If i had 127 tables and each table could fit 4 people how many people could i seat?
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Answer:
23rd term of the arithmetic sequence is 118.
Step-by-step explanation:
In this question we have been given first term a1 = 8 and 9th term a9 = 48
we have to find the 23rd term of this arithmetic sequence.
Since in an arithmetic sequence
here a = first term
n = number of term
d = common difference
since 9th term a9 = 48
48 = 8 + (9-1)d
8d = 48 - 8 = 40
d = 40/8 = 5
Now
= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118
Therefore 23rd term of the sequence is 118.