How do you find the sum of the arithmetic sequence given a1=3, d=-4, n=8?

Question

Serenity Jones · Accepted Answer

#-88#

The sum to n terms of an arithmetic sequence is.

#color(orange)"Reminder"#

#color(red)(bar(ul(|color(white)(a/a)color(black)(S_n=n/2[2a+(n-1)d])color(white)(a/a)|)))# where a is the first term, d the common difference and n, the terms being summed.

here a = 3 , d =- 4 and n = 8

#rArrS_8=8/2[(2xx3)+(7xx-4)]#

#=4(6-28)=-88#

Greyson Brown · Answer

undefined To find the sum of an arithmetic sequence, you can use the formula:

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

Where:
- $ S_n $ is the sum of the first $ n $ terms of the sequence.
- $ a_1 $ is the first term of the sequence.
- $ a_n $ is the $ n $-th term of the sequence.
- $ n $ is the number of terms in the sequence.

Given:
- $ a_1 = 3 $ (the first term)
- $ d = -4 $ (the common difference)
- $ n = 8 $ (the number of terms)

We first need to find the $ n $-th term of the sequence ($ a_n $) using the formula for the $ n $-th term of an arithmetic sequence:

$$ a_n = a_1 + (n - 1)d $$

Substitute the given values:

$$ a_n = 3 + (8 - 1)(-4) $$

$$ a_n = 3 + 7(-4) $$

$$ a_n = 3 - 28 $$

$$ a_n = -25 $$

Now, we can plug $ a_1 = 3 $, $ a_n = -25 $, and $ n = 8 $ into the formula for the sum of the arithmetic sequence:

$$ S_8 = \frac{8}{2}(3 + (-25)) $$

$$ S_8 = \frac{8}{2}(3 - 25) $$

$$ S_8 = \frac{8}{2}(-22) $$

$$ S_8 = 4(-22) $$

$$ S_8 = -88 $$

Therefore, the sum of the arithmetic sequence given $ a_1 = 3 $, $ d = -4 $, and $ n = 8 $ is $ -88 $.