Sequences and Series: Mixed Practice, Worked Solutions

Ques­tion 1: A sequence defined by a recur­rence#

The prob­lem#

Find the $n$th term and the $10$th term of $a_1 = 1$, $a_{n+1} = a_n + 2n$.

Under­stand­ing the prob­lem#

The rule tells you how to get each term from the one before: add $2n$. So $a_2 = a_1 + 2$, $a_3 = a_2 + 4$, $a_4 = a_3 + 6$, and so on. You need a for­mula for $a_n$ directly in terms of $n$, and then its value at $n = 10$.

The idea#

Write $a_n$ as the first term plus all the increases made on the way from $a_1$ to $a_n$. Those increases are $2 \times 1, 2 \times 2, \ldots, 2(n - 1)$, which is $2$ times the sum of the first $n - 1$ nat­ural num­bers.

Step-by-step solu­tion#

Step 1. List the first few terms to see the pat­tern.

$$a_1 = 1,\quad a_2 = 1 + 2 = 3,\quad a_3 = 3 + 4 = 7,\quad a_4 = 7 + 6 = 13.$$

Step 2. In gen­eral,

$$a_n = a_1 + (2 \times 1) + (2 \times 2) + \cdots + 2(n - 1) = 1 + 2\bigl(1 + 2 + \cdots + (n - 1)\bigr).$$

Step 3. Use $\displaystyle 1 + 2 + \cdots + m = \frac{m(m + 1)}{2}$ with $m = n - 1$.

$$\displaystyle a_n = 1 + 2 \cdot \frac{(n - 1)n}{2} = 1 + n^2 - n = n^2 - n + 1.$$

Step 4. Put $n = 10$.

$$a_{10} = 100 - 10 + 1 = 91.$$

Check­ing the answer#

The for­mula gives $a_1 = 1$, $a_2 = 3$, $a_3 = 7$, $a_4 = 13$, match­ing Step 1.

Answer#

$a_n = n^2 - n + 1$, and $a_{10} = 91$.

Ques­tion 2: An AP from two con­di­tions#

The prob­lem#

The $12$th term of an AP is $-13$ and the sum of its first four terms is $24$. Find the sum of its first $10$ terms.

Under­stand­ing the prob­lem#

You know two facts about an AP with first term $a$ and com­mon dif­fer­ence $d$. Two facts give two equa­tions, enough to find $a$ and $d$. Then find $S_{10}$.

The idea#

Use $a_n = a + (n - 1)d$ and $\displaystyle S_n = \frac{n}{2}\bigl[2a + (n - 1)d\bigr]$.

Step-by-step solu­tion#

Step 1. The $12$th term.

$$a + 11d = -13. \qquad (1)$$

Step 2. The sum of the first four terms.

$$\displaystyle S_4 = \frac{4}{2}(2a + 3d) = 4a + 6d = 24 \;\Rightarrow\; 2a + 3d = 12. \qquad (2)$$

Step 3. Dou­ble (1) and sub­tract (2).

$$\begin{aligned} 2a + 22d &= -26 \\ (2a + 22d) - (2a + 3d) &= -26 - 12 \\ 19d &= -38 \\ d &= -2. \end{aligned}$$

Step 4. Sub­sti­tute into (1).

$$a = -13 - 11(-2) = -13 + 22 = 9.$$

Step 5. Find $S_{10}$.

$$\displaystyle S_{10} = \frac{10}{2}\bigl[2(9) + 9(-2)\bigr] = 5(18 - 18) = 0.$$

Check­ing the answer#

The AP is $9, 7, 5, 3, 1, -1, -3, -5, -7, -9, \ldots$ The first four add to $24$, the $12$th term is $9 + 11(-2) = -13$, and the first ten pair off ($9$ with $-9$, $7$ with $-7$, ...) to give $0$.

Answer#

$S_{10} = 0$.

Ques­tion 3: Three-digit num­bers leav­ing remain­der 2 on divi­sion by 5#

The prob­lem#

How many three-digit num­bers leave remain­der $2$ when divided by $5$? Find their sum.

Under­stand­ing the prob­lem#

Such num­bers have the form $5m + 2$ and lie from $100$ to $999$. You need how many there are and their total.

The idea#

These num­bers form an AP with com­mon dif­fer­ence $5$. Find the first and last terms, then use $a_n = a + (n - 1)d$ to count and $\displaystyle S_n = \frac{n}{2}(a + l)$ to add.

