Geometric Progressions and Special Sums, Worked Solutions

Ques­tion 1: Ninth term and sum of a GP with neg­a­tive ratio#

The prob­lem#

Find the $9$th term and the sum of $9$ terms of $2, -6, 18, \ldots$

Under­stand­ing the prob­lem#

The terms alter­nate in sign, which tells you the com­mon ratio is neg­a­tive. You need two things: the sin­gle term in posi­tion $9$, and the total of the first $9$ terms.

The idea#

Iden­tify $a$ and $r$, then use $a_n = ar^{n-1}$ for the term and $\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$ for the sum. The form with $1 - r$ in the denom­i­na­tor is con­ve­nient here because $r < 1$.

Step-by-step solu­tion#

Step 1. Iden­tify the first term and the com­mon ratio (divide any term by the one before it).

$$\displaystyle a = 2, \qquad r = \frac{-6}{2} = -3$$

Check with the next pair: $\displaystyle \tfrac{18}{-6} = -3$. Good.

Step 2. Find the $9$th term. The power is $n - 1 = 8$, which is even, so the sign is pos­i­tive.

$$a_9 = ar^8 = 2 \times (-3)^8 = 2 \times 6561 = 13{,}122$$

Step 3. Find $(-3)^9$ for the sum. An odd power keeps the minus sign.

$$(-3)^9 = -19{,}683$$

Step 4. Sub­sti­tute into the sum for­mula.

$$\displaystyle \begin{aligned} S_9 &= \frac{a(1 - r^9)}{1 - r} = \frac{2\left(1 - (-19{,}683)\right)}{1 - (-3)} \\ &= \frac{2 \times 19{,}684}{4} = \frac{39{,}368}{4} = 9842 \end{aligned}$$

Check­ing the answer#

Since the signs alter­nate, the sum should be smaller than the last term $13{,}122$ but still pos­i­tive (the last term is pos­i­tive and largest). $9842$ fits. Also $S_9 = S_8 + a_9$, where $\displaystyle S_8 = \tfrac{2(1 - 6561)}{4} = -3280$, and $-3280 + 13{,}122 = 9842$.

Answer#

$a_9 = 13{,}122$ and $S_9 = 9842$.

Com­mon mis­take to avoid#

Writ­ing $1 - r = 1 - 3$ instead of $1 - (-3) = 4$. Keep brack­ets around a neg­a­tive ratio.

Ques­tion 2: Which term equals a given num­ber#

The prob­lem#

Which term of $\sqrt{2}, 2, 2\sqrt{2}, \ldots$ is $64$?

Under­stand­ing the prob­lem#

You know the sequence and one value, $64$. You must find the posi­tion $n$ at which $64$ appears.

The idea#

Write the gen­eral term $a_n = ar^{n-1}$, set it equal to $64$, and solve for $n$ by writ­ing both sides as pow­ers of the same base, as in Exam­ple 2.

Step-by-step solu­tion#

Step 1. Iden­tify $a$ and $r$.

$$\displaystyle a = \sqrt{2}, \qquad r = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Step 2. Write the gen­eral term. Since $a$ and $r$ are equal, the pow­ers com­bine.

$$a_n = \sqrt{2}\,(\sqrt{2})^{n-1} = (\sqrt{2})^{n}$$

Step 3. Write $64$ as a power of $\sqrt{2}$. Since $(\sqrt{2})^2 = 2$ and $64 = 2^6$,

$$64 = 2^6 = \left((\sqrt{2})^2\right)^6 = (\sqrt{2})^{12}$$

Step 4. Com­pare expo­nents.

$$(\sqrt{2})^n = (\sqrt{2})^{12} \;\Longrightarrow\; n = 12$$

Check­ing the answer#

$a_{12} = \sqrt{2} \times (\sqrt{2})^{11} = (\sqrt{2})^{12} = 2^6 = 64$. Cor­rect.

