Q) Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.
1. Which of the following terms are in AP for the given situation
a) 51,53,55…. b) 51, 49, 47…. c) -51, -53, -55…. d) 51, 55, 59…
2. What is the minimum number of days he needs to practice till his goal is achieved
a) 10 b) 12 c) 11 d) 9
3. Which of the following term is not in the AP of the above given situation
a) 41 b) 30 c) 37 d) 39
4. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is
a) 2 b) 3 c) 5 d) 1
5. The value of x, for which 2x, x+ 10, 3x + 2 are three consecutive terms of an AP
a) 6 b) -6 c) 18 d) -18
Ans:
VIDEO SOLUTION
STEP BY STEP SOLUTION
1. Terms of the AP:
Since the performance is being improved every day regularly, it makes an AP.
Given that the first-day performance is 51 seconds, therefore the first term of AP, ∴ a = 51
Each day’s practice reduces the time by 2 seconds, ∴ d = – 2
Given that the performance needs to be improved to 31, ∴ Tn = 31
∴ the AP is 51, 49, 47,……. 33, 31
Therefore, the answer is (b).
2. Minimum number of days to achieve the goal:
Since the goal is 31 seconds, it is the last term of the AP.
We know that the nth term of an AP is given by: Tn = a + (n – 1) d
Therefore, 31 = 51 + (n – 1) x (- 2)
∴ -20 = – 2 (n – 1)
∴ n – 1 = 10 or n = 11
Now n = 11 means that 31 is the 11th term of the AP
Starting with 51 seconds in 1st term, 49 seconds in 2nd term is achieved after 1 day of practice, similarly 47 in 3rd term is achieved after 2 days of practice and so on….similarly, 31 in the 11th term will be achieved after 10 days of the practice.
Therefore, the answer is (a).
3. Term not in the AP:
Since the first term of AP is 51, the end term is 31, and common difference is – 2, therefore, all terms of the AP will be odd numbers
Hence, even number will not be the term in the AP.
Therefore, the answer is (b).
4. Common difference of an AP if Tn is 2n + 3:
Since Tn = 2n + 3:
Therefore, Tn+1 =2 (n + 1) + 3 = 2n + 5
Since the common difference is the difference of the two consecutive terms,
∴ d = Tn+1 – Tn
∴ d = (2n + 5) – (2n + 3)
∴ d = 2
Therefore, the answer is (a).
5. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP
Since the common difference of an AP is the difference between two consecutive terms,
Therefore, 2nd term – 1st term = 3rd term – 2nd term
∴ (x + 10) – (2 x) = (3 x + 2) – ( x + 10)
∴ 10 – x = 2 x – 8
∴ 18 = 3 x
∴ x = 6
Therefore, the answer is a).
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