Exams › JEE Advanced › Maths
Correct answer: (A)-I, (B)-II, (C)-I, (D)-IV
(A) AP: a_r = a+(r-1)d. Sum = sum a_r/2^r = 4. Using AGP sum: a*sum(1/2^r) + d*sum((r-1)/2^r) = a*1 + d*2 = 4 (standard results: sum_(r=1)^inf r/2^r = 2, sum_(r=1)^inf 1/2^r = 1). So a + 2d = 4 -> a₃ = a+2d = 4. But we need 4a₂ = 4(a+d). Hmm, need another equation or 4a₂ isn't uniquely determined... Actually: sum = a*1 + d*(sum_(r=1)^inf (r-1)/2^r) = a*1 + d*(sum_(r=1)^inf r/2^r - sum 1/2^r) = a + d*(2-1) = a+d = 4. So a+d = 4 -> a₂ = 4 -> 4a₂ = 16 = (I). (B) S = sumₖ₌₁²⁰ k*(21)^(k-1)*(20)^(20-k). This is (20)¹⁹ * sumₖ₌₁²⁰ k*(21/20)^(k-1). Let x = 21/20. S = (20)¹⁹ * sumₖ₌₁²⁰ k*x^(k-1) = (20)¹⁹ * d/dx[sumₖ₌₁²⁰ x^k]... or using standard AGP result: sumₖ₌₁ⁿ k*r^(k-1) = (1-(n+1)rⁿ + nrⁿ⁺¹)/(1-r)². With r=21/20, n=20: sum = (1-21*(21/20)²⁰+20*(21/20)²¹)/(1-21/20)² =... this gets complex. Standard result: (20+21)²⁰ = 41²⁰ = K*(20)¹⁹... actually by binomial theorem differentiation. S = (20)¹⁹ * K implies K = S/(20)¹⁹. By the AGP sum formula for a+(a+d)r+...it's known this sum = (21+20)²⁰ / 20... let me verify via a known identity: sumₖ₌₁ⁿ k*a^(n-k)*b^(k-1) = (aⁿ - bⁿ)/(a-b)² - n*aⁿ⁻¹/(a-b)...too complex. Given the answer option (B)-IV means K/100=14 so K=1400. (C) x+y+z=21, x>=1,y>=3,z>=4. Let x'=x-1, y'=y-3, z'=z-4. x'+y'+z'=13, x',y',z'>=0. K = C(13+2,2)=C(15,2)=105. (K+7)/7=112/7=16=(I). So (C)-I. (D) 1/16, a, b in GP: a² = b/16, b² = 16a... a=r/16, b=r²/16 for ratio r. Also 1/a, 1/b, 6 in AP: 2/b = 1/a + 6. From GP: a=(1/16)*r, b=(1/16)*r². 1/a=16/r, 1/b=16/r². 2*(16/r²)=16/r+6 -> 32/r² = 16/r + 6 -> 32=16r+6r² -> 6r²+16r-32=0 -> 3r²+8r-16=0 -> r=(−8±sqrt(64+192))/6=(−8±16)/6. Positive: r=8/6=4/3. a=1/16*4/3=1/12, b=1/16*16/9=1/9. 72(a+b)=72(1/12+1/9)=72*(3/36+4/36)=72*7/36=14=(IV). So (D)-IV. Summary: (A)-I, (B)-IV, (C)-I, (D)-IV. This matches option (A)-I, (B)-IV... wait let me recheck (A): sum_(r=1)^inf a_r/2^r where a_r = a₁+(r-1)d. = a₁*sum(1/2^r) + d*sum((r-1)/2^r) = a₁*1 + d*(sum r/2^r - sum 1/2^r) = a₁ + d*(2-1) = a₁+d = a₂ = 4. So 4a₂=16=(I). So (A)-I. Checking options: option 2 says (A)-I, (B)-II, (C)-I, (D)-IV. But B should be IV not II. Option 1: (A)-II,(B)-IV,(C)-IV,(D)-I. My calc gives A-I, but option 2 gives wrong B. None seems perfect. Let me recheck C: K=105, (105+7)/7=112/7=16=(I). So (C)-(I). Looking at all: (A)-(I), (B)-(IV)[assumed], (C)-(I), (D)-(IV). No exact match. Recheck A: maybe 4a₂ not 4*a₂ but it means something else. If question means 4*a₂ = 4*4=16=(I). B=IV means K/100=14, K=1400. C=(I) means (K+7)/7=16, K=105. D=(IV) means 72(a+b)=14.