Question

Case study based questions are compulsory.Attempt any four subparts of each question.
Each subpart carries 1 mark.
Q17. Case Study based -1
Sprint events in track and field usually consists 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
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 in the A.P of the above given situation.
i) 31
i) 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) iv)n + nd
e) Ifx,y,z are in AP, then is equal to
12
Z
i) 0
ii) - 1
i
iv) 1
2​

Answer 1

Answer:

45,43,41,39,..............

Step-by-step explanation:

Girish can run currently 200m in 45 sec

a=45(first term)

with each day of practise it takes 2 sec less

d= -2

he wants to do job in 30 sec

so AP is 45,43,41,39,37,35,33,31,29,27.......

a)

ans) (i)

b)

ans (i)

c)

ans (ii) i.e 30

d)

ans) (i)

e)

ans (iv)

Answer 2

Given:

Time at which Girish can complete 200m = 45s

Time required by him to complete 200m = 30s

Difference in time he attains each day by practice = 2s

To find:

The AP formed.

Minimum no. of days he needs to practise till his goal is accomplished.

Solution:

(a) When Girish reduces his time by 2s each day through practice, he attain a series of timings from day one in the following manner:

45, 43, 41...

This is an AP because the common difference between two consecutive terms is always -2. Hence, option (iii) is the correct answer.

(b) To find the minimum no. of days to achieve his goal, determine the value of $n$ when he achieves 30s which is the nth term. Thus,

$a=45, d=-2, a_{n}=30$

$a_{n}=a+(n-1)d$

$30=45+(n-1)(-2)$

$-15=-2n+2$

$-17=-2n$

∴ $n=\frac{17}{2}=8.5$ days

Hence, Girish accomplishes his goal of 30 seconds for 200m sprint in 8.5 days. Thus, option (ii) is correct.

(c) The term 30 will not be present in the above AP because the series follows an odd number pattern whereas 30 is even. Hence, option (ii) is the correct answer.

(d) The nth term of a sequence $a,a+d,a+2d...$ is given by the formula $a_{n}=a+(n-1)d$ where $a, d, n$ are the first term, common difference and no. of terms in the sequence respectively. Hence, option (iii) is correct.

The following are the answers:

(a) 45, 43, 41... Option (iii) is the correct answer.

(b) Girish accomplishes his goal of 30 seconds for 200m sprint in 8.5 days. Option (ii) is correct.

(c) The term 30 will not be present in the above AP. Option (ii) is the correct answer.

(d) $a_{n}=a+(n-1)d$ Option (iii) is correct.