Let [tex]{a_n}_{n=1}^{\infty}[/tex] be a sequence such that [tex]a_n = \frac{1}{2}a_{n-1}[/tex], and let the sixth term of the sequence be [tex]\frac{1}{96}[/tex]. Then what will be the tenth term of the sequence?
Answer (2)
The tenth term of the sequence is 1536 1 . This was determined by using the recurrence relation to backtrack from the known sixth term, 96 1 , through previous terms until reaching the tenth term. Each term is half of the previous term in the sequence.
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To find the tenth term of the sequence { a n } n = 1 ∞ , we are given that a n = 2 1 a n − 1 and the sixth term a 6 = 96 1 . We want to find a 10 .
The sequence can be described as a geometric sequence where each term is half of the previous term. We know:
a 6 = 96 1
Since each term is half of the previous term, we can write it in terms of a 1 :
a 6 = a 1 ⋅ ( 2 1 ) 5
Thus, a 1 ⋅ 32 1 = 96 1 .
To find a 1 , solve:
a 1 = 96 1 × 32 = 96 32 = 3 1
Now that we have a 1 , we can find a 10 :
Since a n = a 1 ⋅ ( 2 1 ) n − 1 , we have:
a 10 = a 1 ⋅ ( 2 1 ) 9
Substitute a 1 = 3 1 :
a 10 = 3 1 ⋅ 512 1 = 1536 1
Therefore, the tenth term of the sequence is 1536 1 .
