try some examples to verify it.

 

[Curious3141]

I am beginning to understand. The difference would need to be even, since there are TWO differences involved among the m and the n terms.

Sorry, I just noticed a typo in my post (since edited and corrected). I meant that since there are 2n terms between the two terms, the difference is 2nd.

There is only one common difference d in an arithmetic progression.

 

[Government$]

Thanks for help, everybody.