Answer#

$64$ is the $12$th term.

Ques­tion 3: A GP from its sum and prod­uct#

The prob­lem#

The sum of the first three terms of a GP is $\displaystyle \tfrac{13}{12}$ and their prod­uct is $-1$. Find the terms.

Under­stand­ing the prob­lem#

You do not know $a$ or $r$. You are given two facts about three con­sec­u­tive terms: their sum and their prod­uct. Two equa­tions for two unknowns is enough.

The idea#

Choose the three terms as $\displaystyle \frac{b}{r}, \; b, \; br$. This is a clever choice: when you mul­ti­ply them, $r$ can­cels, and the prod­uct equa­tion gives $b$ at once. Then the sum equa­tion gives $r$.

Step-by-step solu­tion#

Step 1. Use the prod­uct. The $r$'s can­cel.

$$\displaystyle \frac{b}{r} \cdot b \cdot br = b^3 = -1 \;\Longrightarrow\; b = -1$$

Step 2. Use the sum with $b = -1$.

$$\displaystyle -\frac{1}{r} - 1 - r = \frac{13}{12}$$

Step 3. Clear frac­tions by mul­ti­ply­ing every term by $12r$.

$$\begin{aligned} -12 - 12r - 12r^2 &= 13r \\ 0 &= 12r^2 + 25r + 12 \end{aligned}$$

Step 4. Fac­torise. We need two num­bers with prod­uct $12 \times 12 = 144$ and sum $25$: they are $16$ and $9$.

$$12r^2 + 16r + 9r + 12 = 4r(3r + 4) + 3(3r + 4) = (4r + 3)(3r + 4) = 0$$

So $\displaystyle r = -\frac{3}{4}$ or $\displaystyle r = -\frac{4}{3}$.

Step 5. Find the terms for $\displaystyle r = -\tfrac{3}{4}$.

$$\displaystyle \frac{b}{r} = \frac{-1}{-3/4} = \frac{4}{3}, \qquad b = -1, \qquad br = \frac{3}{4}$$

Step 6. For $\displaystyle r = -\tfrac{4}{3}$ you get $\displaystyle \tfrac{3}{4}, -1, \tfrac{4}{3}$: the same three num­bers in reverse order.

Check­ing the answer#

Sum: $\displaystyle \tfrac{4}{3} - 1 + \tfrac{3}{4} = \tfrac{16 - 12 + 9}{12} = \tfrac{13}{12}$. Prod­uct: $\displaystyle \tfrac{4}{3} \times (-1) \times \tfrac{3}{4} = -1$. Both con­di­tions hold.

Answer#

The terms are $\displaystyle \tfrac{4}{3}, -1, \tfrac{3}{4}$ (or $\displaystyle \tfrac{3}{4}, -1, \tfrac{4}{3}$).

Com­mon mis­take to avoid#

Start­ing with $a, ar, ar^2$ also works but leads to a harder sys­tem. The sym­met­ric choice $\displaystyle \tfrac{b}{r}, b, br$ is worth remem­ber­ing when­ever a prod­uct is given.

Ques­tion 4: Sums to infin­ity and a recur­ring dec­i­mal#

The prob­lem#

(a) Find the sum to infin­ity: $\displaystyle 1 - \tfrac{1}{3} + \tfrac{1}{9} - \cdots$. (b) Write $0.\overline{6}$ as a frac­tion.

Under­stand­ing the prob­lem#

(a) is an infi­nite GP. (b) is a recur­ring dec­i­mal $0.666\ldots$, which you can see as an infi­nite sum. In both you need the value of the whole infi­nite sum.

The idea#

If $\lvert r \rvert < 1$, the infi­nite sum exists and $\displaystyle S_\infty = \frac{a}{1 - r}$. Always check $\lvert r \rvert < 1$ first.

