TS Police Constable Mains 2023 · Question 44
Question 44 of the TSLPRB Police Constable Final Written Examination 2023 (General Studies) (30 April 2023). Try it first, then open the answer and explanation.
If x = √(3080 + √(3080 + √(3080 + ... to ∞))), y = √(200√(200√(200 ... to ∞))) and x + y = z, then √(z√(z√z)) =
- 64
- 128
- 256
- 512
x = √(3080 + √(3080 + √(3080 + ... to ∞))), y = √(200√(200√(200 ... to ∞))) మరియు x + y = z. అయితే √(z√(z√z)) =
- 64
- 128
- 256
- 512
Show answer
Answer: (2) 128
సమాధానం: (2) 128
x = √(3080 + x) gives x² − x − 3080 = 0, so x = 56 (56 × 55 = 3080); y = √(200y) gives y = 200. Then z = 256 = 2⁸ and √(z√(z√z)) = z^(1/2 + 1/4 + 1/8) = z^(7/8) = 2⁷ = 128.
x = √(3080 + x) నుండి x² − x − 3080 = 0, కాబట్టి x = 56 (56 × 55 = 3080); y = √(200y) నుండి y = 200. అప్పుడు z = 256 = 2⁸ మరియు √(z√(z√z)) = z^(1/2 + 1/4 + 1/8) = z^(7/8) = 2⁷ = 128.