plz solve according to sequence and series pattern

a man was trying to swim to a buoy placed at a distance of 200m out in the sea. it took him 1 min to swim 20m. then a wave pushed him back 10m and he rested for about 1min before swimming again. he continued in his way for the rest of their journey. how long would it take the man to swim to the buoy?

37 minutes

see previous solution for the explanation.

I have no idea what a sequence and series pattern involves, but surely you do. Try fitting Reiny's solution into your framework.

took me 40

http://www.jiskha.com/display.cgi?id=1420913304

At a rate of 5 m/min, after 36 minutes he has gone 180m. On the 37th minute he swims 20m and is done. No need to let go and float backwards again.

To solve this question using the sequence and series pattern, we need to break down the swim into individual steps and find a pattern.

Step 1: The man swims 20m in 1 minute.
Step 2: The man is pushed back 10m.
Step 3: The man rests for 1 minute.

Let's analyze step by step:

Step 1: The man swims 20m in 1 minute.
This step implies the man's swimming speed is 20m per minute. We can represent this as the first term of an arithmetic sequence, with a first term (a₁) of 20.

Step 2: The man is pushed back 10m.
This step implies the man loses 10m of progress made from step 1. The distance to the buoy reduces from 200m to 190m.

Step 3: The man rests for 1 minute.
During this step, there is no progress in the distance, so the man's position remains at 190m.

Now, let's summarize the steps:

Step 1: Distance covered = 20m in 1 min (a₁ = 20).
Step 2: Distance remaining = 190m (after being pushed back by the wave).
Step 3: Rest for 1 min.

To determine how many times the man will repeat these steps, we can set up an equation based on the sum of an arithmetic series.

The sum of the first n terms of an arithmetic series is given by the formula:

Sn = (n/2)(2a₁ + (n-1)d)

Where Sn is the sum of the first n terms, a₁ is the first term, and d is the common difference.

In this case, a₁ = 20 (distance covered in step 1) and d = 0 (there is no change in distance covered in each step).

So, the equation becomes:

Sn = (n/2)(2 × 20 + (n-1) × 0)
= n × 20
= 20n

We want to find the value of n so that the sum of the first n terms (Sn) is equal to or greater than 200 (the distance to the buoy).

20n ≥ 200

Dividing both sides by 20:

n ≥ 10

Therefore, the man needs to repeat the steps at least 10 times to cover a distance of 200m.

Since each cycle of steps takes 2 minutes (1 minute swimming and 1 minute resting), we can multiply the number of cycles (10) by 2 to find the total time:

Total time = Number of cycles × Time per cycle
= 10 × 2
= 20 minutes

So, it will take the man approximately 20 minutes to swim to the buoy.