sequence of numbers 11,28,45.... in which each number is 17 more than its predeccessor. what is the 100th number in sequence.
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
"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
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
To find the 100th number in the sequence starting from 11, where each number is 17 more than its predecessor, you can use a simple arithmetic formula.
First, let's represent the first term in the sequence as a₁. In this case, a₁ = 11. We also know that each term is 17 more than its predecessor, so the common difference (d) is 17.
We can use the formula for the nth term of an arithmetic sequence:
aₙ = a₁ + (n-1)d
Substituting the values into the formula, we have:
aₙ = 11 + (100-1) * 17
Simplifying the equation:
aₙ = 11 + 99 * 17
aₙ = 11 + 1683
aₙ = 1694
Therefore, the 100th number in the sequence is 1694.