(i) Form an A.P representing the number of push-ups per day and hence find the minimum number of days he needs to practice

Question

Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.

Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour. But he wants to achieve a target of 3900 push-ups in 1 hour for which he practices regularly. With each day of practice, he is able to make 5 more push-ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved. Keeping the above situation in mind answer the following questions:

(i) Form an A.P representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished?

(ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved.

Answer 1

(i) By the given situation the A.P. to be formed is:

3000, 3005, 3010, ...... 3900

First term = a = 3000

Common Difference = d = 3005 – 3000

d = 5

According to the problem

n^th term (a_n) = 3900

To Find n

n^th term = a_n

= a + (n – 1)d

3900 = 3000 + (n – 1)5

3900 – 3000 = (n – 1)5

900 = (n – 1)5

900/5 = n - 1

180 + 1 = n

or n = 181

Minimum number of days he needs to practice before his goal is accomplished

= 181 – 1 [Excluding the last day]

= 180

(ii) Total Number of push-ups performed means sum of all push-ups he did in 181 days.

∴ S_n = n/2[a + l] ...(i)

S_n = Sum of all push-ups in 181 days.

n = number of days = 181

a = first term of the A.P.

= 3000

l = last term = 3900

Put all values in (i)

⇒ S_n = (181/2) [3000 + 3900]

= (181/2)[6900]

= 624450

Total number of Push-ups performed in 181 days = 624450.

Answer 2

(i) 3000, 3005, 3010, ...,3900.

a_n = a + (n - 1)d

3900 = 3000 + (n - 1)5

⇒ 900 = 5n − 5

⇒ 5n = 905

⇒ n = 181

Minimum number of days of practice = n - 1 = 180 days

(ii) S_n = \(\frac{n}{2}\)(a + l)

= \(\frac{181}{2}\) × (3000 + 3900) = 624450 pushups