Exams › JEE Advanced › Maths
Correct answer: 4
Denominators: 15, 30, 50, 75,... Second differences are constant (5), so denominators follow a quadratic pattern. The nth term denominator is (n*(5n+5))/2... Let us find it: aₙ = n*(5n+5)/2 = 5n(n+1)/2. Check: n=1: 5, n=2: 15... doesn't fit. Try aₙ = (5n² + 5n)/2: n=1: 5, n=2: 15. Doesn't match 15, 30, 50. Actual denominators: 15=3*5, 30=5*6, 50=5*10... Let us try 5*n*(n+2)/2: n=1: 5*3/2 not integer. Try: Tₙ = 5n(n+1)/2 + 5: n=1: 5+5=10 no. Rethink. Differences of denominators: 30-15=15, 50-30=20, 75-50=25. So differences form AP: 15,20,25,... with d=5 from d₁=15. Sum of first n differences = 15n + 5n(n-1)/2 = n(25+5n)/2... so aₙ = 15 + sumₖ₌₁ⁿ⁻¹(15+5k) for n>=2. a₁=15. aₙ = 15 + 15(n-1) + 5*n(n-1)/2 = 15n + 5n(n-1)/2 = 5n(6+n-1)/2 = 5n(n+5)/2. Check: n=1: 5*6/2=15 yes. n=2: 5*2*7/2=35 no, should be 30. Let me redo: aₙ = 15 + sumₖ₌₁ⁿ⁻¹(10+5k). For n=2: a₂ = 15 + (10+5) = 30 yes. For n=3: a₃ = 15 + 15 + 20 = 50 yes. For n=4: a₄ = 15+15+20+25=75 yes. So the increment from term n to n+1 is 10+5n. Thus aₙ = 15 + sumₖ₌₁ⁿ⁻¹(10+5k) = 15 + 10(n-1) + 5*(n-1)n/2 = 5 + 10n + 5n(n-1)/2 = 5 + 10n + 5n²/2 - 5n/2 = 5 + 15n/2 + 5n²/2 = (10+15n+5n²)/2 = 5(n²+3n+2)/2 = 5(n+1)(n+2)/2. Check: n=1: 5*2*3/2=15 yes. n=2: 5*3*4/2=30 yes. n=3: 5*4*5/2=50 yes. n=4: 5*5*6/2=75 yes. So Tₙ = 2/(5(n+1)(n+2)) = (2/5)*[1/(n+1) - 1/(n+2)]. Sum = (2/5)*[1/2 - 0] (telescoping as n->inf) = (2/5)*(1/2) = 1/5 = 4/20. So k=4.