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

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(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 + (

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

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(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

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