A question about AP.

Question

How is the encircled step justifiable? According to my knowledge I can substitute m=any variable but how can I substitute m=2m-1, isn't it the same as assuming m=1?

Answer 1

$$\dfrac mn=\dfrac{a+d\cdot\dfrac{m-1}2}{a+d\cdot\dfrac{n-1}2}$$

Substitute $\dfrac{m-1}2=M-1\iff m=2M-1,\dfrac{n-1}2=N-1$

to find the ratio of $M,N$ terms

No confusion, I hope

Answer 2

In equation 1 we have obtained a relationship that any 2 quantities, m and n when taken in ratio, $\frac{(2a + (m-1)d)}{(2a + (n-1)d)} = \frac{(m)}{(n)}$ are of the form of LHS in eq1. After that we see what happens when the two quantities are of the form 2m-1 and 2n-1 we just replace m and n with the corresponding new forms. We don't actually consider them equal. For example 3 is a number of the form 2m-1 where m is 2. But 4 isn't of the form 2m-1, not for m being an integer.

Answer 3

If a statement is true for every m and n, it is also true for 2m-1 and 2n-1. For example from (m + n)^2 = m^2 + n^2 + 2mn, we get (2m+2n-2)^2 = (2m-1)^2 + (2n-1)^2 + 2(2m-1)(2n-1).

Answer 4

After you are convinced with

$$ \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n} $$

Simply, Cross multiply it to get,

$$ 2an + n(m-1)d = 2am + m(n-1)d $$

$$ \implies \require{cancel} \cancel{ (m-n)} d = 2a\cancel{(m-n)} $$

$$ \implies d= 2a $$

Now it is obvious to find the ratio of $m^{th}:n^{th}$ terms' ratio.

$ \frac{m^{th} \ \text{term}}{n^{th} \ \text{term}} = \frac{a+(m-1)2a}{a+(n-1)2a} $ [ Since $d=2a$]

$$ = \frac{1+(2m-2)}{1+(2n-2)} = \frac{(2m-1)}{(2n-1)} \square $$

Answer 5

The question itself is confusing. What it means is that the condition is true for any $m,n$ you care to choose, rather than a specific $m,n$.

Given that it is true for any $m,n$ it is somewhat confusing to use the same symbols in the suggested answer, as has been done.

Maybe clearer to say that if it is true for any $m,n$ then we can pick the specific values $m=2r-1, n=2t-1$ and go from there.

It doesn't matter what you call your variables - you can give them what names you like. So having worked out an expression in $r,t$, which is true for any $r$ and $t$, it is true in particular for $r=m, t=n$.

The essential thing here is that the expression in $r$ and $t$ has been shown to be true for any choice of $r$ and $t$ (or any choice of interest) and does not depend on the original relation with $m$ and $n$.