37.8k views
3 votes
What's the 111st element of a sequence that starts at 1.2 and increments by .3?

User AhabLives
by
7.7k points

1 Answer

7 votes

Answer:

34.2

Step-by-step explanation:

To find the 111st element of the sequence that starts at 1.2 and increments by 0.3, we can use the formula for the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

where:

an = the nth term of the sequence

a1 = the first term of the sequence

d = the common difference between terms (in this case, 0.3)

n = the term number we want to find

Substituting the given values, we have:

a111 = 1.2 + (111 - 1)0.3

a111 = 1.2 + 33.0

a111 = 34.2

Therefore, the 111st element of the sequence that starts at 1.2 and increments by 0.3 is 34.2.

User Pavel Uvarov
by
8.0k points