Let x and y be your two consecutive whole numbers
x < sqrt(142) < y
x^2 < 142 < y
So, we are looking for x and y such that x^2 < 142 and y^2 > 142.
The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.
11 and 12 are consecutive.
11^2 = 121 < 142 < 144 = 12^2
Thus, 11 and 12 are your numbers
The Answer to the question is B
The 400th term is 425.There are floor(√400) = 20 squares in the range 1..400, so the 400th term will be at least 420. There are floor(∛420) = 7 cubes in the range 1..400, so the 400th term may be as high as 427. However, there are
![\lfloor\sqrt[6]{427}\rfloor=2](https://tex.z-dn.net/?f=%5Clfloor%5Csqrt%5B6%5D%7B427%7D%5Crfloor%3D2)
numbers that are both squares and cubes. Consequently, the 400th term will be 427-2 =
425.