Arithmetic Sequence

IshaanM8

New member

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!

Subhotosh Khan

Super Moderator
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[SUB]n[/SUB] = a + (n-1)*d = m and ...................(1)

t[SUB]m[/SUB] = a + (m-1)*d = n and ...................(2)

t[SUB](n+m)[/SUB] = a + (m+n-1)*d =? ..................(3)

Using (1) and (2), you can solve for 'a' and 'd' and evaluate t[SUB](n+m) [/SUB]from (3).

Jomo

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

t[SUB]n[/SUB] = a + (n-1)*d = m and ...................(1)

t[SUB]m[/SUB] = a + (m-1)*d = n and ...................(2)

t[SUB](n+m)[/SUB] = a + (m+n-1)*d =? ..................(3)

Using (1) and (2), you can solve for 'a' and 'd' and evaluate t[SUB](n+m) [/SUB]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[SUB]n[/SUB] = m. Since d> 0 it follows that t[SUB]m[/SUB] > t[SUB]n[/SUB] = m. But t[SUB]m[/SUB] = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So t[SUB]n[/SUB]=a = m and t[SUB]m[/SUB] =a =n. Then m=n, another contraction. How do I finish this up?

Harry_the_cat

Member
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[SUB]n[/SUB] = m. Since d> 0 it follows that t[SUB]m[/SUB] > t[SUB]n[/SUB] = m. But t[SUB]m[/SUB] = n< m which is a contradiction. Since d is not positive (and not negative), then d = 0. So t[SUB]n[/SUB]=a = m and t[SUB]m[/SUB] =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 $$\displaystyle d = -1$$

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

Jomo

New member
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!

Jomo

New member
What else is new?
Go to he double hockey sticks. It's actually in Michigan.