If frac{S_n}{S_m} = frac{n^4}{m^4} (where S_k | vedclass

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

Question

(A) Given $\frac{S_n}{S_m} = \frac{n^4}{m^4}$.Since $S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have $\frac{\frac{n}{2}[2a_1 + (n-1)d]}{\frac{m}{2}[2a_1 + (

Vedclass · Accepted Answer

$\frac{(2m+1)^3}{(2n+1)^3}$

Vedclass · Answer

(A) Given $\frac{S_n}{S_m} = \frac{n^4}{m^4}$.Since $S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have $\frac{\frac{n}{2}[2a_1 + (n-1)d]}{\frac{m}{2}[2a_1 + (m-1)d]} = \frac{n^4}{m^4}$.Simplifying,$\frac{2a_1 + (n-1)d}{2a_1 + (m-1)d} = \frac{n^3}{m^3}$.To find the ratio of terms,we substitute $n = 2m-1$ and $m = 2n-1$ or use the property that $\frac{a_n}{a_m} = \frac{S_n - S_{n-1}}{S_m - S_{m-1}}$.For the ratio $\frac{a_{m+1}}{a_{n+1}}$,we set the index such that $2n-1 = 2(m+1)-1 = 2m+1$. Thus,the ratio becomes $\frac{(2m+1)^3}{(2n+1)^3}$.

If $\frac{S_n}{S_m} = \frac{n^4}{m^4}$ (where $S_k$ is the sum of the first $k$ terms of an $A$.$P$. $a_1, a_2, \dots, \infty$),then the value of $\frac{a_{m+1}}{a_{n+1}}$ in terms of $m$ and $n$ will be

Similar Questions

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is

The $4$th term of an $A.P.$ is equal to $3$ times the first term and the seventh term exceeds twice the third term by $1$. Find the first term and the common difference.

$a, g, h$ are the arithmetic mean,geometric mean,and harmonic mean between two positive numbers $x$ and $y$ respectively. Identify the correct statement among the following:

Find the sum of infinite terms of the series $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots$

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module