Question

sequence of numbers 11,28,45.... in which each number is 17 more than its predeccessor. what is the 100th number in sequence.

Answer 1

Arithmetic progression:

an = a1 + ( n - 1 ) d

a1 = initial term of an arithmetic progression

d = common difference

In this case:

a1 = 11 , d = 17


an = a1 + ( n - 1 ) d

a100 = a1 + ( 100 - 1 ) d

a100 = a1 + 99 d

a100 = 11 + 99 ∙ 17

a100 = 11 + 1683

a100 = 1694

Answer 2

"in which each number is 17 more than its predecessor "
I hope you realize that is a great hint.

so a = 11, d = 17 of an arithmetic sequence
use your formula for the 100th term to find it

Answer 3

Or, simply:

an = a1 + ( n - 1 ) d

an = 11 + ( n - 1 ) ∙ 17

an = 11 + 17 n - 17

an = 17 n - 6


a100 = 17 ∙ 100 - 6

a100 = 1700 - 6

a100 = 1694