Let a, b and c be the 7^{th}, 11^{th} and 13^ | vedclass

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(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d 
eq 0$.Given $a = A + 6d$,$b = A +

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$4$

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(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d 
eq 0$.Given $a = A + 6d$,$b = A + 10d$,and $c = A + 12d$.Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Substituting the values: $(A + 10d)^2 = (A + 6d)(A + 12d)$.$A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2$.$2Ad = -28d^2$.Since $d 
eq 0$,we divide by $2d$ to get $A = -14d$,or $\frac{A}{d} = -14$.Now,$\frac{a}{c} = \frac{A + 6d}{A + 12d} = \frac{\frac{A}{d} + 6}{\frac{A}{d} + 12}$.Substituting $\frac{A}{d} = -14$: $\frac{-14 + 6}{-14 + 12} = \frac{-8}{-2} = 4$.

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to

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