Question

A snail covered 3 mm in the first second and 4 mm more in each successive second, for a certain number of seconds. However, if it had covered I mm in the first and 8 mm more in each successive second for some seconds, then difference benween the length of path it would have covered during the same time and the actual
path 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:

snail moved​ for 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

Distance 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