In the infinite sequence 13, 17, 21, 25, 29...

[rkafc81]

In the infinite sequence 13, 17, 21, 25, 29, . . . , where each term is 4 greater than the previous term, the 46th term is?

A. 183
B. 187
C. 191
D. 193
E. 197


OA = D... i chose E.

my reasoning was this:

the terms are 4 apart, so the 46th term is 46*4 = 184 bigger than the 1st term in the sequence, which is 13. so the 46th term is 13 + 184 = 197.

what exactly have I done wrong here? I know this is something to do with consecutive multiples and not including the last term... but i'm not sure how that works...

in my question review. i wrote out some example terms to help understand the concept:
(t = the term # in the sequence)-->

t1 = 13
(+4)
t2 = 17
(+4)
t3 = 21
(+4)
t4 = 24

so t4 is actually 12 greater than t1 and NOT 16 greater than t1.

i really can't see how this works!! please help me understand so i wont make this mistake again...

[jlucero]

For the same reason that the 46th positive integer (46) is not 46 larger than the first positive integer (1).

Think of it in terms of steps- to go from the first to the forty-sixth number in a sequence, you are making 45 jumps. If you did the same math with 45 jumps, you'd have the right answer here.