sequence problems

[kaedwards10]

Pg 114 of study guide Book 1 Ch. 5 #8
If each number of a sequence is 4 more than the previous number, and the third number in the sequence is 13, what is the 114th number in the sequence?

I don't understand how to solve this problem with such a large jump? How do I find a pattern?

[n00bpron00bpron00b]

If each number of a sequence is 4 more than the previous number, and the third number in the sequence is 13, what is the 114th number in the sequence?

3rd number in the sequence is 13 and we the know the quantity by which each iteration increases (i.e by 4 : 4 more than the previous number)

If 3rd value is 13
2nd value is 13-4 = 9
1st value is 9-4 = 5

If you notice, value in the sequence increases by a fixed amount (a case of Arithmetic Progression)

tn = nth element in the sequence
a = starting or 1st element in the sequence
d = common difference between two consecutive values
n = total number of elements in the sequence

The formula to find any "nth" element using Arithmetic Progression is as follows -

tn = a + (n-1)*d

tn = we need to find the 114th value ; so tn = t114
a = starting or first element in sequence = 5 (derived above)
d = common difference = 4 (given)
n = total number of elements in the sequence = 114 (given)

t114 = 5 + (114-1)*4
t114 = 5 + (113)*4
t114 = 5 + 452
t114 = 457

[tommywallach]

Noob is absolutely right.

Just to keep it a bit simpler, the term number always represents one more than the number of jumps.

For example, if I'm starting at 3 and going up by 2s, the second term will be 5. In this case, there was ONE jump of 2. The third term will be 7. In this case, there were TWO jumps of 2 (from 3 to 5, and then from 5 to 7). So you see how the jumps are always one less than the term number?

So in this case, we want the 114th term, so we need 113 jumps.

-t