If x, y, z are in Arithmetic Progression (AP) | vedclass

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(A) Given that $x, y, z$ are in $AP$,we have $2y = x + z$. This implies $z = 2y - x$. Given that $x, y, t$ are in $GP$,we have $y^2 = xt$. This implie

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Geometric Progression

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(A) Given that $x, y, z$ are in $AP$,we have $2y = x + z$. This implies $z = 2y - x$. Given that $x, y, t$ are in $GP$,we have $y^2 = xt$. This implies $t = \frac{y^2}{x}$. Now,consider the sequence $x, x - y, t - z$. Substitute $t$ and $z$: $t - z = \frac{y^2}{x} - (2y - x) = \frac{y^2 - 2xy + x^2}{x} = \frac{(x - y)^2}{x}$. For the sequence $x, x - y, t - z$ to be in $GP$,the square of the middle term must equal the product of the first and third terms: $(x - y)^2 = x 	imes (t - z)$. Substituting the expression for $(t - z)$: $x 	imes \frac{(x - y)^2}{x} = (x - y)^2$. Since the condition holds,$x, x - y, t - z$ are in $GP$.

If $x, y, z$ are in Arithmetic Progression $(AP)$ and $x, y, t$ are in Geometric Progression $(GP)$,then in which progression are $x, x - y, t - z$?

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