# What is the 25th term of the arithmetic sequence where a1 = 8 and a9 = 48?

Jul 9, 2018

${a}_{25} = 128$

#### Explanation:

We know that ,

color(blue)(n^(th)term " of the arithmetic sequence is :"

color(blue)(a_n=a_1+(n-1)d...to(I)

We have ,

color(red)(a_1=8 and a_9=48

Using $\left(I\right)$ ,we get

$\implies {a}_{9} = {a}_{1} + \left(9 - 1\right) d = 48$

$\implies 8 + 8 d = 48$

$\implies 8 d = 48 - 8 = 40$

=>color(red)(d=5

Again using $\left(I\right)$ ,we get

${a}_{25} = {a}_{1} + \left(25 - 1\right) d$

$\implies {a}_{25} = 8 + 24 \left(5\right)$

$\implies {a}_{25} = 128$

Jul 9, 2018

${a}_{25} = 128$

#### Explanation:

$\text{the n th term of an arithmetic sequence is}$

•color(white)(x)a_n=a_1+(n-1)d

$\text{where d is the common difference}$

$\text{given "a_1=8" then}$

${a}_{9} = 8 + 8 d = 48 \Rightarrow 8 d = 40 \Rightarrow d = 5$

${a}_{25} = 8 + \left(24 \times 5\right) = 128$