If α and β are the roots of the equation 375x² - 25x - 2 = 0, then limn→∞Σr=1n αr + limn→∞Σr=1n βr is equal to?

Correct option is C. 112
Step 1: Find the sum of infinite G.P's:
(α + α² + ..upto infinite terms) + (β + β² +..upto infinite terms) = α/(1 - α) + β/(1 - β)
S = α + β - 2αβ/(1 - (α + β) + αβ)
Step 2: Find the relation between roots and coefficient's of quadratic:
375x² - 25x - 2 = 0
α + β = 25/375, αβ = -2/375
Step 3: Find the value of sum using above relations:
S = 25/375 + (-2/375) / (1 - 25/375 -2/375)
S = 25/375 - 4/375 / (350/375)
S = 21/375 / (350/375)
S = 21/350
S = 3/50
S = 0.06
Let's correct the solution
Step 1: Find the sum of infinite G.P's:
(α + α² + ..upto infinite terms) + (β + β² + ..upto infinite terms) = α/(1 - α) + β/(1 - β)
This simplifies to (α + β - 2αβ)/(1 - (α + β) + αβ)
Step 2: Find the relation between roots and coefficients of quadratic:
375x² - 25x - 2 = 0
α + β = 25/375 = 1/15
αβ = -2/375
Step 3: Find the value of sum using above relations:
S = (1/15 + 2/375) / (1 - 1/15 + 2/375) = (25 + 2) / 375 / (375 - 25 + 2)/375 = 27/352
S = (1/15 - 4/375)/(1 - 1/15 + 2/375) = (25 - 4)/375/(375 - 25 + 2)/375 = 21/352 ≈ 0.0596
Let's re-examine the sum of infinite GP's:
Sum = α/(1-α) + β/(1-β) = (α(1-β) + β(1-α))/(1-α)(1-β) = (α - αβ + β - αβ)/(1 - α - β + αβ) = (α + β - 2αβ)/(1 - (α + β) + αβ)
Substituting the values:
Sum = (1/15 + 4/375)/(1 - 1/15 + 2/375) = (25 + 4)/375 / (375 - 25 -2)/375 = 29/348 ≈ 0.0833
However, the sum of infinite geometric series is given by a/(1-r), where 'a' is the first term and 'r' is the common ratio. Thus we have:
S = α/(1-α) + β/(1-β) = (α - αβ + β - αβ)/(1 -α -β +αβ) = (α + β -2αβ)/(1 - (α+β) + αβ) = (1/15 + 4/375) / (1 - 1/15 + 2/375) = (29/375) / (352/375) = 29/352
This is approximately 0.0826. None of the options match this value. Let's check the problem statement and the options once again. There appears to be an error in either the question or the provided solution.
There might be a mistake in the question or the given options. The calculation seems correct.