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The 4th and 13th terms of an arithmetic progression (AP) are 5 and -1, respectively. Find the 8th term of the AP.

Answer :

now, we know the 4th term is 5.... ok... now, what's the common difference? well, we dunno, but notice, from the 4th to the 13th term, you do have to use it 9 times to hop over to the 13th term, let's say is "d", then

[tex]\bf \begin{array}{llll}
term&value\\
------&------\\
a_4&5\\
a_5&5+d\\
a_6&(5+d)+d\\
a_7&(5+d+d)+d\\
a_8&(5+d+d+d)+d\\
a_9&(5+d+d+d+d)+d\\
a_{10}&(5+d+d+d+d+d)+d\\
a_{11}&(5+d+d+d+d+d+d)+d\\
a_{12}&(5+d+d+d+d+d+d+d)+d\\
a_{13}&(5+d+d+d+d+d+d+d+d)+d\\
&5+9d
\end{array}[/tex]

we also know that the 13th term is -1

[tex]\bf \stackrel{a_{13}}{5+9d}=-1\implies 9d=-6\implies d=\cfrac{-6}{9}\implies \boxed{d=-\cfrac{2}{3}}[/tex]

now, recall above what the 8th term is

[tex]\bf a_8=5+4d\implies a_8=5+4\left(-\frac{2}{3} \right)\implies a_8=5-\cfrac{8}{3}\implies a_8=\cfrac{7}{3}[/tex]

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