# Arithmetic progression problem

1. Jan 3, 2013

### Government$1. The problem statement, all variables and given/known data Let $a_{m+n}=A$ and $a_{m-n}=B$ be members of arithmetic progression then $a_{m}$ and $a_{n}$ are? (m>n). 3. 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 text book solution is $a_{n}=\frac{(2n-m)A + mB}{2}$ How did they arrived to that solution? 2. Jan 3, 2013 ### tiny-tim Hi Government$!
an = am + (n-m)∆

3. Jan 3, 2013

### Government$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 cant get rid of. 4. Jan 3, 2013 ### symbolipoint How did you calculate that? 5. Jan 4, 2013 ### Curious3141 I get the same answer you do. I think the textbook has a typo. 6. Jan 4, 2013 ### Curious3141 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: Jan 4, 2013 7. Jan 4, 2013 ### symbolipoint 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. 8. Jan 4, 2013 ### SammyS Staff Emeritus $\displaystyle a_{n}=\frac{(2n-m)A + mB}{2n}\ \$ looks right to me. Try some examples to verify it. 9. Jan 4, 2013 ### Curious3141 Sorry, I just noticed a typo in my post (since edited and corrected). I meant that since there are 2n terms between the two terms, the difference is 2nd. There is only one common difference d in an arithmetic progression. 10. Jan 4, 2013 ### Government$

Thanks for help, everybody.