Write an arithmetic sequence with first term 7/6 and common difference 2/3.
Prove that the terms of this sequence can't be integers
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Step-by-step explanation:
First term of the sequence is 10 and common difference is 3.
a
1
= 10 and d = 3
Next term = a
2
=a
1
+d=10+3=13
a
3
=a
2
+d=13+3=16
Thus, first three terms of the sequence are 10, 13 and 16.
Let 100 be the nth term of the sequence.
a
n
=a
1
+(n−1)d
100=10+(n−1)3
90=(n−1)3
n−1=30
n=31, which is a whole number.
Therefore, 100 is the 31
st
term of the sequence.
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