Let $<a_n>$ and $<b_n>$ be the arithmetic progression sequences each with common difference 2 such that $a_1<b_1$ and let $c_n=\sum ^n_{k=1} a_k$ and $d_n=\sum ^n _{k=1} b_k$. Suppose the points $A_n (a_n, c_n)$ and $B_n(b_n,d_n)$ are all lying on the parabola, $y=px^2+qx+r$, then find value of $p,q$
Starting off, I am really confused on what the question is trying to say. I typed it out as it is.
I can find out $c_n$ and $d_n$ in terms of n and $a_n$
$$c_n=\frac n2 (a_1+a_n)$$
But that doesn’t get me anywhere. I just want some insight.
The points $A_n(a_n,c_n)$ and $B_n(b_n,d_n)$ lie on the parabola. This means that $$c_n=pa_n^2+qa_n+r \\ d_n=pb_n^2+qb_n+r$$ for each $n$. Subtract the two equations to obtain $$d_n-c_n=p(b_n-a_n)(b_n+a_n)+q(b_n-a_n) $$ Note that $$b_n-a_n=(b_{n-1}+2)-(a_{n-1}+2)=b_{n-1}-a_{n-1}=\ldots=b_1-a_1 $$ and $$d_n-c_n=(b_n-a_n)+(b_{n-1}-a_{n-1})+\cdots+(b_1-a_1)=n(b_1-a_1) $$ so we have $$n(b_1-a_1)=p(b_1-a_1)(b_n+a_n)+q(b_1-a_1) \\ \Rightarrow n=p(b_n+a_n)+q $$ where we divided by $b_1-a_1\ne 0$. This holds for each $n$ so you can take $n=1$ and $n=2$ to obtain $$1=p(a_1+b_1)+q \\ 2=p(a_2+b_2)+q $$ Then subtract the two equations (note that $a_2-a_1=2$ and $b_2-b_1=2$), and we have $4p=1\Leftrightarrow p=1/4$. To find $q$, use that $(a_1,a_1)$ and $(a_2,a_1+a_2)$ lie on the parabola: $$a_1=p(a_1)^2+qa_1+r \\ a_1+a_2=p(a_2)^2+qa_2+r $$ Again, subtract the two equations, substitute $a_2=a_1+2$, and then plug in $p=1/4$. You'll arrive at $2q=1\Leftrightarrow q=1/2$.