Question

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

Answer 1

Answer: $a_{25} = 128$

Step-by-step explanation:

You need to use this formula:

$a_n = a_1 + (n - 1)d$

Where $a_n$ is the nth] term,  $a_1$ is the first term,"n" is the term position and  "d" is the common diference.

You must find the value of "d". Substitute $a_1=8$, $a_9=48$ and $n=9$ into the formula and solve for "d":

$48 = 8 + (9 - 1)d\\48=8+8d\\48-8=8d\\40=8d\\d=5$

Now, you can calculate the 25th term substituting into the formula these values:

$a_1=8$

$d=5$ and $n=25$

Then you get:

$a_{25} = 8 + (25 - 1)5$

$a_{25} = 8 + 120$

$a_{25} = 128$