Question

Algebra II:

The 4th term of a geometric sequence is 48, and the common ratio is 2. What is the first term of the sequence?

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Answer 1

6

Explanation:

geometric sequence is given as Tn=ar^n-1

where Tn is the last term,a is the first term, r is the common ratio, n is the nth term

hence 48=a2^(4-1)

48=a2^3

48=a8 (divide both by 8 to ramain with a)

6=a

therefore the first term is 6 .

Answer 2

6

Explanation:

`nth \: term \: of \: geometric \: \\ sequence \: is \: given \: as : \\ a_n = a {r}^(n - 1) \\ (a = ?, \: \: r = 2)\\ \\ a_4 =48......(given) \\ \\ \therefore \: a {(2)}^(4 - 1) = 48 \\ \\ \therefore \: a {(2)}^(3) = 48\\ \\ \therefore \: 8a = 48 \\ \\ \therefore \: a = (48)/(8) \\ \\ \therefore \:a = 6`