Sequences and Arithmetic Progressions, Worked Solutions

Ques­tion 1: First four terms of a sequence#

The prob­lem#

Write the first four terms: (a) $\displaystyle a_n = \frac{2n - 1}{n + 1}$; (b) $a_n = (-1)^n 3^n$; (c) $a_1 = 4$, $\displaystyle a_{n+1} = \frac{a_n}{2} + 1$.

Under­stand­ing the prob­lem#

In (a) and (b) the gen­eral term is given by a for­mula in $n$, so you sub­sti­tute $n = 1, 2, 3, 4$. In (c) the sequence is recur­sive: each term is built from the one before it, start­ing from $a_1 = 4$.

The idea#

Sub­sti­tute directly for (a) and (b); apply the rule repeat­edly for (c).

Step-by-step solu­tion#

Part (a)

Step 1. Put $n = 1, 2, 3, 4$ into $\displaystyle \frac{2n - 1}{n + 1}$.

$$\displaystyle a_1 = \frac{1}{2}, \quad a_2 = \frac{3}{3} = 1, \quad a_3 = \frac{5}{4}, \quad a_4 = \frac{7}{5}$$

Part (b)

Step 1. The fac­tor $(-1)^n$ is $-1$ for odd $n$ and $+1$ for even $n$, so the signs alter­nate.

$$a_1 = -3, \quad a_2 = 9, \quad a_3 = -27, \quad a_4 = 81$$

Part (c)

Step 1. Start with $a_1 = 4$ and apply $\displaystyle a_{n+1} = \frac{a_n}{2} + 1$ three times.

$$\displaystyle \begin{aligned} a_2 &= \frac{4}{2} + 1 = 3 \\ a_3 &= \frac{3}{2} + 1 = \frac{5}{2} \\ a_4 &= \frac{5/2}{2} + 1 = \frac{5}{4} + 1 = \frac{9}{4} \end{aligned}$$

Check­ing the answer#

In (c) the terms $4, 3, 2.5, 2.25$ move steadily towards $2$, and $2$ is the num­ber that the rule leaves unchanged ($\displaystyle \tfrac{2}{2} + 1 = 2$), which makes sense.

Answer#

(a) $\displaystyle \tfrac12, 1, \tfrac54, \tfrac75$; (b) $-3, 9, -27, 81$; (c) $\displaystyle 4, 3, \tfrac52, \tfrac94$.

Ques­tion 2: The 25th term and the sum of 25 terms#

The prob­lem#

Find the $25$th term and the sum of $25$ terms of $3, 8, 13, \ldots$

Under­stand­ing the prob­lem#

The terms increase by $5$ each time, so this is an AP with first term $a = 3$ and com­mon dif­fer­ence $d = 5$. You need $a_{25}$ and $S_{25}$.

The idea#

Use $a_n = a + (n - 1)d$, then $\displaystyle S_n = \tfrac{n}{2}(a + l)$ with $l = a_{25}$.

Step-by-step solu­tion#

Step 1. Find the 25th term.

$$a_{25} = 3 + 24 \times 5 = 3 + 120 = 123$$

Step 2. Use the first and last terms to find the sum.

$$\displaystyle S_{25} = \frac{25}{2}(3 + 123) = \frac{25}{2} \times 126 = 25 \times 63 = 1575$$

Check­ing the answer#

Using the other for­mula: $\displaystyle S_{25} = \tfrac{25}{2}[2(3) + 24(5)] = \tfrac{25}{2}(126) = 1575$. ✓

Answer#

$a_{25} = 123$; $S_{25} = 1575$.

Ques­tion 3: Which term is a given num­ber?#

The prob­lem#

Which term of $\sqrt{3}, 3\sqrt{3}, 5\sqrt{3}, \ldots$ is $39\sqrt{3}$?

Under­stand­ing the prob­lem#

You must find the posi­tion $n$ at which the term equals $39\sqrt3$. Here $a = \sqrt3$ and $d = 3\sqrt3 - \sqrt3 = 2\sqrt3$.

The idea#

Set $a_n = a + (n - 1)d$ equal to $39\sqrt3$ and solve for $n$; it must come out a pos­i­tive whole num­ber.

Step-by-step solu­tion#

Step 1. Write the equa­tion.

$$\sqrt3 + (n - 1)\,2\sqrt3 = 39\sqrt3$$

Step 2. Divide every term by $\sqrt3$.

$$1 + 2(n - 1) = 39 \;\Rightarrow\; 2n - 1 = 39$$

Step 3. Solve.

$$2n = 40 \;\Rightarrow\; n = 20$$

Check­ing the answer#

$a_{20} = \sqrt3 + 19 \times 2\sqrt3 = 39\sqrt3$. ✓

Answer#

$39\sqrt3$ is the $20$th term.

