If the first term of the AP (arithmatic progression or sequence) is 'a' and the common differnce is 'd', then:View attachment 11016
That's the Question ^
I've just started learning arithmetic sequences in school. I don't know how to even approach this questions. Need some Assistance. Thanks!
Khan Academy, can you please help me finish my proof. Assume d> 0 (you can do a separate case where d< 0) and wolog assume n<m. Now tn = m. Since d> 0 it follows that tm > tn = m. But tm = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So tn=a = m and tm =a =n. Then m=n, another contraction. How do I finish this up?If the first term of the AP (arithmatic progression or sequence) is 'a' and the common differnce is 'd', then:
tn = a + (n-1)*d = m and ...................(1)
tm = a + (m-1)*d = n and ...................(2)
t(n+m) = a + (m+n-1)*d =? ..................(3)
Using (1) and (2), you can solve for 'a' and 'd' and evaluate t(n+m) from (3).
Khan Academy, can you please help me finish my proof. Assume d> 0 (you can do a separate case where d< 0) and wolog assume n<m. Now tn = m. Since d> 0 it follows that tm > tn = m. But tm = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So tn=a = m and tm =a =n. Then m=n, another contraction. How do I finish this up?
I did say that you can do a separate case where d< 0. I thought the case with d<0 it would be symmetrical to the case where d>0. Apparently I was wrong!Very sloppy thinking on my part!How can you conclude that d is not negative?
(1)-(2): (n−m)∗d=m−n
So d=−1
tn+m=a+(n+m−1)∗d=(a+(n−1)∗d)+m∗d=m+m∗d=m+m∗−1=0
What else is new?Very sloppy thinking on my part!
Go to he double hockey sticks. It's actually in Michigan.What else is new?