Arithmetic progression problem

AI Thread Summary
In the discussion about an arithmetic progression problem, participants explore the values of a_m and a_n given a_m+n = A and a_m-n = B. One contributor correctly identifies that a_m = (A + B) / 2 but struggles to derive a_n, while the textbook solution states a_n = [(2n - m)A + mB] / 2. Confusion arises regarding the calculation of a_n, with some suggesting a typo in the textbook. The key point is that the difference A - B equals 2nd, where d is the common difference, leading to insights about the evenness of differences in the sequence. Overall, the discussion emphasizes understanding arithmetic progressions and verifying solutions through examples.
Government$
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Homework Statement



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

The Attempt at a Solution


I fugured that a_{m}=\frac{A+B}{2} but i have no idea what a_{n} is.
In my textbook solution is a_{n}=\frac{(2n-m)A + mB}{2}
How did they arrived to that solution?
 
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Hi Government$! :smile:
Government$ said:
In my textbook solution is a_{n}=\frac{(2n-m)A + mB}{2}
How did they arrived to that solution?

an = am + (n-m)∆ :wink:
 
Hi,

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

How did you calculate that?
 
Government$ said:
Hi,

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

I get the same answer you do. I think the textbook has a typo.
 
symbolipoint said:
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.
 
Last edited:
Curious3141 said:
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.
 
Government$ said:
Hi,

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

Try some examples to verify it.
 
symbolipoint said:
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.
 
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Thanks for help, everybody.
 
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