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mario62 [17]
3 years ago
10

PLS help simple math properties ASAP needing 5 and 6 that’s all :)

Mathematics
1 answer:
RoseWind [281]3 years ago
4 0
They both seem to be substitution
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riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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In the spring, the owner of a sporting goods store decreases the price of winter gloves from $10.00 to $8.00. She increases the
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2

Step-by-step explanation:

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