Step-by-step solu­tion#

Step 1. The first one: $100 = 5 \times 20$, so $100$ leaves remain­der $0$; add $2$ to get $102$.

Step 2. The last one: $995 = 5 \times 199$, so $997$ leaves remain­der $2$ (and $1002$ is too big).

Step 3. Count the terms of $102, 107, \ldots, 997$.

$$\begin{aligned} 997 &= 102 + (n - 1)5 \\ 895 &= 5(n - 1) \\ n - 1 &= 179 \;\Rightarrow\; n = 180. \end{aligned}$$

Step 4. Sum.

$$\displaystyle S = \frac{180}{2}(102 + 997) = 90 \times 1099 = 98\,910.$$

Check­ing the answer#

Among any $5$ con­sec­u­tive num­bers exactly one leaves remain­der $2$. There are $900$ three-digit num­bers, and $900 / 5 = 180$. Cor­rect.

Answer#

$180$ num­bers, with sum $98\,910$.

Ques­tion 4: Ratio of 18th terms from a ratio of sums#

The prob­lem#

The sums of $n$ terms of two APs are in the ratio $(5n + 4) : (9n + 6)$. Find the ratio of their $18$th terms.

Under­stand­ing the prob­lem#

Let the APs have first terms $a, a'$ and com­mon dif­fer­ences $d, d'$. You know $\displaystyle \frac{S_n}{S'_n}$ for every $n$. You want $\displaystyle \frac{a + 17d}{a' + 17d'}$.

The idea#

Write the ratio of sums:

$$\displaystyle \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'}.$$

If you choose $n$ so that $\displaystyle \frac{n - 1}{2} = 17$, the right side becomes exactly the ratio of the $18$th terms.

Step-by-step solu­tion#

Step 1. Solve $\displaystyle \frac{n - 1}{2} = 17$: $n = 35$.

Step 2. So

$$\displaystyle \frac{a + 17d}{a' + 17d'} = \frac{S_{35}}{S'_{35}}.$$

Step 3. Put $n = 35$ in the given ratio.

$$\displaystyle \frac{5(35) + 4}{9(35) + 6} = \frac{175 + 4}{315 + 6} = \frac{179}{321}.$$

Check­ing the answer#

$179$ is prime and does not divide $321 = 3 \times 107$, so the frac­tion is in low­est terms. The gen­eral rule is: ratio of $m$th terms $=$ ratio of $S_{2m - 1}$; with $m = 18$, $2m - 1 = 35$. Con­sis­tent.

Answer#

The $18$th terms are in the ratio $179 : 321$.

Ques­tion 5: Three terms of a GP#

The prob­lem#

The $5$th, $8$th and $11$th terms of a GP are $p$, $q$, $s$. Show $q^2 = ps$.

Under­stand­ing the prob­lem#

Let the GP have first term $a$ and com­mon ratio $r$. You must prove a rela­tion between three of its terms.

The idea#

Write each term with $a_n = ar^{n - 1}$ and com­pare $q^2$ with $ps$.

Step-by-step solu­tion#

Step 1. Write the three terms.

$$p = ar^4, \qquad q = ar^7, \qquad s = ar^{10}.$$

Step 2. Square $q$.

$$q^2 = a^2 r^{14}.$$

Step 3. Mul­ti­ply $p$ and $s$.

$$ps = ar^4 \cdot ar^{10} = a^2 r^{14}.$$

Step 4. The two are equal, so $q^2 = ps$. $\blacksquare$

Check­ing the answer#

The posi­tions $5, 8, 11$ are equally spaced, so $p, q, s$ form a GP them­selves (with ratio $r^3$), and in any GP the mid­dle term squared equals the prod­uct of its neigh­bours.

Answer#

$q^2 = a^2r^{14} = ps$.

Ques­tion 6: The series $7 + 77 + 777 + \cdots$#

The prob­lem#

Find the sum of $n$ terms of $7 + 77 + 777 + \cdots$

Under­stand­ing the prob­lem#

This is nei­ther an AP nor a GP. The $k$th term is a string of $k$ sev­ens.

The idea#

Write a string of $k$ nines as $10^k - 1$. A string of $k$ sev­ens is $\displaystyle \frac{7}{9}$ of that. This turns the series into a GP minus a con­stant.

