Verified Answer :

caylus

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Answer C: 4

\(u_{6}=128\)
\(u_{5}=64\)
\(u_{4}=32\)
\(u_{3}=16\)
\(u_{2}=8\)
\(u_{1}=4\)

Answer: The first term of the sequence is 4.

Step-by-step explanation:

Let a be the first term of the sequence,

Here, the common ratio of the sequence is 2,

Thus, the sequence is,

a, 2a, 4a, 8a, 16a, 32a, .............., so on,...

Since, the sixth term of the sequence = 32a

According to the question,

32 a = 128

⇒ a = 4

Thus, the first term of the sequence is 4.