-6,0,6,12,...;S²⁴
need ko din ng solution

Solution: Identify the sum of the first 24 terms in the sequence. Since the last term [tex](a_n) [/tex] isn't given, we will be using the another form 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 [tex] (d). [/tex]

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

[tex] d = a_2 - a_1 = 0 - (\text-6) = 6 [/tex]
[tex] d = a_3 - a_2 = 6 - 0 = 6 [/tex]
[tex] d = a_4 - a_3 = 12 - 6 = 6 [/tex]

» Now, identify the sum of the first 24 terms [tex] (S_{24}) [/tex] of the sequence where the first term [tex] (a_1) [/tex] is -6; the common difference [tex] (d) [/tex] is 6; and the number of terms [tex] (n) [/tex] is 24.

[tex] S_{24} = \small \frac{24}{2} \normalsize \big[2(\text-6) + 6(24 - 1) \big] \\ [/tex]

[tex] S_{24} = \small \frac{24}{2} \normalsize \big[2(\text-6) + 6(23) \big] \\ [/tex]

[tex] S_{24} = 12 \big[\text-12 + 138 \big] [/tex]

[tex] S_{24} = 12 \big[126\big] [/tex]

[tex] S_{24} = 1,\!512[/tex]

[tex] \therefore [/tex] The sum of the first 24 terms in the sequence is...

[tex] \Large \underline{\boxed{\tt \purple{1,\!512}}} [/tex]

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