If a, b, c are in AP,and b-a, c-b, a are in G | vedclass

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(A) Since $a, b, c$ are in $AP$,we have $2b = a + c$,which implies $c - b = b - a$. Given that $b-a, c-b, a$ are in $GP$,the square of the middle term

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$1: 2: 3$

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(A) Since $a, b, c$ are in $AP$,we have $2b = a + c$,which implies $c - b = b - a$. Given that $b-a, c-b, a$ are in $GP$,the square of the middle term equals the product of the extremes: $(c-b)^2 = (b-a)a$. Substituting $c-b = b-a$ into the equation,we get $(b-a)^2 = (b-a)a$. Assuming $b 
eq a$,we divide by $(b-a)$ to get $b-a = a$,which simplifies to $b = 2a$. Substituting $b = 2a$ into $2b = a + c$,we get $2(2a) = a + c$,so $4a = a + c$,which means $c = 3a$. Thus,the ratio $a: b: c = a: 2a: 3a = 1: 2: 3$.

Column-$I$	Column-$II$
$1. m = -2$	$a$. Convex mirror
$2. m = -\frac{1}{2}$	$b$. Concave mirror
$3. m = +2$	$c$. Real Image
$4. m = +\frac{1}{2}$	$d$. Virtual image

If $a, b, c$ are in $AP$,and $b-a, c-b, a$ are in $GP$,then $a: b: c$ is

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Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]

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$1. m = -2$ $a$. Convex mirror

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$4. m = +\frac{1}{2}$ $d$. Virtual image

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