Question

A snail covered 3 mm in the first second and 4 mm more in each successive second, for a certain number a
seconds. However, if it had covered 1 mm in the first and 8 mm more in each successive second for some
seconds, then difference between the length of path it would have covered during the same time and the actua
bath it actually took would have been more than 6 mm but less than 30 mm. Find the time for which the snail moved​

Answer 1

Answer:

4 sec

Step-by-step explanation:

A snail covered 3 mm in the first second and 4 mm more in each successive second,

3  , 7  ,  11   ,  15  

a = 3 d = 4

Distnace covered in n sec

= (n/2)(3 + 3 + (n-1)4)

= n(3 + 2(n-1))

= n(2n + 1)

= 2n² + n

1  , 9  ,  17   ,  25  

a = 1 d = 8

Distance covered in n sec

= (n/2)(1 + 1 + (n-1)8)

= n(1 + 4(n-1))

= n(4n - 3)

= 4n² - 3n

Difference = 4n² - 3n - (2n² + n)

= 2n² - 4n

30 > 2n² - 4n > 6  

n = 3  => 2n² - 4n = 6

n =5 => 2n² - 4n = 30

n = 4 sec   2n² - 4n = 16 mm