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

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(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$Since the $(m + 1)^{th}$,$(n + 1)^{th}$,and $(r + 1)^{th}$ terms are

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$-\frac{2}{n}$

Vedclass · Answer

(A) Let $a$ be the first term and $d$ be the common difference of the given $A.P.$Since the $(m + 1)^{th}$,$(n + 1)^{th}$,and $(r + 1)^{th}$ terms are in $G.P.$,we have:$(a + md), (a + nd), (a + rd)$ are in $G.P.$Therefore,$(a + nd)^2 = (a + md)(a + rd)$.Expanding both sides: $a^2 + 2and + n^2d^2 = a^2 + ard + amd + mrd^2$.$2and - ard - amd = mrd^2 - n^2d^2$.$ad(2n - r - m) = d^2(mr - n^2)$.Thus,$\frac{d}{a} = \frac{2n - (m + r)}{mr - n^2}$ ... $(i)$.Given $m, n, r$ are in $H.P.$,we have $n = \frac{2mr}{m + r}$,which implies $m + r = \frac{2mr}{n}$.Substituting $m + r$ into equation $(i)$:$\frac{d}{a} = \frac{2n - \frac{2mr}{n}}{mr - n^2} = \frac{\frac{2n^2 - 2mr}{n}}{mr - n^2} = \frac{-2(mr - n^2)}{n(mr - n^2)} = -\frac{2}{n}$.

If the $(m + 1)^{th}$,$(n + 1)^{th}$ and $(r + 1)^{th}$ terms of an $A.P.$ are in $G.P.$ and $m, n, r$ are in $H.P.$,then the value of the ratio of the common difference to the first term of the $A.P.$ is

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