What is the 32nd term of the arithmetic sequence where a1 = 4 and a6 = 24?

Jan 20, 2016

32nd term ${a}_{32} = 128$

Explanation:

If ${a}_{1}$ is first term, $d$ is the common difference, and ${a}_{n}$ is the $n t h$ term of an arithmetic sequence, then relationship between these is

a_n = a_1 + (n – 1)d

To find $d$, plug-in the given values
a_6 = a_1 + (6 – 1)d
24 = 4 + (6 – 1)d
Solving for $d$ we obtain
$5 d = 20$
or $d = 4$

To find 32nd term, plug-in appropriate values
a_32 = 4 + (32 – 1)4
${a}_{32} = 128$