CAN SOMEONE HELP ME PLEASE? Find the sixth term of a geometric sequence for which a1 = -3 and r = -2.?

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answers:
Raymond say: The "easy" way is to simply multiply the "seed" (a1) by the ratio (r) and just count the number of times you do it:

a2 = a1 * r = (-3) * (-2) = +6
a3 = a2 * r = 6 * (-2) = -12
a4 = a3 * r = -12 * -2 = +24
and so on...
You can get to a6 faster than I can type all this

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The "real" way is to raise the ratio to the power of the number of steps, then multiply that by the seed.

There are 5 steps from a1 to a6

a6 = a1 * (r)^5
a6 = -3 * (-2)^5
a6 = -3 * (-32) = +something
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Krishnamurthy say: -3, 6, -12, 24, ... ( a1 = -3 and r = -2)
an = 3 (-1)^n 2^(n - 1)
a6 = 96
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la console say: For a geometric sequence:

a₁

a₂ = a₁ * r → where r is the common ratio

a₃ = a₂ * r = a₁ * r * r = a₁ * r²

a₄ = a₃ * r = a₁ * r² * r = a₁ * r³

…and you can generalize writing: a(n) = a₁ * r^(n - 1)


a(n) = a₁ * r^(n - 1) → where: n = 6

a₆ = a₁ * r⁵ → you know that: a₁ = - 3

a₆ = (- 3) * r⁵ → given that: r = - 2

a₆ = (- 3) * (- 2)⁵

a₆ = 96
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Puzzling say: Here's the definition:
The first term (a1) is -3.
Every subsequent term is found by multiplying by the common ratio (r) of -2.

First term --> -3
Second term --> -3 * -2 = 6
Third term --> 6 * -2 = -12
Fourth term --> -12 * -2 = 24
Fifth term --> 24 * -2 = -48
Sixth term --> -48 * -2 = 96

Answer:
96

You could also use the explicit formula for the nth term.
a[n] = a1 * r^(n-1)

So the 6th term is:
a[6] = -3 * (-2)^5
a[6] = -3 * -32
a[6] = 96
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-j. say: For a geometric sequence:
a1 = a constant
a2 = a1 * r
a3 = a2 * r, or a1 * r^2
a4 = a3 * 4, or a1* r^3

And so on. So for any a(n), where n is the number of the term,
a(n) = a1 * r^(n-1)

We want to find a6, the sixth term. So
a6 = a1 * r^(6 - 1)
a6 = -3 * -2^(5)
a6 = -3 * -32
a6 = 96

Hope that helps!
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