Step-by-step solu­tion#

Part (a)

Step 1. Iden­tify $a$ and $r$.

$$\displaystyle a = 1, \qquad r = \frac{-1/3}{1} = -\frac{1}{3}$$

Step 2. Check the con­di­tion: $\displaystyle \lvert r \rvert = \tfrac{1}{3} < 1$, so the sum to infin­ity exists.

Step 3. Apply the for­mula.

$$\displaystyle S_\infty = \frac{1}{1 - \left(-\frac{1}{3}\right)} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$$

Part (b)

Step 1. Break the dec­i­mal into place val­ues.

$$0.\overline{6} = 0.6 + 0.06 + 0.006 + \cdots$$

Step 2. This is a GP: each term is $\displaystyle \tfrac{1}{10}$ of the one before.

$$\displaystyle a = \frac{6}{10}, \qquad r = \frac{1}{10}, \qquad \lvert r \rvert < 1$$

Step 3. Apply the for­mula.

$$\displaystyle S_\infty = \frac{\frac{6}{10}}{1 - \frac{1}{10}} = \frac{\frac{6}{10}}{\frac{9}{10}} = \frac{6}{9} = \frac{2}{3}$$

Check­ing the answer#

(a) Par­tial sums $1, 0.667, 0.778, 0.741, \ldots$ close in on $0.75$ from both sides. (b) $2 \div 3 = 0.666\ldots$ Cor­rect.

Answer#

(a) $\displaystyle \tfrac{3}{4}$; (b) $\displaystyle 0.\overline{6} = \tfrac{2}{3}$.

Ques­tion 5: Num­ber of terms for a given sum#

The prob­lem#

Find $n$ so that $3 + 6 + 12 + \cdots$ to $n$ terms equals $765$.

Under­stand­ing the prob­lem#

The sum is known ($765$); the num­ber of terms $n$ is unknown.

The idea#

Write $S_n$ with $a = 3$, $r = 2$; since $r > 1$, use $\displaystyle S_n = \frac{a(r^n - 1)}{r - 1}$. Then solve the result­ing equa­tion for $2^n$.

Step-by-step solu­tion#

Step 1. Iden­tify $a = 3$ and $\displaystyle r = \tfrac{6}{3} = 2$.

Step 2. Set up the equa­tion.

$$\displaystyle \frac{3(2^n - 1)}{2 - 1} = 765$$

Step 3. Sim­plify: the denom­i­na­tor is $1$; divide both sides by $3$.

$$2^n - 1 = 255 \;\Longrightarrow\; 2^n = 256$$

Step 4. Write $256$ as a power of $2$: $256 = 2^8$. So $n = 8$.

Check­ing the answer#

List the terms: $3, 6, 12, 24, 48, 96, 192, 384$. Their sum is $765$. Cor­rect.

Answer#

$n = 8$.

Ques­tion 6: Insert­ing geo­met­ric means#

The prob­lem#

Insert $3$ geo­met­ric means between $2$ and $162$.

Under­stand­ing the prob­lem#

You must find three num­bers $G_1, G_2, G_3$ so that $2, G_1, G_2, G_3, 162$ is a GP.

The idea#

With $3$ means inserted, the GP has $5$ terms, so $162$ is the $5$th term: $ar^4 = 162$. This is the lesson's rule $\displaystyle r = \left(\tfrac{b}{a}\right)^{1/(n+1)}$ with $n = 3$.

Step-by-step solu­tion#

Step 1. Use $a_5 = ar^4$ with $a = 2$.

$$2r^4 = 162 \;\Longrightarrow\; r^4 = 81$$

Step 2. Solve. $81 = 3^4$, so the real val­ues are $r = 3$ or $r = -3$. Take $r = 3$ (the usual answer, giv­ing pos­i­tive means).

Step 3. Build the terms by mul­ti­ply­ing by $3$ each time.

$$G_1 = 2 \times 3 = 6, \quad G_2 = 6 \times 3 = 18, \quad G_3 = 18 \times 3 = 54$$

Step 4. Note the other option: $r = -3$ gives $-6, 18, -54$, which also makes a GP. The stan­dard answer uses the pos­i­tive ratio.

