Arithmetic mean b/w two terms of an A.P

Question

I came across the following question in my textbook today:

If A.M b/w $p$th and $q$th terms of an A.P be equal to the A.M b/w the $r$th and $s$th terms of the A.P, then prove that $p+q=r+s$

My first instinct was to assume that there is but a single A.M b/w $p$ and $q$, and similarly for $r$ and $s$. Proceeding in this manner, we get

$(p+q)/2$=$(r+s)/2$ which gives us the required condition almost too easily.

I dunno how to proceed in any other way because I'm unable to interpret the question in any other way correctly; if there are actually more than one A.Ms b/w $p$ & $q$ and $r$ & $s$, then, we still don't know if they have the same number of A.Ms b/w them or different.

Supposing that there really are more than one A.Ms b/w them, what does the statement "A.M b/w $p$th and $q$th terms of an A.P be equal to the A.M b/w $r$th and $s$th terms of the A.P" even mean? Do they mean to say that the arithmetic mean of the arithmetic means are equal?

Please help! MUCH thanks in advance :) Regards.

Answer 1

The pth term of an arithmetic progression is given by the formula $a + (p-1)d$, where $a$ is the first term and $d$ is the difference.

So, we can write down the other terms:

$q$th term $= a + (q-1)d$

$r$th term $= a + (r-1)d$

$s$th term $= a + (s-1)d$

If $d=0$ the problem is obvious, and I'll leave it to you. We will assume that $d \neq 0$.

Now, the arithmetic mean of two quantities is just their average. So the arithmetic mean of the $p$th and the $q$th term is $\frac{a + (p-1)d+a + (q-1)d}{2}$. Similarly, the arithmetic mean of the $r$th and the $s$th term is $\frac{a + (r-1)d+a + (s-1)d}{2}$.

Now the question says that these are equal hence, we see that: $$ \frac{a + (r-1)d+a + (s-1)d}{2} = \frac{a + (p-1)d+a + (q-1)d}{2} $$

Subtract the quantity $\frac{2a-2d}{2}$ from both sides, and you will get:

$$ \frac{rd+sd}{2} = \frac{pd+ qd}{2} $$ Finally, divide by $d/2$ on both sides: $$ r+s=p+q $$ This should do it.

Answer 2

it is given $$a_p=a_1+(p-1)d,a_q=a_1+(q-1)d$$ and $$a_r=a_1+(r-1)d,a_s=(s-1)d$$ thus we obtain $$\frac{a_p+a_q}{2}=\frac{2a_1+(p-1)d+(q-1)d}{2}$$ and $$\frac{a_r+a_s}{2}=\frac{2a_1+(r-1)d+(s-1)d}{2}$$ and we get $$pd+qd=rd+sd$$ if $d\ne 0$ we get $$p+q=r+s$$ what was to show