If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

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If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

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    2021-09-08T21:37:06+00:00

    this is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the “common difference” “d”.

    now, we know the 17th term is 34, let’s go to the 15term then

    34 + d
    34 + d + d
    34 + d + d + d
    34 + d + d + d + d
    34 + d + d + d + d + d
    34 + d + d + d + d + d + d
    34 + d + d + d + d + d + d + d
    34 + d + d + d + d + d + d + d + d

    now, notice, we got to the 15th term, which is 34  + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

    \bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}

    ok, now that we know what the common difference is, let’s find the first term in the sequence using the 7th term of 34,

    \bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad  \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ n=7\\ d=5\\ a_7=34 \end{cases} \\\\\\ a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5 \\\\\\ 34=a_1+30\implies \boxed{4=a_1}

    and now let’s use those 2 found folks, to get the 21st term,

    \bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad  \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=4\\ d=5\\ n=21 \end{cases} \\\\\\ a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104

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