The first term of the sequence 3, 7, 11,... that exceeds 200 is 203.
Analyzing the Sequence
The given sequence is an arithmetic progression with the first term a = 3 and a common difference d = 4. We want to find the smallest term in this sequence that is greater than 200.
The general formula for the nth term of an arithmetic sequence is:
an = a + ( n - 1) d
In our case, an = 3 + ( n - 1) 4. We want to find the smallest n such that an* > 200.
Finding the Term
We can set up the inequality:
3 + ( n - 1) *4 > 200
Simplifying this:
3 + 4n - 4 > 200
4n - 1 > 200
4n > 201
n > 201/4
n > 50.25
Since n must be an integer, the smallest integer value for n that satisfies this inequality is n = 51.
Now, we can find the 51st term of the sequence:
a51 = 3 + (51 - 1) 4
a51 = 3 + (50) 4
a51 = 3 + 200
a51 = 203
Odd Number Verification
As mentioned in the provided reference, all terms in this sequence will be odd. We can see this directly from the formula: 3 + (n-1)*4 = 3 + 4n -4 = 4n -1. Since 4n is always even, 4n-1 is always odd. Since 203 is an odd number, it's a potential member of the sequence.
