Question

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

• A. \( a = 1, r = -\frac{3}{4} \)
• B. \( a = 3, r = \frac{1}{4} \)
• C. \( a = -3, r = -\frac{1}{4} \)
• D. \( a = -1, r = \frac{3}{4} \)

Hint:

When dealing with geometric series, use the sum formula \( S = \frac{a}{1 - r} \) and the formula for the nth term to relate the terms and solve for unknowns.

Answer 1

The Correct Option is B

The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r|<1 \)) is given by the formula: \[ S = \frac{a}{1 - r} \] We are given that the sum of the infinite series is 4, so: \[ \frac{a}{1 - r} = 4 \] (Equation 1) We are also given that the second term of the series is \( \frac{3}{4} \). The second term is given by: \[ a r = \frac{3}{4} \] (Equation 2) From Equation 2, we can express \( a \) as: \[ a = \frac{3}{4r} \] Now, substitute this value of \( a \) into Equation 1: \[ \frac{\frac{3}{4r}}{1 - r} = 4 \] Multiply both sides by \( 4r(1 - r) \): \[ 3 = 16r(1 - r) \] Expanding the right-hand side: \[ 3 = 16r - 16r^2 \] Rearranging the terms: \[ 16r^2 - 16r + 3 = 0 \] This is a quadratic equation in \( r \). Using the quadratic formula: \[ r = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(16)(3)}}{2(16)} \] \[ r = \frac{16 \pm \sqrt{256 - 192}}{32} \] \[ r = \frac{16 \pm \sqrt{64}}{32} \] \[ r = \frac{16 \pm 8}{32} \] So, we have two possible values for \( r \): \[ r = \frac{16 + 8}{32} = \frac{24}{32} = \frac{3}{4} \] or \[ r = \frac{16 - 8}{32} = \frac{8}{32} = \frac{1}{4} \] Now, substituting \( r = \frac{1}{4} \) into Equation 2: \[ a \times \frac{1}{4} = \frac{3}{4} \] \[ a = 3 \] Thus, the correct answer is \( a = 3 \) and \( r = \frac{1}{4} \).