Question

What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?

Answer 1

$a_1=8$
$a_2=a_1+d=8+d$
$a_3=a_2+d=(8+d)+d=8+2d$
$a_4=a_3+d=(8+d)+2d=8+3d$
...
$a_9=8+8d=48\implies d=5$
$a_{10}=8+9d$
...
$a_{23}=8+22d=118$

Answer 2

Answer:

23rd term of A.P is 118.

Step-by-step explanation:

Given that the first and ninth term of the arithmetic sequence which is 8 and 48 respectively.

we have to find 23rd term of A.P

The recursive formula for A.P is

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

$a=8, a_9=48$

Put n=9, we get

$a_9=a+8d$

$48=8+8d$

$d=\frac{40}{8}=5$

Now, twenty-third term is

$a_{23}=a+22d=8+22(5)=118$

Hence, 23rd term of A.P is 118.

Option 2 is correct