If the first and (2n-1)-th terms of an AP, GP | vedclass

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(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively, they are the Arithmet

Vedclass · Accepted Answer

$ac-b^2=0$

Vedclass · Answer

(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively, they are the Arithmetic Mean, Geometric Mean, and Harmonic Mean of $x$ and $y$.We know that for any two positive numbers $x$ and $y$, the relationship between their means is $AM \geq GM \geq HM$.Therefore, $a \geq b \geq c$.Also, the relationship between $AM, GM,$ and $HM$ is $AM \cdot HM = GM^2$, which implies $a \cdot c = b^2$ or $ac - b^2 = 0$.Since the question asks for the relationship that is 'always' true, both $a \geq b \geq c$ and $ac = b^2$ are correct. However, in standard competitive mathematics, $ac = b^2$ is the specific identity derived from the definition of these means.

If the first and $(2n-1)$-th terms of an $AP$, $GP$, and $HP$ are equal and their $n$-th terms are respectively $a, b, c$, then always

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