✏️ARITHMETIC SERIES ============================== Directions: Find S, given the following conditions: \begin{gathered} \begin{aligned} & \bold{Formula:} \\ & \boxed{S_n = \frac{n}{2}\big[ 2a_1 + d(n-1) \big]} \end{aligned} \end{gathered}Formula:Sn=2n[2a1+d(n−1)] #1: \sf a_1 = 2 \:;\:\: d = 6 \:;\:\: n = 18a1=2;d=6;n=18 \begin{gathered} S_{18} = \frac{18}{2} \big[ 2(2) + 6(18-1) \big] \\ \end{gathered}S18=218[2(2)+6(18−1)] \begin{gathered} S_{18} = \frac{18}{2} \big[ 2(2) + 6(17) \big] \\ \end{gathered}S18=218[2(2)+6(17)] S_{18} = 9 \big[4 + 102\big]S18=9[4+102] S_{18} = 9 \big[106\big]S18=9[106] S_{18} = 954S18=954 \therefore∴ The sum of the first 18 terms of the given arithmetic series is... \Large \underline{\boxed{\tt \purple{\,954\,}}}954 \: #2: \sf a_1 = 5 \:;\:\: d = 9 \:;\:\: n = 20a1=5;d=9;n=20 \begin{gathered} S_{20} = \frac{20}{2} \big[ 2(5) + 9(20-1) \big] \\ \end{gathered}S20=220[2(5)+9(20−1)] \begin{gathered} S_{20} = \frac{20}{2} \big[ 2(5) + 9(19) \big] \\ \end{gathered}S20=220[2(5)+9(19)] S_{20} = 10 \big[10 + 171\big]S20=10[10+171] S_{20} = 10 \big[181\big]S20=10[181] S_{20} = 1,\!810S20=1,810 \therefore∴ The sum of the first 20 terms of the given arithmetic series is... \Large \underline{\boxed{\tt \purple{\,1,\!810\,}}}1,810 \: #3: \sf a_1 = 2 \:;\:\: d = 6 \:;\:\: n = 10a1=2;d=6;n=10 \begin{gathered} S_{10} = \frac{10}{2} \big[ 2(2) + 6(10-1) \big] \\ \end{gathered}S10=210[2(2)+6(10−1)] \begin{gathered} S_{10} = \frac{10}{2} \big[ 2(2) + 6(9) \big] \\ \end{gathered}S10=210[2(2)+6(9)] S_{10} = 5 \big[4 + 54\big]S10=5[4+54] S_{10} = 5 \big[58\big]S10=5[58] S_{10} = 290S10=290 \therefore∴ The sum of the first 10 terms of the given arithmetic series is... \Large \underline{\boxed{\tt \purple{\,290\,}}}290 \: #4: \sf a_1 = 4 \:;\:\: d = 3 \:;\:\: n = 12a1=4;d=3;n=12 \begin{gathered} S_{12} = \frac{12}{2} \big[ 2(4) + 3(12-1) \big] \\ \end{gathered}S12=212[2(4)+3(12−1)] \begin{gathered} S_{12} = \frac{12}{2} \big[ 2(4) + 3(11) \big] \\ \end{gathered}S12=212[2(4)+3(11)] S_{12} = 6 \big[8 + 33\big]S12=6[8+33] S_{12} = 6 \big[41\big]S12=6[41] S_{12} = 246S12=246 \therefore∴ The sum of the first 12 terms of the given arithmetic series is... \Large \underline{\boxed{\tt \purple{\,246\,}}}246 \: #5: \sf a_1 = \text-3 \:;\:\: d = 2 \:;\:\: n = 10a1=-3;d=2;n=10 \begin{gathered} S_{10} = \frac{10}{2} \big[ 2(\text-3) + 2(10-1) \big] \\ \end{gathered}S10=210[2(-3)+2(10−1)] \begin{gathered} S_{10} = \frac{10}{2} \big[ 2(\text-3) + 2(9) \big] \\ \end{gathered}S10=210[2(-3)+2(9)] S_{10} = 5 \big[\text-6 + 18\big]S10=5[-6+18] S_{10} = 5 \big[12\big]S10=5[12] S_{10} = 60S10=60 \therefore∴ The sum of the first 1p terms of the given arithmetic series is... \Large \underline{\boxed{\tt \purple{\,60\,}}}60 ============================== #CarryOnLearning (ノ^_^)ノ
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