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(D) For an infinite $G.P.$,the sum $S = \frac{a}{1-r} = 4$ and the second term $ar = \frac{3}{4}$.From the first equation,$a = 4(1-r)$.Substituting $a

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$a = 3, r = \frac{1}{4}$

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(D) For an infinite $G.P.$,the sum $S = \frac{a}{1-r} = 4$ and the second term $ar = \frac{3}{4}$.From the first equation,$a = 4(1-r)$.Substituting $a$ into the second equation: $4(1-r)r = \frac{3}{4}$.$r(1-r) = \frac{3}{16} \implies r - r^2 = \frac{3}{16} \implies 16r^2 - 16r + 3 = 0$.Factoring the quadratic: $(4r - 3)(4r - 1) = 0$.This gives $r = \frac{3}{4}$ or $r = \frac{1}{4}$.If $r = \frac{1}{4}$,then $a = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.If $r = \frac{3}{4}$,then $a = 4(1 - \frac{3}{4}) = 4(\frac{1}{4}) = 1$.The possible pairs $(a, r)$ are $(3, \frac{1}{4})$ and $(1, \frac{3}{4})$.Comparing with the options,$(a, r) = (3, \frac{1}{4})$ is correct.

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. Its sum is $4$ and the second term is $3/4$. Then:

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