Question

For an AP, $\dfrac{{{S}_{kn}}}{{{S}_{n}}}$ is independent of n. The value of $\dfrac{d}{a}$ for this AP is
a) 1
b) 2
c) 3
d) 4

Answer 1

Hint: Let us assume that we have the following Arithmetic Progression (AP) series:
$a,a+d,a+2d,a+3d,........,a+\left( n-1 \right)d$ , where a is first term, d is common difference and n is number of terms. Since, sum of n terms of an AP is given as:${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, us the formula to find ${{S}_{n}}$ and ${{S}_{kn}}$ and then use the values to get a ratio between them in terms of d and a.

Complete step by step answer:
Since, we have an Arithmetic Progression (AP) series as: $a,a+d,a+2d,a+3d,........,a+\left( n-1 \right)d$
So, the sum of n terms of the AP is given as:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]......(1)$
Similarly, for $kn$ terms, we can write:
${{S}_{kn}}=\dfrac{kn}{2}\left[ 2a+\left( kn-1 \right)d \right]......(2)$
Now, we need to find $\dfrac{{{S}_{kn}}}{{{S}_{n}}}$.
So, divide equation (2) by equation (1), we get:
$\dfrac{{{S}_{kn}}}{{{S}_{n}}}=\dfrac{\dfrac{kn}{2}\left[ 2a+\left( kn-1 \right)d \right]}{\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]}.......(3)$
Now, by simplifying equation (3), we get:
$\dfrac{{{S}_{kn}}}{{{S}_{n}}}=\dfrac{k\left[ 2a+\left( kn-1 \right)d \right]}{\left[ 2a+\left( n-1 \right)d \right]}......(4)$
Let us assume that $\dfrac{{{S}_{kn}}}{{{S}_{n}}}=x$
So, we can write equation (4) as:
$x=\dfrac{k\left[ 2a+\left( kn-1 \right)d \right]}{\left[ 2a+\left( n-1 \right)d \right]}......(5)$
Now, simplify equation (5), we get:
$\begin{align}
  & \Rightarrow x\left[ 2a+\left( n-1 \right)d \right]=k\left[ 2a+\left( kn-1 \right)d \right] \\
 & \Rightarrow 2xa+xnd-xd=2ka+{{k}^{2}}nd-kd \\
 & \Rightarrow 2ak-2ax+xd-kd+\left( {{k}^{2}}d-xd \right)n=0......(6) \\
\end{align}$
So, we get:
Either $\left( {{k}^{2}}d-xd \right)=0......(7)$
Or $2ak-2ax+xd-kd=0......(8)$
From equation (7), we can write:
$\begin{align}
  & \Rightarrow {{k}^{2}}d-xd=0 \\
 & \Rightarrow {{k}^{2}}d=xd \\
 & \Rightarrow {{k}^{2}}=x......(9) \\
\end{align}$
Now, substitute equation (9) in equation (8), we get:
$\begin{align}
  & \Rightarrow 2ak-2ax+xd-kd=0 \\
 & \Rightarrow 2ak-2a{{k}^{2}}+{{k}^{2}}d-kd=0.....(10) \\
\end{align}$
Now, get all the terms of “a” on one side and terms of “d” on other side, we get:
$\begin{align}
  & \Rightarrow 2ak-2a{{k}^{2}}=-{{k}^{2}}d+kd \\
 & \Rightarrow 2ak\left( 1-k \right)=-kd\left( k-1 \right) \\
 & \Rightarrow 2ak\left( 1-k \right)=kd\left( 1-k \right) \\
 & \Rightarrow 2a=d......(11) \\
\end{align}$
Now, to get a ratio between d and a, divide equation (11) by a.
We get:
$\dfrac{d}{a}=2$

So, the correct answer is “Option B”.

Note: The sum of n terms of an AP is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
This is deduced as follows:
Sum of n terms of an AP ${{S}_{n}}=\dfrac{n}{2}\left[ \text{first term + last term} \right]$
Since, first term of AP is a, and last term of AP is ${{a}_{n}}=a+\left( n-1 \right)d$
So, we can write sum of n terms of AP as:
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ a+a+\left( n-1 \right)d \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
\end{align}$