Arithmetic Sequence

[IshaanM8]

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!

 

[Deleted member 4993]

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!

If the first term of the AP (arithmatic progression or sequence) is 'a' and the common differnce is 'd', then:

t_n = a + (n-1)*d = m and ...................(1)

t_m = 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).

 

[Steven G]

If the first term of the AP (arithmatic progression or sequence) is 'a' and the common differnce is 'd', then:

t_n = a + (n-1)*d = m and ...................(1)

t_m = 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 t_n = m. Since d> 0 it follows that t_m > t_n = m. But t_m = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So t_n=a = m and t_m =a =n. Then m=n, another contraction. How do I finish this up?

 

[Harry_the_cat]

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 t_n = m. Since d> 0 it follows that t_m > t_n = m. But t_m = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So t_n=a = m and t_m =a =n. Then m=n, another contraction. How do I finish this up?

How can you conclude that d is not negative?

(1)-(2): \(\displaystyle (n-m)*d = m-n\)

So \(\displaystyle d = -1\)

\(\displaystyle t_{n+m} = a + (n+m-1)*d =(a + (n-1)*d) +m*d = m +m*d = m + m*-1 =0 \)

 

[Steven G]

How can you conclude that d is not negative?

(1)-(2): \(\displaystyle (n-m)*d = m-n\)

So \(\displaystyle d = -1\)

\(\displaystyle t_{n+m} = a + (n+m-1)*d =(a + (n-1)*d) +m*d = m +m*d = m + m*-1 =0 \)

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!

 

[Denis]

Very sloppy thinking on my part!

What else is new?

 

[Steven G]

What else is new?

Go to he double hockey sticks. It's actually in Michigan.