find the sum of the first 20 terms of arirhmetic sequence -1, -5, -9

Solution: Since the last term [tex] (a_n) [/tex] isn't given, we will be using another type of arithmetic series formula where the common difference [tex] (d) [/tex] is needed.

[tex] \begin{align} & \bold{Formula:} \\ & \boxed{S_n = \small \frac{n}{2} \normalsize \big[ 2a_1 + d(n - 1) \big]} \end{align} [/tex]

» Find the common difference of the sequence.

[tex] \begin{align} & \bold{Formula:} \\ & \boxed{d = a_n - a_{n-1}} \end{align} [/tex]

[tex] d = a_2 - a_1 = \text-5 - (\text-1) = \text-4 [/tex]
[tex] d = a_3 - a_2 = \text-9 - (\text-5) = \text-4 [/tex]

» Find the sum of the first 20 terms where the first term [tex] (a_1) [/tex] is -1, the common difference [tex] (d) [/tex] is -4, and the number of terms [tex] (n) [/tex] is 20.

[tex] S_{20} = \small \frac{20}{2} \normalsize \big[ 2(\text-1) + (\text-4)(20-1) \big] \\ [/tex]

[tex] S_{20} = \small \frac{20}{2} \normalsize \big[ 2(\text-1) + (\text-4)(19) \big] \\ [/tex]

[tex] S_{20} = 10 \big[ \text-2 - 76 \big] [/tex]

[tex] S_{20} = 10 \big[ \text-78 \big] [/tex]

[tex] S_{20} = \text-780 [/tex]

[tex] \therefore [/tex] The sum of the first 20 terms in the sequence is...

[tex] \Large \underline{\boxed{\tt \purple{\text-780}}} [/tex]

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