Big Bob is trying to shape up for the summer. He begins on the first day of his exercise program with 3 sit ups. Each day he will do 4 more sit ups than the day before. On what day will Big Bob do 115 sit ups?
looks like an arithmetic sequence with terms:
3, 7, 11, ....
a = 3, d=4 , n = ? so that term(n) ≥ 115
term(n) = a + (n-1)d
6 + 4(n-1) ≥ 115
4n - 4 ≥ 109
4n ≥ 113
n ≥ 28.25
on day 28 he will do 6 + 4(27) or 114 push-ups
so he will need 29 days to reach 114 push-ups
To determine on which day Big Bob will do 115 sit-ups, we can use a simple arithmetic progression formula.
The formula for an arithmetic progression is:
an = a1 + (n-1)d
Where:
an is the nth term of the arithmetic progression
a1 is the first term of the arithmetic progression
n is the number of terms
d is the common difference between consecutive terms
In this case, Big Bob starts with 3 sit-ups and adds 4 sit-ups each day, so the common difference (d) is 4. We want to find the day (n) when he does 115 sit-ups, so an = 115.
Substituting the values into the formula, we have:
115 = 3 + (n-1)4
Simplifying the equation:
112 = (n-1)4
Dividing both sides of the equation by 4:
28 = n - 1
Adding 1 to both sides of the equation:
n = 29
Therefore, Big Bob will do 115 sit-ups on the 29th day of his exercise program.