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Question: For an AP, \(\dfrac{{{S}_{kn}}}{{{S}_{n}}}\) is independent of n. The value of \(\dfrac{d}{a}\) for ...

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

Explanation

Solution

Let us assume that we have the following Arithmetic Progression (AP) series:
a,a+d,a+2d,a+3d,........,a+(n−1)da,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:Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right], us the formula to find Sn{{S}_{n}} and Skn{{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+(n−1)da,a+d,a+2d,a+3d,........,a+\left( n-1 \right)d
So, the sum of n terms of the AP is given as:
Sn=n2[2a+(n−1)d]......(1){{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]......(1)
Similarly, for knkn terms, we can write:
Skn=kn2[2a+(kn−1)d]......(2){{S}_{kn}}=\dfrac{kn}{2}\left[ 2a+\left( kn-1 \right)d \right]......(2)
Now, we need to find SknSn\dfrac{{{S}_{kn}}}{{{S}_{n}}}.
So, divide equation (2) by equation (1), we get:
SknSn=kn2[2a+(kn−1)d]n2[2a+(n−1)d].......(3)\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:
SknSn=k[2a+(kn−1)d][2a+(n−1)d]......(4)\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 SknSn=x\dfrac{{{S}_{kn}}}{{{S}_{n}}}=x
So, we can write equation (4) as:
x=k[2a+(kn−1)d][2a+(n−1)d]......(5)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:
⇒x[2a+(n−1)d]=k[2a+(kn−1)d] ⇒2xa+xnd−xd=2ka+k2nd−kd ⇒2ak−2ax+xd−kd+(k2d−xd)n=0......(6) \begin{aligned} & \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{aligned}
So, we get:
Either (k2d−xd)=0......(7)\left( {{k}^{2}}d-xd \right)=0......(7)
Or 2ak−2ax+xd−kd=0......(8)2ak-2ax+xd-kd=0......(8)
From equation (7), we can write:
⇒k2d−xd=0 ⇒k2d=xd ⇒k2=x......(9) \begin{aligned} & \Rightarrow {{k}^{2}}d-xd=0 \\\ & \Rightarrow {{k}^{2}}d=xd \\\ & \Rightarrow {{k}^{2}}=x......(9) \\\ \end{aligned}
Now, substitute equation (9) in equation (8), we get:
⇒2ak−2ax+xd−kd=0 ⇒2ak−2ak2+k2d−kd=0.....(10) \begin{aligned} & \Rightarrow 2ak-2ax+xd-kd=0 \\\ & \Rightarrow 2ak-2a{{k}^{2}}+{{k}^{2}}d-kd=0.....(10) \\\ \end{aligned}
Now, get all the terms of “a” on one side and terms of “d” on other side, we get:
⇒2ak−2ak2=−k2d+kd ⇒2ak(1−k)=−kd(k−1) ⇒2ak(1−k)=kd(1−k) ⇒2a=d......(11) \begin{aligned} & \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{aligned}
Now, to get a ratio between d and a, divide equation (11) by a.
We get:
da=2\dfrac{d}{a}=2

So, the correct answer is “Option B”.

Note: The sum of n terms of an AP is given as Sn=n2[2a+(n−1)d]{{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 Sn=n2[first term + last term]{{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 an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d
So, we can write sum of n terms of AP as:
⇒Sn=n2[a+a+(n−1)d] ⇒Sn=n2[2a+(n−1)d] \begin{aligned} & \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{aligned}