Help me Please!!!!!!!!!!

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

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?

7 0

2 years ago

Helppppppp mathhhhhhhh

aivan3 [116]

Again using the Pythagorean Theorem:

h^2=x^2+y^2, where h is the hypotenuse of a right triangle and x and y are the side lengths. You are given that the hypotenuse is 34 ft and the base side is 16ft. Let h=34, x=16, and y=height, then you have:

34^2=16^2+y^2

y^2=34^2-16^2

y^2=1156-256

y^2=900

y=√900

y=30 ft

So the ladder will reach a spot 30 feet high on the building.

8 0

3 years ago

Justine enrolls in a music program. The cost of the proram includes a one time registration fee of $45 and a fee of $9.

Anton [14]

Im not sure but i think its 45+9 x X

small x= times
big X = scenario

3 0

3 years ago

Read 2 more answers

Ms. Barclay orders birthday cupcakes for the month of June from an online vendor. Each cupcake costs $1.25 and there is a one-ti

artcher [175]

Answer:

Ms. Barclay ordered 9 cupcakes.

Step-by-step explanation:

$1.25x9=11.25

11.25+3.25=$14.50

8 0

3 years ago

Solve the absolute value 4=1+(2-1/4w)

Vikentia [17]

1+2= 3, this gives us 4=3-1/4w, subtract 3 on both sides is, 1=1/4w, divide 1/4 on each side. ANSWER: 4

3 0

3 years ago