If $p^\text{th}$ term of an A.P is $q$ and $q^\text{th}$ term of the A.P is $p$, how do you find its $r^\text{th}$ term.
This is the solution where I'm stuck at.
From the question we get $a_p=q$ & value of $a_q=p$, where $a$ is the Arithmetic equation.
So,
$a_p=a_1+(p-1)d \rightarrow q$
$a_q=a_1+(q-1)d \rightarrow p$
By substituting the value of $p$ in the first equation ie, $a_p$. We get:-
$a_p=a_1+(a_1+(q-1)d)d=q$
$a_p=a_1+(a_1+dq-d)d=q$
$a_p=a_1+a_1d+d^2q-d^2=q$
Hoping I'm doing this correctly till now, but can't seem to know what do next.
And how do you find the $r^\text{th}$ term?
You have $$a_p = a_1 + (p - 1)d = q$$ and $$a_q = a_1 + (q - 1)d = p.$$
Subtracting the two equations, you get: $$a_p - a_q = (p - q)d = q - p.$$
Assuming WLOG that $p \neq q$, then we obtain that the common difference $$d = -1.$$
Substituting $d=-1$ into the first two equations, we obtain $$a_p = a_1 - (p - 1) = q$$ and $$a_q = a_1 - (q - 1) = p.$$
Thus, we have $$a_1 + 1 = p + q,$$ or in other words $$a_1 = (p + q) - 1.$$
Consequently, the $r$th term is given by the expression $$a_r = a_1 + (r - 1)d = (p + q) - 1 + (r - 1)\cdot{-1} = p + q - 1 + 1 - r = p + q - r.$$