Okay, I need to find the 101th term of the sequence 5,9,13,17,21...
Do I do this:
a1 + (n-1)d
SO...
5+(101-1)4
5+(100)4
105x4
420
Did I do this problem correctly?
Almost!
First, check your formula against values:
a1+(n-1)d
where a1=5, d=4, and n=term number
For the first term, n=1
5+(1-1)*4=5 correct
For the third term:
5+(3-1)*4=5+2*4=13 correct.
So there is a good chance the formula.
Now do the calculations for the 101th term:
5+(101-1)*4
=5+100*4
Remembering that in all calculations, the priority of operations is in the order:
()
* or ÷
+ or -
So we need to multiply before add, or
5+100*4
=5 + (100*4)
=5 + 400
= 405
Yes, you have applied the correct formula to find the 101st term of the sequence. The formula for the nth term of an arithmetic sequence is a1 + (n-1)d, where a1 is the first term and d is the common difference between consecutive terms. In this case, the first term is 5 and the common difference is 4.
Therefore, to find the 101st term, you substituted n = 101 into the formula: 5 + (101 - 1) * 4. Simplifying this expression gives you 5+(100)4, which equals 420.
So, your calculations are correct, and the 101st term of the sequence 5,9,13,17,21... is indeed 420.
To find the 101st term of the sequence, you can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
Where:
an is the nth term
a1 is the first term
n is the position of the term
d is the common difference between consecutive terms.
In this case, the first term (a1) is 5, and the common difference (d) is 4.
So, substituting these values into the formula:
a101 = 5 + (101-1)4
= 5 + (100)4
= 5 + 400
= 405
Therefore, the correct answer is 405, not 420.