a) $$ 1+2+3+4+\ldots $$ to 15 terms b) $$ 4+8+12 $$

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To find the sum of the first 15 terms of the arithmetic series $$1+2+3+4+\ldots$$, we can use the formula for the sum of an arithmetic series:

$$ S_n = \frac{n}{2} \times (a_1 + a_n) $$

where:
- $$ S_n $$ is the sum of the first $$ n $$ terms,
- $$ n $$ is the number of terms,
- $$ a_1 $$ is the first term,
- $$ a_n $$ is the last term.

Given that the first term $$ a_1 = 1 $$ and the last term $$ a_n = 15 $$, we can substitute these values into the formula to find the sum of the first 15 terms.

For part b), the sum of the series $$4+8+12$$ is simply the sum of the individual terms.

Let's calculate the sum for both parts.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{15}{2}\left(1+15\right)$$
- step1: Add the numbers:
  $$\frac{15}{2}\times 16$$
- step2: Reduce the numbers:
  $$15\times 8$$
- step3: Multiply the numbers:
  $$120$$
a) The sum of the first 15 terms of the arithmetic series $$1+2+3+4+\ldots$$ is 120.

b) The sum of the series $$4+8+12$$ is 24.
a) The sum of the first 15 terms is 120. b) The sum of $$4+8+12$$ is 24.

a) \( 1+2+3+4+\ldots \) to 15 terms b) \( 4+8+12 \)

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