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(B) The sum of an infinite geometric series is given by $ S = \frac{a}{1-r} $,where $ |r| < 1 $. Given $ S = 4 $,we have $ \frac{a}{1-r} = 4 \Rightarr

Vedclass · Accepted Answer

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

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(B) The sum of an infinite geometric series is given by $ S = \frac{a}{1-r} $,where $ |r| Given $ S = 4 $,we have $ \frac{a}{1-r} = 4 \Rightarrow a = 4(1-r) = 4 - 4r \Rightarrow a + 4r = 4 \dots(1) $. The second term of a geometric series is $ t_2 = ar $. Given $ t_2 = \frac{3}{4} $,we have $ ar = \frac{3}{4} \Rightarrow a = \frac{3}{4r} \dots(2) $. Substituting $ a $ from $(2)$ into $(1)$: $ \frac{3}{4r} + 4r = 4 $. Multiplying by $ 4r $: $ 3 + 16r^2 = 16r \Rightarrow 16r^2 - 16r + 3 = 0 $. Factoring the quadratic equation: $ 16r^2 - 12r - 4r + 3 = 0 \Rightarrow 4r(4r - 3) - 1(4r - 3) = 0 \Rightarrow (4r - 1)(4r - 3) = 0 $. Thus,$ r = \frac{1}{4} $ or $ r = \frac{3}{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 $. Comparing with the given options,the pair $ (a=3, r=\frac{1}{4}) $ matches option $ B $.

Consider an infinite geometric series with first term $ a $ and common ratio $ r $. If the sum is $ 4 $ and the second term is $ \frac{3}{4} $,then find the values of $ a $ and $ r $.

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