Step-by-step solu­tion#

Step 1. Rewrite the $k$th term.

$$\displaystyle \underbrace{77\ldots7}_{k} = \frac{7}{9} \times \underbrace{99\ldots9}_{k} = \frac{7}{9}(10^k - 1).$$

Step 2. Add the $n$ terms.

$$\displaystyle S_n = \frac{7}{9}\bigl[(10 - 1) + (10^2 - 1) + \cdots + (10^n - 1)\bigr] = \frac{7}{9}\bigl[(10 + 10^2 + \cdots + 10^n) - n\bigr].$$

Step 3. The bracket con­tains a GP with first term $10$, ratio $10$, $n$ terms.

$$\displaystyle 10 + 10^2 + \cdots + 10^n = \frac{10(10^n - 1)}{10 - 1} = \frac{10^{n+1} - 10}{9}.$$

Step 4. Sub­sti­tute and sim­plify.

$$\displaystyle S_n = \frac{7}{9}\left[\frac{10^{n+1} - 10}{9} - n\right] = \frac{7}{9} \cdot \frac{10^{n+1} - 10 - 9n}{9} = \frac{7}{81}\left(10^{n+1} - 10 - 9n\right).$$

Check­ing the answer#

$n = 2$: $\displaystyle \frac{7}{81}(1000 - 10 - 18) = \frac{7 \times 972}{81} = 7 \times 12 = 84 = 7 + 77$. Cor­rect.

Answer#

$\displaystyle S_n = \frac{7}{81}\left(10^{n+1} - 10 - 9n\right)$.

Ques­tion 7: An infi­nite GP from two sums#

The prob­lem#

The sum of an infi­nite GP is $15$ and the sum of the squares of its terms is $45$. Find the GP.

Under­stand­ing the prob­lem#

Let the GP be $a, ar, ar^2, \ldots$ with $\lvert r \rvert < 1$ (so the infi­nite sum exists). The squares $a^2, a^2r^2, a^2r^4, \ldots$ also form a GP, with ratio $r^2$.

The idea#

Use $\displaystyle S_\infty = \frac{a}{1 - r}$ for both GPs, giv­ing two equa­tions. Divid­ing them removes much of the alge­bra because $1 - r^2 = (1 - r)(1 + r)$.

Step-by-step solu­tion#

Step 1. Write the two equa­tions.

$$\displaystyle \frac{a}{1 - r} = 15 \quad (1), \qquad \frac{a^2}{1 - r^2} = 45 \quad (2).$$

Step 2. Divide (2) by (1).

$$\displaystyle \frac{a^2}{(1 - r)(1 + r)} \cdot \frac{1 - r}{a} = \frac{45}{15} \;\Rightarrow\; \frac{a}{1 + r} = 3. \qquad (3)$$

Step 3. From (1), $a = 15(1 - r)$; from (3), $a = 3(1 + r)$. Equate.

$$\displaystyle \begin{aligned} 15 - 15r &= 3 + 3r \\ 12 &= 18r \\ r &= \frac{2}{3}. \end{aligned}$$

Step 4. Find $a$.

$$\displaystyle a = 3\left(1 + \frac{2}{3}\right) = 5.$$

Step 5. Write the GP.

$$\displaystyle 5,\; \frac{10}{3},\; \frac{20}{9},\; \frac{40}{27},\; \ldots$$

Check­ing the answer#

$\displaystyle \frac{5}{1 - 2/3} = 15$ and $\displaystyle \frac{25}{1 - 4/9} = \frac{25}{5/9} = 45$. Both con­di­tions hold, and $\lvert r \rvert < 1$.

Answer#

The GP is $\displaystyle 5, \frac{10}{3}, \frac{20}{9}, \ldots$ (first term $5$, ratio $\displaystyle \frac{2}{3}$).

Ques­tion 8: The series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots$#

The prob­lem#

Find the sum of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots$ to $n$ terms.

Under­stand­ing the prob­lem#

The $k$th term is $k(k + 1)$. You need $\displaystyle \sum_{k=1}^{n} k(k + 1)$.

