Properties of an A.P. and selecting terms

Arith­metic pro­gres­sions have a hand­ful of prop­er­ties that save a great deal of work. Some tell you when three num­bers are in A.P., some tell you what hap­pens when every term is changed in the same way, and one gives a clever way of nam­ing unknown terms so that the alge­bra stays short. This les­son col­lects six prop­er­ties and then shows how to choose three, four or five unknown terms of an A.P.

Prop­erty 1: three num­bers in A.P.#

$$a, b, c \text{ are in A.P.} \iff b - a = c - b \iff 2b = a + c$$

The mid­dle term is the aver­age of its neigh­bours. This is the quick­est test for three num­bers, and it turns a con­di­tion "in A.P." into an equa­tion.

Exam­ple 1#

Find $k$ if $2k - 1$, $k + 4$ and $3k - 2$ are in A.P.

$$\displaystyle \begin{aligned} 2(k + 4) &= (2k - 1) + (3k - 2) \\ 2k + 8 &= 5k - 3 \\ 11 &= 3k \\ k &= \frac{11}{3} \end{aligned}$$

Prop­erty 2: terms equidis­tant from the two ends#

In a finite A.P., the sum of two terms equidis­tant from the begin­ning and the end is always the same, and it equals the sum of the first and last terms.

Why. The $k$th term from the begin­ning is $a + (k - 1)d$. The $k$th term from the end is $l - (k - 1)d$. Their sum is $a + l$, what­ever $k$ is.

For exam­ple, in $2, 5, 8, 11, 14, 17$: $2 + 17 = 19$, $5 + 14 = 19$, $8 + 11 = 19$. This is exactly why the for­mula $\displaystyle S_n = \frac{n}{2}(a + l)$ works.

Exam­ple 2#

An A.P. has 10 terms and $t_3 + t_8 = 25$. Find the sum of all 10 terms.

The 3rd term from the begin­ning and the 3rd term from the end (the 8th term) are equidis­tant from the ends, so $t_1 + t_{10} = t_3 + t_8 = 25$. Hence

$$\displaystyle S_{10} = \frac{10}{2}(t_1 + t_{10}) = 5 \times 25 = 125$$

Prop­erty 3: adding or sub­tract­ing a con­stant#

If the same num­ber is added to (or sub­tracted from) every term of an A.P., the result is again an A.P. with the same com­mon dif­fer­ence. For exam­ple, adding 10 to each term of $1, 4, 7, 10$ gives $11, 14, 17, 20$, still with $d = 3$. The dif­fer­ences are not affected because the extra 10 can­cels in each sub­trac­tion.

Prop­erty 4: mul­ti­ply­ing or divid­ing by a con­stant#

If every term of an A.P. is mul­ti­plied by a non-zero num­ber $k$, the result is an A.P. with com­mon dif­fer­ence $kd$. If every term is divided by $k$, the com­mon dif­fer­ence becomes $\displaystyle \frac{d}{k}$. For exam­ple, dou­bling $1, 4, 7, 10$ gives $2, 8, 14, 20$, with $d = 6$.

Prop­erty 5: each term is the aver­age of terms equally spaced around it#

$$\displaystyle t_n = \frac{t_{n-k} + t_{n+k}}{2} \quad \text{for } k < n$$

This gen­er­alises Prop­erty 1: you can step out by any num­ber of places on each side. For exam­ple, in $3, 7, 11, 15, 19, 23, 27$, the 4th term 15 is the aver­age of 11 and 19, of 7 and 23, and of 3 and 27.

Prop­erty 6: terms taken at reg­u­lar inter­vals#

If terms of an A.P. are taken at reg­u­lar inter­vals, for exam­ple every sec­ond term or every third term, they form another A.P. Tak­ing every $k$th term mul­ti­plies the com­mon dif­fer­ence by $k$. From $1, 4, 7, 10, 13, 16, 19$, every sec­ond term gives $1, 7, 13, 19$, an A.P. with $d = 6$.

Select­ing terms in an A.P.#

When a ques­tion says "three num­bers in A.P. have a given sum", do not call them $a$, $a + d$, $a + 2d$. Choose them so that the $d$'s can­cel when you add:

In each case the sum of the terms is just the num­ber of terms times $a$, so the sum gives $a$ at once.

Exam­ple 3#

Three num­bers in A.P. have sum 27 and prod­uct 693. Find them.

Let the num­bers be $a - d$, $a$, $a + d$. Their sum is $3a = 27$, so $a = 9$. Their prod­uct is

$$(9 - d) \times 9 \times (9 + d) = 693 \implies 81 - d^2 = 77 \implies d^2 = 4 \implies d = \pm 2$$

With $d = 2$ the num­bers are $7, 9, 11$; with $d = -2$ they are $11, 9, 7$. Either way the num­bers are 7, 9 and 11. Check: $7 \times 9 \times 11 = 693$.

Exam­ple 4#

Four num­bers in A.P. have sum 20 and the sum of their squares is 120. Find them.

Let the num­bers be $a - 3d$, $a - d$, $a + d$, $a + 3d$. Their sum is $4a = 20$, so $a = 5$. The sum of squares:

$$(5 - 3d)^2 + (5 - d)^2 + (5 + d)^2 + (5 + 3d)^2 = 120$$

When expanded, the terms in $d$ can­cel in pairs, leav­ing $4 \times 25 + 20d^2 = 120$. So $20d^2 = 20$, $d = \pm 1$, and the num­bers are $2, 4, 6, 8$. Check: $4 + 16 + 36 + 64 = 120$.

Exam­ple 5#

The angles of a tri­an­gle are in A.P. and the largest is twice the small­est. Find the angles.

Let the angles be $a - d$, $a$, $a + d$. Their sum is $3a = 180^\circ$, so $a = 60^\circ$. Also $60 + d = 2(60 - d)$, which gives $3d = 60$ and $d = 20$. The angles are $40^\circ$, $60^\circ$ and $80^\circ$.

Exam­ple 6#

Five num­bers in A.P. have sum 35, and the prod­uct of the first and last is 13. Find them.

Let them be $a - 2d, a - d, a, a + d, a + 2d$. Then $5a = 35$, so $a = 7$, and $(7 - 2d)(7 + 2d) = 13$ gives $49 - 4d^2 = 13$, so $d^2 = 9$ and $d = \pm 3$. The num­bers are $1, 4, 7, 10, 13$.

Prac­tice#

1. Find $x$ if $5$, $x$ and $17$ are in A.P.
2. Find $k$ if $k + 2$, $4k - 6$ and $3k - 2$ are in A.P.
3. In $4, 9, 14, 19, 24, 29, 34$, check Prop­erty 2 for three dif­fer­ent pairs.
4. Each term of the A.P. $6, 10, 14, 18$ is divided by 2. Write the new sequence and its com­mon dif­fer­ence.
5. An A.P. has 12 terms and $t_4 + t_9 = 30$. Find $S_{12}$.
6. Three num­bers in A.P. have sum 24 and prod­uct 440. Find them.
7. Three num­bers in A.P. have sum 15 and the sum of their squares is 83. Find them.
8. Four num­bers in A.P. have sum 32 and the prod­uct of the first and last is 55. Find them.
9. The angles of a quadri­lat­eral are in A.P. and the small­est is $75^\circ$. Find all four angles.
10. Five num­bers in A.P. have sum 25 and the sum of their squares is 165. Find them.

Answers#