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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 th

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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 this 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: $(4r - 3)(4r - 1) = 0$.Thus,$r = \frac{3}{4}$ or $r = \frac{1}{4}$.If $r = \frac{3}{4}$,then $a = 4(1 - 3/4) = 4(1/4) = 1$.If $r = \frac{1}{4}$,then $a = 4(1 - 1/4) = 4(3/4) = 3$.The possible pairs $(a, r)$ are $(1, 3/4)$ and $(3, 1/4)$.Comparing with the given options,option $(d)$ is correct.

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,then:

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