The idea#

Expand the gen­eral term as $k^2 + k$ and use the stan­dard sums

$$\displaystyle \sum k = \frac{n(n + 1)}{2}, \qquad \sum k^2 = \frac{n(n + 1)(2n + 1)}{6}.$$

Step-by-step solu­tion#

Step 1. Split the sum.

$$\displaystyle S_n = \sum_{k=1}^{n}(k^2 + k) = \sum k^2 + \sum k.$$

Step 2. Sub­sti­tute the stan­dard results.

$$\displaystyle S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}.$$

Step 3. Take out the com­mon fac­tor $\displaystyle \frac{n(n + 1)}{6}$.

$$\displaystyle S_n = \frac{n(n + 1)}{6}\bigl[(2n + 1) + 3\bigr] = \frac{n(n + 1)(2n + 4)}{6} = \frac{n(n + 1)(n + 2)}{3}.$$

Check­ing the answer#

$n = 3$: $2 + 6 + 12 = 20$, and $\displaystyle \frac{3 \times 4 \times 5}{3} = 20$. Cor­rect.

Answer#

$\displaystyle S_n = \frac{n(n + 1)(n + 2)}{3}$.

Ques­tion 9: Sum of squares of odd num­bers#

The prob­lem#

Find the sum of $n$ terms of $1^2 + 3^2 + 5^2 + \cdots$

Under­stand­ing the prob­lem#

The $k$th odd num­ber is $2k - 1$, so the $k$th term is $(2k - 1)^2$.

The idea#

Expand the square and use $\displaystyle \sum k^2$, $\displaystyle \sum k$ and $\displaystyle \sum 1 = n$.

Step-by-step solu­tion#

Step 1. Expand the gen­eral term.

$$(2k - 1)^2 = 4k^2 - 4k + 1.$$

Step 2. Sum each piece.

$$\displaystyle S_n = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} - 4 \cdot \frac{n(n + 1)}{2} + n.$$

Step 3. Sim­plify, tak­ing $n$ out­side.

$$\displaystyle \begin{aligned} S_n &= n\left[\frac{2(n + 1)(2n + 1)}{3} - 2(n + 1) + 1\right] \\ &= \frac{n}{3}\bigl[2(2n^2 + 3n + 1) - 6(n + 1) + 3\bigr] \\ &= \frac{n}{3}\bigl[4n^2 + 6n + 2 - 6n - 6 + 3\bigr] \\ &= \frac{n}{3}(4n^2 - 1) = \frac{n(2n - 1)(2n + 1)}{3}. \end{aligned}$$

Check­ing the answer#

$n = 3$: $1 + 9 + 25 = 35$, and $\displaystyle \frac{3 \times 5 \times 7}{3} = 35$. Cor­rect.

Answer#

$\displaystyle S_n = \frac{n(2n - 1)(2n + 1)}{3}$.

Ques­tion 10: Link­ing an AP, a GP and an HP-type con­di­tion#

The prob­lem#

If $a, b, c$ are in AP and $b, c, d$ in GP, and $\displaystyle \tfrac{1}{c}, \tfrac{1}{d}, \tfrac{1}{e}$ in AP, prove $a, c, e$ are in GP.

Under­stand­ing the prob­lem#

You must show $c^2 = ae$, which is what "$a, c, e$ in GP" means. Assume all the num­bers are non-zero (so the rec­i­p­ro­cals exist).

The idea#

Turn each con­di­tion into an equa­tion, then elim­i­nate $b$ and $d$ so that only $a, c, e$ remain.

Step-by-step solu­tion#

Step 1. Write the three con­di­tions.

$$\displaystyle \begin{aligned} &a, b, c \text{ in AP:} && 2b = a + c \;\Rightarrow\; b = \frac{a + c}{2} \\ &b, c, d \text{ in GP:} && c^2 = bd \\ &\tfrac1c, \tfrac1d, \tfrac1e \text{ in AP:} && \frac{2}{d} = \frac{1}{c} + \frac{1}{e} = \frac{c + e}{ce} \;\Rightarrow\; d = \frac{2ce}{c + e} \end{aligned}$$

Step 2. Sub­sti­tute $b$ and $d$ into $c^2 = bd$.

$$\displaystyle c^2 = \frac{a + c}{2} \cdot \frac{2ce}{c + e} = \frac{(a + c)\,ce}{c + e}.$$

Step 3. Divide both sides by $c$ and mul­ti­ply by $c + e$.

$$c(c + e) = (a + c)e.$$

Step 4. Expand and can­cel $ce$.

$$c^2 + ce = ae + ce \;\Rightarrow\; c^2 = ae.$$

So $a, c, e$ are in GP. $\blacksquare$

Check­ing the answer#

Take $a = 1, b = 2, c = 3$ (AP). Then $\displaystyle d = \frac{c^2}{b} = \frac92$. From $\displaystyle \frac{2}{d} = \frac13 + \frac1e$: $\displaystyle \frac49 - \frac13 = \frac19$, so $e = 9$. Indeed $1, 3, 9$ is a GP.

