Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms?

Question

Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms?

Answer 1

30/2 * (a1 + a30)

Answer 2

All the previous answers assume the first term is 1 and so is the common difference. If we let the first term be a, and the common difference be d, then: the first term is a 30th term is a+29d the sum of the 30 terms is a + (a+d) + (a+2d) + ....+(a+29) that equals 30a + d(0+1+2+3+...+29) that equals 30a + 435d and that is the general expression for the sum of 30 terms of any arithmetic progression. If a=1 and also d=1, then the sum is 465, which is what everyone else is saying.

Answer 3

((Number of Terms) / 2) * (First Term + Last Term)

For example, the sum of 1+2+3+4+5+6+7+8+9+10 is

(10 / 2) * (10+1) = 5*11 = 55

You can also calculate it in this way:

(Last Term * (Last Term + 1)) / 2
(10 * 11) / 2 = 110 / 2 = 55


Good luck.

Answer 4

Depends on the first term and how much is being added each time.

Basically the formula is (first term + last term) x (# of terms) and then take that total and divide it by 2.

Answer 5

the formula for the sum of arithmetic sequnce is
(n/2)(a1+nth)

n-number of terms
a1-first term
nth-the nth term

Answer 6

yeah again i will assume that this too is the continuation of your previous questions.. right?

sum of the first 30 terms = [n(2a1+(n-1)d)/]2
sum=30[2(2)+29(2)]/2
sum=930

Answer 7

let the 1st term be a and the difference between terms be d.

30th term is a + 29d.
sum[30] is 30/2 ( 2a + 29d) = 30a + 435d