Check­ing the answer#

$54 \times 3 = 162$, so the GP $2, 6, 18, 54, 162$ ends cor­rectly.

Answer#

$6, 18, 54$.

Ques­tion 7: Two num­bers from their AM and GM#

The prob­lem#

The AM and GM of two pos­i­tive num­bers are $25$ and $15$. Find the num­bers.

Under­stand­ing the prob­lem#

Let the num­bers be $a$ and $b$ (both pos­i­tive). "AM is $25$" gives their sum; "GM is $15$" gives their prod­uct.

The idea#

From the sum and prod­uct of two num­bers, form the qua­dratic $t^2 - (\text{sum})t + (\text{product}) = 0$, whose roots are the two num­bers. This is the method of Exam­ple 8.

Step-by-step solu­tion#

Step 1. Turn the AM into a sum.

$$\displaystyle \frac{a + b}{2} = 25 \;\Longrightarrow\; a + b = 50$$

Step 2. Turn the GM into a prod­uct by squar­ing.

$$\sqrt{ab} = 15 \;\Longrightarrow\; ab = 225$$

Step 3. Form the qua­dratic.

$$t^2 - 50t + 225 = 0$$

Step 4. Fac­torise: two num­bers with prod­uct $225$ and sum $50$ are $45$ and $5$.

$$(t - 45)(t - 5) = 0 \;\Longrightarrow\; t = 45 \text{ or } 5$$

Check­ing the answer#

AM $\displaystyle = \tfrac{45 + 5}{2} = 25$ and GM $= \sqrt{45 \times 5} = \sqrt{225} = 15$. Also AM $\ge$ GM ($25 \ge 15$), as it must be.

Answer#

The num­bers are $45$ and $5$.

Ques­tion 8: Using the for­mu­las for sums of pow­ers#

The prob­lem#

Eval­u­ate (a) $\displaystyle \sum_{k=1}^{12} (3k^2 - k)$ and (b) $1^3 + 2^3 + \cdots + 10^3$.

Under­stand­ing the prob­lem#

In (a) you must add $3k^2 - k$ for $k = 1, 2, \ldots, 12$. In (b) you must add the first ten cubes. Adding term by term is slow; the for­mu­las are faster.

The idea#

A sum splits over addi­tion and con­stants come out­side: $\displaystyle \sum (3k^2 - k) = 3\sum k^2 - \sum k$. Then use

$$\displaystyle \sum_{k=1}^n k = \frac{n(n + 1)}{2}, \quad \sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}, \quad \sum_{k=1}^n k^3 = \left[\frac{n(n + 1)}{2}\right]^2.$$

Step-by-step solu­tion#

Part (a)

Step 1. Split the sum.

$$\displaystyle \sum_{k=1}^{12}(3k^2 - k) = 3\sum_{k=1}^{12}k^2 - \sum_{k=1}^{12}k$$

Step 2. Sum of squares with $n = 12$ (so $2n + 1 = 25$).

$$\displaystyle \sum_{k=1}^{12}k^2 = \frac{12 \times 13 \times 25}{6} = 2 \times 13 \times 25 = 650$$

Step 3. Sum of the num­bers.

$$\displaystyle \sum_{k=1}^{12}k = \frac{12 \times 13}{2} = 78$$

Step 4. Com­bine.

$$3 \times 650 - 78 = 1950 - 78 = 1872$$

Part (b)

Step 1. Use the sum of cubes with $n = 10$.

$$\displaystyle \sum_{k=1}^{10}k^3 = \left(\frac{10 \times 11}{2}\right)^2 = 55^2 = 3025$$

Check­ing the answer#

For (b), the sum of the first $10$ num­bers is $55$, and the sum of cubes is always the square of that sum. For (a), the first few terms are $2, 10, 24, \ldots$; the for­mula with $n = 3$ gives $3 \times 14 - 6 = 36 = 2 + 10 + 24$, so the method is right.

