Question

Let $< a _{ n }>$ and $< b _{ n }>$ be the arithmetic sequences each with common difference 2 such that $a _1< b _1$ and let $c_n=\displaystyle\sum_{k=1}^n a_k, d n=\displaystyle\sum_{k=1}^n b_k$. 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 $C: y= px ^2+ qx + r$ 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 $c_n=a_1+a_2+a_3+\ldots .+a_n$
where $a _1, a _2, \ldots \ldots . a _{ n }$ are in A.P. with $d =2$
and $d _{ n }= b _1+ b _2+ b _3+\ldots . .+ b _{ n }$ are in A.P. with $d =2$.
also $\left(a_n, c_n\right)$ lies on $y=p^2+q x+r$
Now $c _{ n }= pa _{ n }+ qa a _{ n }+ r$....(1)
$c _{ n -1}= pa ^2{ }_{ n -1}+ qa _{ n -1}+ r$....(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

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