Infinite Geometric Series: Additional Practice (Set 2)

Two conditions have to hold at once: $\dfrac{a}{1-r} = 27$ and $ar = 6$. From the second, $a = \dfrac{6}{r}$, and substituting into the first gives an equation in $r$ alone.

$\frac{6/r}{1-r} = 27 \;\Longrightarrow\; 6 = 27r(1-r) \;\Longrightarrow\; 9r^2 - 9r + 2 = 0$

Factor it: $(3r - 1)(3r - 2) = 0$, so $r = \tfrac13$ or $r = \tfrac23$. Both are genuinely valid (they give $a = 18$ and $a = 9$, both with positive terms and $|r| < 1$), which is why the question asks which value could be the ratio. Only $\tfrac23$ appears among the choices.

The repeating block is two digits long, so the blocks form a geometric series with ratio $\tfrac{1}{100}$ and first term $\tfrac{72}{100}$.

$0.\overline{72} = \frac{\tfrac{72}{100}}{1 - \tfrac{1}{100}} = \frac{72}{99}$

Both numbers share a factor of $9$, so reduce.

$\frac{72}{99} = \frac{8}{11}$

Watch the two decoys: $\tfrac{18}{25}$ and $\tfrac{72}{100}$ both equal $0.72$ exactly, which is a terminating decimal and a different number from $0.727272\ldots$ .

Find the ratio of areas first. Joining the midpoints of a square of side $s$ produces a square whose side is the hypotenuse of a right triangle with legs $\tfrac{s}{2}$ and $\tfrac{s}{2}$, so that side is $\tfrac{s}{\sqrt2}$ and its area is $\tfrac{s^2}{2}$. Each square therefore has exactly half the area of the one before it.

The first area is $8^2 = 64$, so the areas form a geometric series with $a = 64$ and $r = \tfrac12$.

$S = \frac{64}{1 - \tfrac12} = \frac{64}{\tfrac12} = 128$

All the squares together cover twice the area of the first one, which is the signature of a ratio of $\tfrac12$.

The ratio is $r = -1$, so the sum formula is off limits: it requires $|r| < 1$. Go back to the definition and compute the partial sums.

$S_1 = 9, \quad S_2 = 0, \quad S_3 = 9, \quad S_4 = 0, \quad \ldots$

They hop between two distinct values and close in on nothing, so the series diverges. They also stay bounded, so the fourth option is wrong as well. Cancelling in pairs looks tempting, but it is a choice of grouping, and a different grouping ($9 + (-9 + 9) + (-9 + 9) + \cdots$) would suggest $9$ instead. That contradiction is precisely why the sum has to be defined through partial sums, and why this series simply has no sum.

Compute the ratio at this value of $x$.

$r = 2(0.6) - 1 = 0.2$

That lies inside the interval $0 < x < 1$ found above, so the series converges. Its first term is $1$.

$S = \frac{1}{1 - 0.2} = \frac{1}{0.8} = \frac54$

The answer $5$ comes from dividing by $r$ instead of $1 - r$, which is the standard slip.

Call the first term $a$. Since $\left|\tfrac25\right| < 1$ the series converges, and its sum is $\dfrac{a}{1 - 2/5} = \dfrac{a}{3/5} = \dfrac{5a}{3}$. The condition says that sum exceeds $a$ by $10$.

$\frac{5a}{3} = a + 10$

Multiply through by $3$ and collect the $a$ terms.

$5a = 3a + 30 \;\Longrightarrow\; 2a = 30 \;\Longrightarrow\; a = 15$

Check it: the sum is $\dfrac{15}{3/5} = 25$, which is indeed $10$ more than $15$.

Only one of these survives every choice of $a$, and the case that breaks the other three is $a = 0$. Then every term is $0 \cdot r^{\,n-1} = 0$, every partial sum is $0$, and the series converges to $0$ no matter how large $r$ is.

So take $a = 0$ and $r = 10$. That series converges while $|r| \ge 1$, which refutes the second statement, the fourth statement, and the "for no other $r$" half of the first.

