Question

Let $<a_n>$ and $<b_n>$ be the arithmetic sequences each with common difference 2 such that $a_1<b_1$ and let $c_n=\sum _{k=1}^n{a}_k,dn=\sum _{k=1}^n{b}_k$. Suppose that the points $A_n\left(a_n,c_n\right),B_n\left(b_n,d_n\right)$ are all lying on the parabola $C:y=p{x}^2+qx+r$ where $p,q,r$ are constants.
The value of $p$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 2

Answer 1

Answer: (A)

Given $c_n=a_1+a_2+a_3+\dots .+a_n$
where $a_1,a_2,\dots \dots .a_n$ are in A.P. with $d=2$
and $d_n=b_1+b_2+b_3+\dots ..+b_n$ are in A.P. with $d=2$.
also $\left(a_n,c_n\right)$ lies on $y=p^2+qx+r$
Now $c_n=p{a}_n+qa{a}_n+r$....(1)
$c_{n-1}=p{a}^2{}_{n-1}+q{a}_{n-1}+r$....(2)
$\therefore$ from(1) and (2), we get
$c_n-c_{n-1}=p\left(a^2-a^2{}_{n-1}\right)+q\left(a_n-a_{n-1}\right)$
$\Rightarrow a_n=p\left(a_n+a_{n-1}\right)\left(a_n-a_{n-1}\right)+q\left(a_n-a_{n-1}\right)$
$a_n=\left(a_n-a_{n-1}\right)\left[p\left(a_n+a_{n-1}\right)+q\right]$....(3)
$\left(a_n-a_{n-1}=d\right)$
on putting $n=2$ and 3 in equation (3), we get
$a_2=d\left[p\left(a_2+a_1\right)+q\right]$
$a_3=d\left[p\left(a_3+a_2\right)+q\right]$
Now (5)-(4), we get

$4p=1\Rightarrow p=\frac{1}{4}$
$\mathrm{adding}(4)\text{ and }(5)\Rightarrow q=\frac{1}{2}$