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6 + 10 + 14 + ... + 38
Find the sum of the arithmetic series

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6 + 10 + 14 + ... + 38

Find the sum of the arithmetic series

  • [tex]\Large\sf{\underline{The\:sum\:of\:the\: series = \green{198}}}[/tex]

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» In the given arithmetic series:

  • The first term [tex]\rm{(a) = 6}[/tex]

  • Common difference [tex]\rm{(d) = (a_n - a_{n - 1}) = 10 - 6 = 4}[/tex]

  • Last term [tex]\rm{(a_n) = 38}[/tex]

» To find the sum of the series, we need to find the number of terms (n) at first. So,

  • [tex]\rm{a_n = a + (n - 1)d}[/tex]

  • [tex]\rm{38 = 6 + (n - 1)4}[/tex]

  • [tex]\rm{38 - 6 = 4n - 4}[/tex]

  • [tex]\rm{32 = 4n - 4}[/tex]

  • [tex]\rm{32 ÷ 4 = 4n}[/tex]

  • [tex]\rm{36 = 4n}[/tex]

  • [tex]\rm{36 ÷ 4 = n}[/tex]

  • [tex]\rm{9 = n}[/tex]

» Now, let's find the sum of the arithmetic series [tex]\rm{(S_n)}[/tex]:

  • [tex]\rm{S_n = \dfrac{n}{2}[2a + (n - 1)d]}[/tex]

  • [tex]\rm{S_n = \dfrac{9}{2}[2 × 6 ÷ (9 - 1)4]}[/tex]

  • [tex]\rm{S_n = \dfrac{9}{2}[12 + (8 × 4)]}[/tex]

  • [tex]\rm{S_n = \dfrac{9}{2}[12 + 32]}[/tex]

  • [tex]\rm{S_n = \dfrac{9}{2}(44)}[/tex]

  • [tex]\rm{S_n = 9 × 22}[/tex]

  • [tex]\rm{S_n = 198}[/tex]

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