Consider an infinite geometric series with first term and common ratio . If its sum is 4 and the second term is 3/4, then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The two conditions are: a/(1-r) = 4 and ar = 3/4.
From the first equation, a = 4(1-r). Substituting into ar = 3/4: 4(1-r)r = 3/4, so 16r(1-r) = 3, giving 16r² - 16r + 3 = 0, which factors as (4r-1)(4r-3) = 0. Thus r = 1/4 or r = 3/4.
For r = 1/4: a = 4(1 - 1/4) = 3, and |r| < 1 ✓. Check: ar = 3×(1/4) = 3/4 ✓.
So a = 3, r = 1/4.
From the first equation, a = 4(1-r). Substituting into ar = 3/4: 4(1-r)r = 3/4, so 16r(1-r) = 3, giving 16r² - 16r + 3 = 0, which factors as (4r-1)(4r-3) = 0. Thus r = 1/4 or r = 3/4.
For r = 1/4: a = 4(1 - 1/4) = 3, and |r| < 1 ✓. Check: ar = 3×(1/4) = 3/4 ✓.
So a = 3, r = 1/4.
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