Ques­tion 4: Find­ing a and d from two facts#

The prob­lem#

The $5$th term of an AP is $19$ and the sum of the first $10$ terms is $185$. Find $a$ and $d$.

Under­stand­ing the prob­lem#

There are two unknowns, $a$ and $d$, and two pieces of infor­ma­tion. Each fact gives one equa­tion.

The idea#

Trans­late "5th term is 19" with $a_n = a + (n - 1)d$, and "sum of 10 terms is 185" with $\displaystyle S_n = \tfrac{n}{2}[2a + (n - 1)d]$. Then solve the pair.

Step-by-step solu­tion#

Step 1. From the 5th term.

$$a + 4d = 19 \qquad (1)$$

Step 2. From the sum of 10 terms.

$$\displaystyle \frac{10}{2}[2a + 9d] = 185 \;\Rightarrow\; 2a + 9d = 37 \qquad (2)$$

Step 3. Dou­ble equa­tion (1): $2a + 8d = 38$. Sub­tract this from (2).

$$(2a + 9d) - (2a + 8d) = 37 - 38 \;\Rightarrow\; d = -1$$

Step 4. Sub­sti­tute into (1).

$$a - 4 = 19 \;\Rightarrow\; a = 23$$

Check­ing the answer#

The AP is $23, 22, 21, \ldots$; its 5th term is $19$ ✓ and $S_{10} = 5(23 + 14) = 185$. ✓

Answer#

$a = 23$, $d = -1$.

Ques­tion 5: Sum of odd num­bers between 100 and 300#

The prob­lem#

Find the sum of all odd num­bers between $100$ and $300$.

Under­stand­ing the prob­lem#

The odd num­bers are $101, 103, \ldots, 299$. They form an AP with $a = 101$, $d = 2$ and last term $l = 299$. You first need to know how many terms there are.

The idea#

Find $n$ from $l = a + (n - 1)d$, then use $\displaystyle S_n = \tfrac{n}{2}(a + l)$.

Step-by-step solu­tion#

Step 1. Count the terms.

$$101 + (n - 1)2 = 299 \;\Rightarrow\; 2(n - 1) = 198 \;\Rightarrow\; n = 100$$

Step 2. Add them.

$$\displaystyle S_{100} = \frac{100}{2}(101 + 299) = 50 \times 400 = 20000$$

Check­ing the answer#

The aver­age of 100 evenly spaced num­bers from 101 to 299 is the mid­dle value $200$, and $100 \times 200 = 20000$. ✓

Answer#

$20000$

Ques­tion 6: nth term from the sum for­mula#

The prob­lem#

The sum of $n$ terms of an AP is $3n^2 + 2n$. Find its $n$th term.

Under­stand­ing the prob­lem#

You are told $S_n$, the sum of the first $n$ terms, for every $n$. The $n$th term is what gets added to go from $S_{n-1}$ to $S_n$.

The idea#

$a_n = S_n - S_{n-1}$ for $n \ge 2$, and $a_1 = S_1$.

Step-by-step solu­tion#

Step 1. Write $S_{n-1}$ by replac­ing $n$ with $n - 1$.

$$S_{n-1} = 3(n - 1)^2 + 2(n - 1) = 3n^2 - 6n + 3 + 2n - 2 = 3n^2 - 4n + 1$$

Step 2. Sub­tract.

$$a_n = (3n^2 + 2n) - (3n^2 - 4n + 1) = 6n - 1$$

Step 3. Check it also works for $n = 1$: $S_1 = 3 + 2 = 5$ and $6(1) - 1 = 5$. So $a_n = 6n - 1$ for all $n$.

Check­ing the answer#

The terms are $5, 11, 17, \ldots$; $S_2 = 5 + 11 = 16$ and $3(4) + 2(2) = 16$. ✓

Answer#

$a_n = 6n - 1$

Ques­tion 7: The (p + q)th term is zero#

The prob­lem#

If the $p$th term is $q$ and the $q$th term is $p$ ($p \ne q$), show that the $(p + q)$th term is $0$.

Under­stand­ing the prob­lem#

This is a proof. You know two terms of an AP in let­ters, and must show $a_{p+q} = 0$. The con­di­tion $p \ne q$ lets you divide by $p - q$.

The idea#

Write both facts using $a_n = a + (n - 1)d$, sub­tract to find $d$, then find $a$ and sub­sti­tute into $a_{p+q}$.

