Question

2%7D%2B%5Cleft%28a%20r%5E%7B2%7D%5Cright%29%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D150%24.%20Find%20%24a%20r%5E%7B3%7D%2Ba%20r%5E%7B4%7D%2Ba%20r%5E%7B6%7D%2B%5Cldots%20%5Cinfty%24" id="TexFormula1" title="$a+a r+a r^{2}+\ldots \infty=15$$a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}+\ldots \infty=150$. Find $a r^{3}+a r^{4}+a r^{6}+\ldots \infty$" alt="$a+a r+a r^{2}+\ldots \infty=15$$a^{2}+(a r)^{2}+\left(a r^{2}\right)^{2}+\ldots \infty=150$. Find $a r^{3}+a r^{4}+a r^{6}+\ldots \infty$" align="absmiddle" class="latex-formula">
Options:
$(a) \frac{1}{2} \\(b) \frac{2}{5}$

Answer 1

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?