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Let {aₖ} and {bₖ} (k belonging to natural numbers) be two geometric progressions with common ratios r1 and r2 respectively, where a₁ = b₁ = 4 and r1 < r2. Define cₖ = aₖ + bₖ. Given that c₂ = 5 and c₃ = 13/4, find the value of (sum of cₖ from k=1 to infinity) minus (12*a₆ + 8*b₄).
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Correct answer: 2
Solution
From c₂: r1+r2 = 5/4, and from c₃: r1²+r2² = 13/16, giving r1*r2 = (25/16 - 13/16)/2 = 6/32 = 3/16. So r1 and r2 are roots of 16t² - 20t + 3 = 0, giving r1=1/4, r2=3/4. The infinite sum is 4/(1-1/4)+4/(1-3/4) = 16/3+16 = 64/3, and 12*a₆+8*b₄ = 12*4*(1/4)⁵+8*4*(3/4)³ = 12*(1/256)+32*(27/64) = 3/64+... Let me recompute: a₆=4*(1/4)⁵=4/1024=1/256, 12*a₆=12/256=3/64. b₄=4*(3/4)³=4*27/64=108/64=27/16, 8*b₄=8*27/16=216/16=27/2. So 12a₆+8b₄=3/64+27/2 = 3/64+864/64=867/64. Sum=64/3. 64/3-867/64=(4096-2601)/192=1495/192. That's not integer. Re-examine.
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