Step-by-step solu­tion#

Step 1. Write the two facts as equa­tions.

$$a + (p - 1)d = q \qquad (1)$$
$$a + (q - 1)d = p \qquad (2)$$

Step 2. Sub­tract (2) from (1); $a$ can­cels.

$$(p - q)d = q - p = -(p - q)$$

Step 3. Since $p \ne q$, divide by $p - q$.

$$d = -1$$

Step 4. Put $d = -1$ into (1).

$$a - (p - 1) = q \;\Rightarrow\; a = p + q - 1$$

Step 5. Now find the $(p + q)$th term.

$$a_{p+q} = a + (p + q - 1)d = (p + q - 1) - (p + q - 1) = 0$$

This is what was to be shown.

Check­ing the answer#

Try $p = 2$, $q = 5$: $d = -1$, $a = 6$, the AP is $6, 5, 4, 3, 2, 1, 0, \ldots$ The 2nd term is $5$ ✓, the 5th term is $2$ ✓ and the 7th term is $0$. ✓

Answer#

$d = -1$, $a = p + q - 1$, so $a_{p+q} = 0$.

Ques­tion 8: Insert­ing arith­metic means#

The prob­lem#

Insert $6$ arith­metic means between $-5$ and $30$.

Under­stand­ing the prob­lem#

You need 6 num­bers so that $-5$, the six num­bers, and $30$ together form an AP. That is $8$ terms in all, with first term $-5$ and 8th term $30$.

The idea#

With $n$ means between $a$ and $b$, the com­mon dif­fer­ence is $\displaystyle d = \frac{b - a}{n + 1}$.

Step-by-step solu­tion#

Step 1. Find $d$ with $n = 6$.

$$\displaystyle d = \frac{30 - (-5)}{6 + 1} = \frac{35}{7} = 5$$

Step 2. Add $5$ repeat­edly start­ing from $-5$.

$$-5,\; 0,\; 5,\; 10,\; 15,\; 20,\; 25,\; 30$$

Step 3. The six means are the num­bers strictly between the ends.

Check­ing the answer#

$25 + 5 = 30$, so the sequence lands exactly on the last term. ✓

Answer#

$0, 5, 10, 15, 20, 25$

Ques­tion 9: Monthly sav­ings over two years#

The prob­lem#

A man saves ₹800 in the first month and increases his sav­ing by ₹150 each month. How much has he saved in $2$ years?

Under­stand­ing the prob­lem#

His monthly sav­ings are ₹800, ₹950, ₹1100, …, an AP with $a = 800$, $d = 150$. Two years is $n = 24$ months. You need the total, $S_{24}$.

The idea#

Use $\displaystyle S_n = \tfrac{n}{2}[2a + (n - 1)d]$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $n = 24$, $a = 800$, $d = 150$.

$$\displaystyle S_{24} = \frac{24}{2}[2(800) + 23(150)]$$

Step 2. Work out the bracket.

$$2(800) + 23(150) = 1600 + 3450 = 5050$$

Step 3. Mul­ti­ply.

$$S_{24} = 12 \times 5050 = 60600$$

Check­ing the answer#

In the 24th month he saves $800 + 23 \times 150 = 4250$ rupees. Aver­age monthly sav­ing $\displaystyle = \tfrac{800 + 4250}{2} = 2525$, and $24 \times 2525 = 60600$. ✓

Answer#

He has saved ₹60,600.

Com­mon mis­take to avoid#

Use $n = 24$ (months), not $n = 2$ (years).

Ques­tion 10: Angles of a hexa­gon in AP#

The prob­lem#

The angles of a hexa­gon are in AP with the small­est $95^\circ$. Find the com­mon dif­fer­ence.

Under­stand­ing the prob­lem#

A hexa­gon has 6 angles, and their sum is $(6 - 2) \times 180^\circ = 720^\circ$. The angles form an AP with $a = 95^\circ$, $n = 6$.

The idea#

Set $S_6 = 720^\circ$ and solve for $d$.

Step-by-step solu­tion#

Step 1. Write the sum of the 6 angles.

$$\displaystyle S_6 = \frac{6}{2}[2(95) + 5d] = 3(190 + 5d) = 570 + 15d$$

Step 2. Set this equal to the angle sum.

$$570 + 15d = 720$$

Step 3. Solve.

$$15d = 150 \;\Rightarrow\; d = 10^\circ$$

Check­ing the answer#

The angles are $95^\circ, 105^\circ, 115^\circ, 125^\circ, 135^\circ, 145^\circ$; their sum is $720^\circ$ ✓, and each is less than $180^\circ$, as a hexa­gon's inte­rior angle should be in a con­vex hexa­gon.

Answer#

$d = 10^\circ$