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

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

[tex]a_n=a+(n-1)d[/tex]

[tex]a=8, a_9=48[/tex]

Put n=9, we get

[tex]a_9=a+8d[/tex]

[tex]48=8+8d[/tex]

[tex]d=\frac{40}{8}=5[/tex]

Now, twenty-third term is

[tex]a_{23}=a+22d=8+22(5)=118[/tex]

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

Option 2 is correct

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