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