If the p^{th},q^{th},and r^{th} terms of an A | vedclass

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(B) Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$. Let the first term of the Geometric Progression be $A$ a

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$1$

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(B) Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$. Let the first term of the Geometric Progression be $A$ and the common ratio be $R$.$a + (p - 1)d = A R^{p - 1} = x$$a + (q - 1)d = A R^{q - 1} = y$$a + (r - 1)d = A R^{r - 1} = z$Taking logarithms on both sides for each equation:$\ln(x) = \ln(A) + (p - 1)\ln(R)$$\ln(y) = \ln(A) + (q - 1)\ln(R)$$\ln(z) = \ln(A) + (r - 1)\ln(R)$Consider the expression $E = x^{y - z} \cdot y^{z - x} \cdot z^{x - y}$.Taking $\ln(E) = (y - z)\ln(x) + (z - x)\ln(y) + (x - y)\ln(z)$.Substituting the values of $\ln(x), \ln(y), \ln(z)$:$\ln(E) = (y - z)(\ln(A) + (p - 1)\ln(R)) + (z - x)(\ln(A) + (q - 1)\ln(R)) + (x - y)(\ln(A) + (r - 1)\ln(R))$$\ln(E) = \ln(A)(y - z + z - x + x - y) + \ln(R)((p - 1)(y - z) + (q - 1)(z - x) + (r - 1)(x - y))$Since $(y - z + z - x + x - y) = 0$ and the second term simplifies to $0$ due to the cyclic nature of the terms in an $A$.$P$. and $G$.$P$. relationship:$\ln(E) = 0 \implies E = e^0 = 1$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an Arithmetic Progression are equal to the terms of a Geometric Progression,and these terms are $x, y, z$ respectively,then $x^{y - z} \cdot y^{z - x} \cdot z^{x - y} = \dots$

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