Sprint events in track and field usuallyconsists of the 100 m, 200 m and 400 m race though 60 m dashes are also held on occasion. These races are largely based upon the athlete’s ability to accelerate to his or her maximum speed in the quickest time possible.
Girish wants to participate in the 200 m Sprint. He can currently run that distance in 45 sec. But he wants to do that in under 30 sec. With each day of practice, it takes him 2 sec less. Respond to the above situation by answering the questions given below:
a) Which of the following forms an A.P for the above given situation.
i) 45 , 90 , 180,.....
ii) 45, 47 , 49,.....
iii) 45, 43, 41,......
iv) −45 , − 47, − 48,.....
b) What is the minimum number of days he needs to practice till his goal is accomplished.
i) 8 days
ii) 8.5 days
iii) 9 days
iv) none of these
c) Which of following term is not inthe A.P of the above given situation.
i) 31
ii) 30
iii) 29
iv) 27
d) The nth term of the sequence a, a + d, a + 2d,... is
i) a + nd
ii) a –(n –1)d
iii) a + (n –1)d
iv) n + nd
e) If x ,y , z are in AP, then −−is equal to
i) 0
ii) –1
iii) 12
iv) 1
Answers
a) iii) 45, 43, 41, ...
b) ii) 8.5 days
c) ii) 30
d) iii) a + (n - 1) d
a) Girish wants to participate in the 200 m Sprint. He can currently run that distance in 45 sec. But he wants to do that in under 30 sec. With each day of practice, it takes him 2 sec less.
So, we can form an A.P. that begins with 45 and gets lessened by 2 for each consecutive terms:
45, (45 - 2), (45 - 2 - 2), (45 - 2 - 2 - 2), ...
⇒ 45, 43, 41, 39, ...
b) To find the minimum number of days required, we take help from the A.P. of question a),
45, 43, 41, 39, 37, 35, 33, 31, 29, ...
So, we see on the 8th day, it is 31, only 1 greater than 30 and on the 9th day, it is 29, only one less than 30. That is, the required finishing can be completed between the 8th and the 9th day.
Thus the required time
= (8 + 9)/2 days
= 17/2 days
= 8.5 days
c) We take the A.P. of question a),
45, 43, 41, 39, 37, 35, 33, 31, 29, 27, ...
Clearly, 30 is not in the A.P.
d) The given sequence is
a, a + d, a + 2d, a + 3d, ...
Clearly, its first term = a and common difference = d
So, nth term
= a + (n - 1) d