1. Question :
A light bulb blinks at 12 noon. It then blinks after 4 seconds, then after 8 seconds, then after 12 seconds and so on. How many times shall it have already blinked until it blinks at12.08pm?
Correct Answer: 15
Explanation:
The seconds after which the light bulb blinks are 0, 4, 8, 12 …These form an AP with common difference 4 and first term 0.
Let the number of times it blinks in 8 minutes be n.
Sum of n terms = 8*60 = 480
Sum of n terms in an AP where a is the first term and d is the common difference is given by
Sn = (n/2)[2a+(n-1)d]
(n/2)[2*0+4(n-1)] = 480
2n^2-2n-480=0
n^2-n-240=0
n^2-16n+15n-240=0
n(n-16)+15(n-16)=0
n=16
The bulb had already blinked 15 times before it blinked at 12.08 pm