$|r| < 1 \;\Longrightarrow\; S = \frac{a}{1-r} \quad\text{(true for every } a\text{, including } a = 0)$

The third statement is the safe direction, and it holds for all $a$. To get a genuine "if and only if" you must add the hypothesis $a \neq 0$, and then statements two and four become true as well.

A single repeating digit $d$ gives the geometric series $\tfrac{d}{10} + \tfrac{d}{100} + \cdots$, whose sum is $\tfrac{d}{9}$.

$0.\overline{5} = \frac{\tfrac{5}{10}}{1 - \tfrac{1}{10}} = \frac{\tfrac{5}{10}}{\tfrac{9}{10}} = \frac59$

The others miss: $0.\overline{59} = \tfrac{59}{99}$, and $0.\overline{05} = \tfrac{5}{99}$. The sly one is $0.5\overline{9} = 0.5999\ldots$, which by the same argument as $0.\overline{9} = 1$ equals exactly $0.6 = \tfrac35$, not $\tfrac59$.

Two facts, two equations. The sum gives $\dfrac{a}{1-r} = 12$, so $a = 12(1-r)$, and the first two terms give $a + ar = a(1+r) = 8$. Substitute the first into the second.

$12(1-r)(1+r) = 8 \;\Longrightarrow\; 12\left(1 - r^2\right) = 8 \;\Longrightarrow\; 1 - r^2 = \frac23$

So $r^2 = \tfrac13$, giving two ratios, $r = \tfrac{\sqrt3}{3}$ and $r = -\tfrac{\sqrt3}{3}$. Both have size about $0.577$, comfortably under $1$, so both produce convergent series and both are legitimate answers. That is why the question asks which value could be the ratio.

$r = \pm\frac{1}{\sqrt3} = \pm\frac{\sqrt3}{3}$

Only the positive one appears among the choices. The product $(1-r)(1+r) = 1 - r^2$ is the difference of squares, and spotting it is what keeps this from turning into a messy quadratic.

The series is geometric with $a = \tfrac{9}{10}$ and $r = \tfrac{1}{10}$, and $|r| < 1$, so it converges and the formula applies.

$S = \frac{\tfrac{9}{10}}{1 - \tfrac{1}{10}} = \frac{\tfrac{9}{10}}{\tfrac{9}{10}} = 1$

Every partial sum ($0.9$, $0.99$, $0.999$) is less than $1$, and that is not a contradiction: the sum of a series is the number the partial sums close in on, not one of the partial sums. If you claim the value falls short of $1$ by some positive amount $d$, the gap after $n$ terms is $\left(\tfrac{1}{10}\right)^{n}$, and logarithms produce an $n$ making that smaller than $d$, so the partial sums have already passed your claimed value. There is no room between $0.\overline{9}$ and $1$.

Only one part of $S_n$ depends on $n$, namely $\left(\tfrac23\right)^{\,n}$, and since $\left|\tfrac23\right| < 1$ those powers close in on $0$.

$S_n = 18\left(1 - \left(\tfrac23\right)^{\,n}\right) \quad\text{closes in on}\quad 18(1 - 0) = 18$

The given formula is consistent with the first term: $S_1 = 18\left(1 - \tfrac23\right) = 6$, as it should be. And the sum formula agrees, since $r = \tfrac23$ gives $\dfrac{6}{1 - 2/3} = 18$.

The error after $n$ terms is exactly $S\,r^{\,n} = 4(0.75)^n$, so solve for the smallest $n$ that drives it under the tolerance.

$4(0.75)^n < 0.05 \;\Longleftrightarrow\; (0.75)^n < 0.0125$

Take logarithms, then divide by $\log(0.75) \approx -0.1249$, which flips the inequality.

$n > \frac{\log(0.0125)}{\log(0.75)} = \frac{-1.903}{-0.1249} \approx 15.23$

So the fewest that works is $n = 16$, and every larger $n$ works too, which is why the question asks for the fewest. Checking the boundary: $4(0.75)^{15} \approx 0.0535$, still outside, and $4(0.75)^{16} \approx 0.0401$, inside. A ratio this close to $1$ makes the tail die slowly, which is why so many terms are needed for such a modest tolerance.