Suppose you are given the arithmetic sequence 3,6,9,...,99.What is the sum of the terms of this sequence?

Answer 1

since the difference is 3, there are (99-3)/3 + 1 = 33 terms

S33 = 33/2 (3+99)

Answer 2

Suppose you are given the arithmetic sequence 3,6,9,...,99 . What is the sum of the terms of this sequence?

Answer 3

The formula of an arithmetic sequence is n(a+z)/2 where a is the first term of the sequence and z is the last term of the sequence. n is the number of terms in a sequence. So in this problem, there would be (99-3)/3 + 1 terms. Which is 33. the first term is 3 and the last term is 99 meaning it would be 3 + 99. then plug the numbers in we would get 33(3+99)/2. which is 33(102)/2 which simplifies to 33(51). 33 x 51 is 1683. So 1683 would be our answer.

Answer 4

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