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

Aug 27, 2015

color(blue)(a_32= 128

Explanation:

The data given is :
${a}_{1} = 4$
${a}_{6} = 24$

The formula for $n t h$ term is:
color(blue)(a_n = a_1+(n-1)d

Applying the same:
${a}_{6} = {a}_{1} + \left(6 - 1\right) \cdot d$

$24 = 4 + 5 d$

$5 d = 20$

color(blue)(d=4

Now,
${a}_{32} = {a}_{1} + \left(32 - 1\right) \cdot d$

$= 4 + 31 \left(4\right)$

$= 4 + 124$

color(blue)(a_32= 128