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:

### STEP BY STEP SOLUTION

1. Terms of the AP:

Since the performance is being improved every day on regular basis, it makes an AP.

Given that the first day performance is 51 seconds, therefore first term of AP, ∴ a = 51

Each day’s practice reduces the time by 2 seconds, ∴ d = – 2

Given that the performance need to be improved to 31, ∴ Tn = 31

∴ the AP is 51, 49, 47,……. 33, 31

2. Minimum number of days to achieve the goal:

Since the goal is 31 seconds and hence, 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

Hence it will take 11 days to achieve the goal.

3. Term not in the AP:

Since the first term of AP is 51, 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.

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

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