344,134 views
23 votes
23 votes
Find the sum of the first 33 terms of the following arithmetic

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

User Thsorens
by
2.8k points

2 Answers

16 votes
16 votes

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 :)

User Lawrance
by
3.2k points
18 votes
18 votes

Answer:

4851

Explanation:

the common difference is 9

33rd term = 3 + 32 * 9 = 291

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

User Eric Platon
by
3.4k points
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