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Svetlanka [38]
3 years ago
12

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

Mathematics
2 answers:
yKpoI14uk [10]3 years ago
8 0
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
abruzzese [7]3 years ago
3 0

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

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