Arithmetic Sequence

IshaanM8

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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

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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:

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).
 

Jomo

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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?
 

Harry_the_cat

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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?
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

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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

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Jomo

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