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3 years ago

Find the sum of the geometric series in which a1=160, a6= 5 and r= ½

Mathematics

1 answer:

Luden [163]3 years ago

4 0

Answer:

Sn = 315

The sum of the first six terms of the series is 315

Completed question:

Find the sum of the first six terms of the geometric series in which a1=160, a6= 5 and r= ½

Step-by-step explanation:

The sum of a geometric series in with common ratio

r < 1 is;

Sn = a(1 - r^n)/(1-r)

Where;

r = common ratio

a = first term

n = nth term

Given;

r = 1/2

a = a1 = 160

n = 6

Substituting the values, we have;

Sn = 160(1 - (1/2)^6)/(1 - 1/2)

Sn = 315

The sum of the first six terms of the series is 315

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Answer:

1. A. 1

2. D. $17^9$

Step-by-step explanation:

Properties used:

Power of a Power Property
Product Property
Quotient Property
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1. First, let's deal with the numerator.

$(17^3)^6$ can be turned into $17^1^8$ by using the Power of a Power Property.

And then use the Product Property, $(17^1^8)(17^-^1^0) = 17^8$

So now, our fraction is this: $\frac{17^8}{17^8}$

All number over itself in a fraction is equal to 1. But you can also do this the mathmatic way using the Quotient Property: $\frac{a^m}{a^n} = a^m^-^n$ or $\frac{1}{a^(^n^-^m^)}$ . Which then you plug the numbers in: $\frac{17^8}{17^8} = 17^8^-^8 = 17^0$ . And since we know that in Zero Exponent Property: $a^0 = 1$ , we can see that $17^0 = 1$ . So either way, we get 1.