Answer#

Elim­i­nat­ing $b$ and $d$ gives $c^2 = ae$, so $a, c, e$ are in GP.

Ques­tion 11: Total cost of a trac­tor bought in instal­ments#

The prob­lem#

A farmer buys a trac­tor for ₹6,00,000 and pays ₹1,20,000 down, the rest in yearly instal­ments of ₹60,000 plus $10\%$ inter­est on the unpaid amount. What is the total cost?

Under­stand­ing the prob­lem#

After the down pay­ment, ₹4,80,000 is unpaid. Each year the farmer pays ₹60,000 of the loan, plus inter­est at $10\%$ on what­ever was unpaid dur­ing that year. The total cost is the price plus all the inter­est.

The idea#

The unpaid amounts form an AP: ₹4,80,000, ₹4,20,000, ..., ₹60,000. The total inter­est is $10\%$ of the sum of this AP.

Step-by-step solu­tion#

Step 1. Unpaid bal­ance after the down pay­ment.

$$6\,00\,000 - 1\,20\,000 = 4\,80\,000.$$

Step 2. Num­ber of instal­ments.

$$\displaystyle \frac{4\,80\,000}{60\,000} = 8.$$

Step 3. Inter­est in each year is charged on the bal­ance at the start of that year: $4\,80\,000$, $4\,20\,000$, ..., $60\,000$. This is an AP with $8$ terms, first term $4\,80\,000$, last term $60\,000$.

$$\displaystyle \text{Sum of balances} = \frac{8}{2}(4\,80\,000 + 60\,000) = 4 \times 5\,40\,000 = 21\,60\,000.$$

Step 4. Total inter­est at $10\%$.

$$0.1 \times 21\,60\,000 = 2\,16\,000.$$

Step 5. Total cost.

$$6\,00\,000 + 2\,16\,000 = 8\,16\,000.$$

Check­ing the answer#

Year 1 inter­est is ₹48,000 and Year 8 inter­est is ₹6,000. The aver­age is ₹27,000, and $8 \times 27\,000 = 2\,16\,000$. Cor­rect.

Answer#

The total cost of the trac­tor is ₹8,16,000.

Com­mon mis­take to avoid#

Do not charge inter­est on the full ₹6,00,000 or on the down pay­ment. Inter­est is only on the unpaid bal­ance, which falls by ₹60,000 each year.

Ques­tion 12: $\displaystyle x + \frac{1}{x} \ge 2$#

The prob­lem#

Show that for pos­i­tive $x$, $\displaystyle x + \tfrac{1}{x} \ge 2$.

Under­stand­ing the prob­lem#

$x > 0$, so $\displaystyle \frac1x > 0$ too. You must prove an inequal­ity that holds for every such $x$.

The idea#

Use the AM–GM inequal­ity for two pos­i­tive num­bers: $\displaystyle \frac{p + q}{2} \ge \sqrt{pq}$. Choose $p = x$ and $\displaystyle q = \frac1x$, whose prod­uct is $1$.

Step-by-step solu­tion#

Step 1. Apply AM $\ge$ GM to $x$ and $\displaystyle \frac1x$.

$$\displaystyle \frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1.$$

Step 2. Mul­ti­ply both sides by $2$.

$$\displaystyle x + \frac{1}{x} \ge 2. \qquad \blacksquare$$

Step 3. Equal­ity in AM–GM holds only when the two num­bers are equal: $\displaystyle x = \frac1x$, that is $x = 1$ (as $x > 0$).

Check­ing the answer#

$x = 2$: $2 + 0.5 = 2.5 \ge 2$. $x = 1$: $1 + 1 = 2$, the equal­ity case. A sec­ond proof: $\displaystyle x + \frac1x - 2 = \frac{(x - 1)^2}{x} \ge 0$.

Answer#

By AM $\ge$ GM, $\displaystyle x + \frac{1}{x} \ge 2$ for all $x > 0$, with equal­ity only at $x = 1$.