Arithmetic Progression, Tn and Tm proving sum

Question

How can I prove this?

If in an A.P., Tm=n, Tn=m, prove d=-1.

What are Tm, Tn, n, m, d? Providing context helps people answer your question — lioness99a, May 31 '17 at 12:31
In arithmetic progression, there are sequences, in which first number is referred as a or T1, second is T2, third is T3 and so on..., there is one formula called Tn=a+(n-1)d , and second one is Sn=n/2[2a+(n-1)d], these are for those who are studying in class 10 — Chandan Das, May 31 '17 at 14:13
Then add this to your question, for those people who don't know what they mean. Also, not everyone knows what 'class 10' is (I certainly don't) — lioness99a, May 31 '17 at 14:27

score 0 · Accepted Answer · answered May 31 '17 at 11:03

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Assume it is given that $m\ne n$.

You have $T(m)=n$ and $T(n)=m$. So,

$$a+(m-1)d=n\quad \text{and} \quad a+(n-1)d=m$$

Hint: Subtract the two equalities.

answered May 31 '17 at 11:03

CY Aries

I came up to a+nd-d=a+md-d what to do next? – Chandan Das May 31 '17 at 11:24
I think you have made a mistake. $a+nd-d=m$ and $a+md-d=n$. They are not equal. – CY Aries May 31 '17 at 11:26
Subtract them. You will have $(a+nd-d)-(a+md-d)=m-n$. – CY Aries May 31 '17 at 11:27
OK, SO NOW I'VE GOT m-n=(n-m)d what about next? Should I have to pplut values of m and n or something else? – Chandan Das May 31 '17 at 11:33
Since $n-m\ne0$, you can divide both sides by $n-m$. – CY Aries May 31 '17 at 11:34
what have you written and why are the dollar signs – Chandan Das May 31 '17 at 11:37
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference Mathjax. Try to read the post on computer. – CY Aries May 31 '17 at 11:41
Ok, so after dividing both sides by n-m I've got this m-n/n-m=d – Chandan Das May 31 '17 at 11:41
And that is $-1$ on L.H.S. – CY Aries May 31 '17 at 11:42
And that's my question why m-n/n-m is equal to - 1? – Chandan Das May 31 '17 at 11:43
1

$(-1)(n-m)=-n+m=m-n$. – CY Aries May 31 '17 at 11:44

score 0 · Answer 2 · answered May 31 '17 at 11:04

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Notwithstanding the imprecision of the question, and assuming $m\neq n$:

The slope is $\frac{T_m-m}{T_n-n}=\frac{n-m}{m-n}=-1$

answered May 31 '17 at 11:04

Evargalo

score 0 · Answer 3 · answered May 31 '17 at 11:14

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we have

$(T_i) $ arithmetic sequence

$$\implies T_n=T_m+(n-m)d $$

$$\implies m-n=(n-m)d $$

$$\implies (d+1)(n-m )=0$$

$$\implies d=-1$$

answered May 31 '17 at 11:14

hamam_Abdallah

3 Answers3