Question

Find the sum of the first 33 terms of the following arithmetic

series, to the nearest integer.
3, 12, 21, ...

Answer 1

Answer: Sum of arithmetic terms = n/2 × [2a + (n - 1)×d], where 'a' is the first term, 'd' is the common difference between two numbers, and 'n' is the number of terms.

this is also the same as n×(a1 + an)/2, because

an = a1 + (n-1)×d

anyway, using that on the series above :

clearly d = 9, as every new term is the previous term plus 9.

a1 (or simply a) = 3

and we are adding up the first 33 terms, so, n = 33.

33/2 × (2×3 + 32×9) = 33/2 × (6 + 288) = 33/2 × 294 =

= 33 × 147 = 4851

Explanation:

so therefore your answer is 4851 hope this helps :)

Answer 2

4851

Explanation:

the common difference is 9

33rd term = 3 + 32 * 9 = 291

sum of first 33rd terms = 33/2 ( 3 + 291) = 4851