Answer#

(a) $1872$; (b) $3025$.

Ques­tion 9: Prov­ing three expres­sions are in GP#

The prob­lem#

If $a, b, c$ are in GP, show that $a^2 + b^2, \; ab + bc, \; b^2 + c^2$ are also in GP.

Under­stand­ing the prob­lem#

What you are given: $a, b, c$ are in GP. What you must show: the three new expres­sions are in GP.

The idea#

Three quan­ti­ties $x, y, z$ are in GP exactly when the square of the mid­dle one equals the prod­uct of the outer two: $y^2 = xz$. The given fact trans­lates to $b^2 = ac$. So the strat­egy is: com­pute $(ab + bc)^2$ and $(a^2 + b^2)(b^2 + c^2)$ sep­a­rately, replace $b^2$ by $ac$ in each, and show both become the same expres­sion.

Step-by-step solu­tion#

Step 1. Write the given con­di­tion. Since $a, b, c$ are in GP,

$$b^2 = ac$$

Step 2. Work on the square of the mid­dle term. Take out the com­mon fac­tor $b$ first.

$$(ab + bc)^2 = \left(b(a + c)\right)^2 = b^2(a + c)^2 = ac(a + c)^2$$

(the last step uses $b^2 = ac$).

Step 3. Work on the prod­uct of the outer terms. Replace each $b^2$ by $ac$.

$$(a^2 + b^2)(b^2 + c^2) = (a^2 + ac)(ac + c^2)$$

Step 4. Fac­torise each bracket.

$$(a^2 + ac)(ac + c^2) = a(a + c) \cdot c(a + c) = ac(a + c)^2$$

Step 5. Com­pare Steps 2 and 4: both equal $ac(a + c)^2$. So

$$(ab + bc)^2 = (a^2 + b^2)(b^2 + c^2),$$

which is exactly the con­di­tion for $a^2 + b^2, \; ab + bc, \; b^2 + c^2$ to be in GP.

Check­ing the answer#

Try $a = 1$, $b = 2$, $c = 4$ (a GP). The new num­bers are $5, 10, 20$, which is a GP with ratio $2$.

Answer#

Since $(ab + bc)^2 = ac(a + c)^2 = (a^2 + b^2)(b^2 + c^2)$, the three expres­sions are in GP.

Ques­tion 10: Bac­te­ria that dou­ble every hour#

The prob­lem#

A cul­ture of bac­te­ria dou­bles every hour. Start­ing with $50$, how many after $8$ hours?

Under­stand­ing the prob­lem#

At the start (time $0$) there are $50$ bac­te­ria. After each hour the num­ber is mul­ti­plied by $2$. You want the num­ber after $8$ such hours.

The idea#

The counts $50, 100, 200, \ldots$ form a GP with ratio $2$. After $8$ dou­blings, the start­ing num­ber has been mul­ti­plied by $2$ eight times, so the count is $50 \times 2^8$. (If you count the start­ing value as the $1$st term, this is the $9$th term $ar^8$.)

Step-by-step solu­tion#

Step 1. Iden­tify the start value and ratio: $a = 50$, $r = 2$.

Step 2. After $8$ hours there have been $8$ dou­blings.

$$50 \times 2^8 = 50 \times 256$$

Step 3. Mul­ti­ply.

$$50 \times 256 = 12{,}800$$

Check­ing the answer#

Hour by hour: $100, 200, 400, 800, 1600, 3200, 6400, 12{,}800$. The eighth value is $12{,}800$.

Answer#

$12{,}800$ bac­te­ria.

Com­mon mis­take to avoid#

Using $2^7$ because "the $8$th term is $ar^7$". Here the start­ing value is at hour $0$, so after $8$ hours you need $8$ dou­blings.