If {S_k} denotes the sum of the first k terms | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The sum of the first $k$ terms of an arithmetic progression is given by ${S_k} = \frac{k}{2} \{2a + (k - 1)d\}$.We are given the ratio $\frac{S_{k

Vedclass · Accepted Answer

$2a - d = 0$

Vedclass · Answer

(A) The sum of the first $k$ terms of an arithmetic progression is given by ${S_k} = \frac{k}{2} \{2a + (k - 1)d\}$.We are given the ratio $\frac{S_{kn}}{S_n} = \frac{\frac{kn}{2} \{2a + (kn - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}}$.Simplifying the expression,we get:$\frac{S_{kn}}{S_n} = k \left\{ \frac{2a + (kn - 1)d}{2a + (n - 1)d} \right\} = k \left\{ \frac{(2a - d) + knd}{(2a - d) + nd} \right\}$.For this expression to be independent of $n$,the term containing $n$ in the numerator must be proportional to the term containing $n$ in the denominator,or the constant terms must vanish such that $n$ cancels out.If $2a - d = 0$,then the expression becomes:$\frac{S_{kn}}{S_n} = k \left\{ \frac{knd}{nd} \right\} = k^2$.Since $k^2$ is independent of $n$,the condition is $2a - d = 0$.

If ${S_k}$ denotes the sum of the first $k$ terms of an arithmetic progression whose first term and common difference are $a$ and $d$ respectively,then ${S_{kn}}/{S_n}$ is independent of $n$ if:

Similar Questions

Which term of the sequence $(-8 + 18i), (-6 + 15i), (-4 + 12i), ......$ is purely imaginary (in $^{th}$)?

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

If $a, b, c, d$ are in $H.P.$,then $ab + bc + cd$ is equal to

Let ${S_n}$ denote the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$,then the ratio $\frac{{{S_{3n}}}}{{{S_n}}}$ is equal to:

Let $a_1, a_2, a_3, \ldots$ be terms of an $A.P.$ If $\frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p^2}{q^2}$ for $p \neq q$,then $\frac{a_6}{a_{21}}$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module