Question

The fifth term of an arithmetic sequence is 100, and the common difference is 4. Find the first three terms of the sequence.

Answer 1

$\bf a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ --------------\\ a_5=100\\ n=5\\ d=4 \end{cases} \\\\\\ a_5=a_1+(5-1)4\implies 100=a_1+(5-1)4 \\\\\\ \textit{solve for }a_1\textit{ to find the first term}$

then use the common difference "d", to get the 2nd and 3rd terms

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$\bf a_5=a_1+(5-1)4\implies 100=a_1+(5-1)4\implies 100=a_1+(4)4 \\\\\\ 100=a_1+16\implies 100-16=a_1\\\\ 84=a_1\quad \begin{cases} a_1=&84\\ a_2=84+4\to &88\\ a_3=88+4\to &92 \end{cases}$