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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!

t

t

t

Using (1) and (2), you can solve for 'a' and 'd' and evaluate t

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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 tIf 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).

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How can you conclude that d is not negative?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?

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

So \(\displaystyle d = -1\)

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

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I did say thatHow can you conclude that d is not negative?

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

So \(\displaystyle d = -1\)

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

What else is new?Very sloppy thinking on my part!

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Go toWhat else is new?