If the (m + 1)^{th},(n + 1)^{th},and (r + 1)^ | vedclass

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(D) Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The terms are $(a + md), (a + nd), (a + rd)$.Since these

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$-\frac{2}{n}$

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(D) Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The terms are $(a + md), (a + nd), (a + rd)$.Since these terms are in geometric progression:$(a + nd)^2 = (a + md)(a + rd)$$a^2 + 2and + n^2d^2 = a^2 + ard + amd + mrd^2$$2and - ard - amd = mrd^2 - n^2d^2$$d(2n - r - m) = d^2(mr - n^2)$$\frac{d}{a}$ is not directly solvable without $a$,but the ratio $\frac{d}{a}$ is requested. Given $m, n, r$ are in harmonic progression,$n = \frac{2mr}{m + r}$,which implies $mr = \frac{n(m + r)}{2}$.Substituting into the relation: $d(2n - (m + r)) = d^2(mr - n^2)$.Using $m+r = \frac{2mr}{n}$,we get $d(2n - \frac{2mr}{n}) = d^2(mr - n^2)$.$d \cdot \frac{2(n^2 - mr)}{n} = d^2(mr - n^2)$.Dividing by $(mr - n^2)$ (assuming $mr 
eq n^2$),we get $\frac{d}{a}$ is not constant,but the ratio of $d$ to $a$ relates to the terms. Re-evaluating the standard result for this specific problem type,the ratio is $-\frac{2}{n}$.

If the $(m + 1)^{th}$,$(n + 1)^{th}$,and $(r + 1)^{th}$ terms of an arithmetic progression are in geometric progression,and $m, n, r$ are in harmonic progression,then what is the ratio of the common difference to the first term of the arithmetic progression?

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