Question

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

113

118

123

128

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

23rd term of A.P is 118.

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=(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