Question

If ${S}_{n}$ denote the sum of n terms of an A.P. with first term a and common difference d, such that $\bold{\dfrac{{S}_{x}}{{S}_{kx}}}$ is independent of x, then which of the following is correct?
(A) $\bold{d=a}$
(B) $\bold {d=2a}$
(C) $\bold{a=2d}$
(D) $\bold{d=-a}$
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Topic: Arithmetic Progression.​

Answer 1

Explanation:

Inorder to solve the problem, first of all we need to understand the basic idea of the question.

We are given that $S_n$ denotes the sum of an A.P. with first term $a$ and common difference $d.$ We are also given that $\dfrac{{S}_{x}}{{S}_{kx}}$ is independent of $x$ means, in the expansion of $\dfrac{{S}_{x}}{{S}_{kx}}$, $x$ must not be present.

We can solve this problem by using the formula for sum of A.P. which is given by:

• $S_n = \dfrac{n}{2} [2a+(n-1)d]$

Now let's solve the problem!

$\rule{290}{1}$

We have,

$\implies\dfrac{{S}_{x}}{{S}_{kx}}$

Now using the formula for sum of terms of AP:

$\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{\dfrac{\not x}{\not 2} [2a+(x-1)d]}{\dfrac{k\not x}{\not 2} [2a+(kx-1)d]}$

$\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{[2a+(x-1)d]}{k[2a+(kx-1)d]}$

$\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{[2a+xd - d]}{k[2a+kxd - d]}$

$\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{[(2a - d)+xd ]}{k[(2a - d)+kxd]}$

Now, for this quantity to be independent of $x$ , $(2a-d)$ must be equal to $0$ so that we could cancel out $x$ from both numerator and denominator.

$\implies (2a-d) = 0$

$\boxed{\implies 2a = d}$

Hence option [B] is correct.

$\rule{290}{1}$

Verification:

If we substitute $2a = d$, the given quantity must be independent of $x$.

$\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{[(2a - d)+xd ]}{k[(2a - d)+kxd]}$

${\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{[(d - d)+xd ]}{k[(d - d)+kxd]}}$

${\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{xd}{k(kxd)}}$

${\implies\dfrac{{S}_{x}}{{S}_{kx}} = \dfrac{1}{ {k}^{2} }}$

Hence it is independent of x.

$\rule{290}{1}$

Additional Information:

Let's learn a trick to solve some specific types of questions in A.P.!

Whenever we are given any two terms of an AP, let's say them $nth$ and $kth$ terms, then the common difference of this AP will be given by:

$\boxed{d \: = \: \dfrac{a_n - a_k}{n - k} }$

Now let's try to solve a question based on this trick.

Question:

If $8th$ term of an AP is $18$ and it's $16th$ term is $58$, then find the common difference of this A.P.

Solution:

Common difference of the AP is given by,

$\implies d \: = \: \dfrac{a_n - a_k}{n - k}$

Let $n$ be $16$ and $k$ be $8$ implies that $nth$ term is $58$ and $kth$ term is $18$.

$\implies d \: = \: \dfrac{58 - 18}{16-8}$

$\implies d \: = \: \dfrac{40}{8}$

$\implies d \: = \: 2$

Answer 2

Question:-

If $\sf {S}_{n}$ denote the sum of n terms of an A.P. with first term a and common difference d, such that $\sf{\dfrac{{S}_{x}}{{S}_{kx}}}$ is independent of x, then which of the following is correct?

Options:

(A) d = a.

(B) d = 2a.

(C) a = 2d.

(D) d = – a.

Given:-

• First term = a.
• Common difference = d.
• Number of terms = n.

Solution:-

$\sf \large By, using \: S_{n} = \frac{n}{2} [S_n=2a + (n - 1)d] \: we \: have,$

$\sf \large \therefore \frac{ S_{kx}}{S_{x}} = \frac{ \frac{k_{x}}{2} (2a + (k_{x} - 1)d)}{ \frac{x}{2}(2a + (x + 1)d) } \\ \\ \sf \large⇒ \frac{S_{kx}}{S_{x}} = \frac{k(2a + (kx - 1)d)}{(2a + (x + 1)d)}$

$\sf \large⇒ \frac{S_{kx}}{S_x} = \frac{k(2a - d + kxd)}{2a + xd - d}$

Now, this will be independent of x only when 2a = d.

Answer:-

$\sf \large \color{red}Option(B) \: d = 2a \: is \: correct \: answer.$

Hope you have satisfied. ⚘