Problems on arithmetic progressions

Arith­metic pro­gres­sions describe any quan­tity that changes by a fixed amount at each step: weekly sav­ings, ris­ing instal­ments, rows of seats, stacked logs. This les­son works through prob­lems of that kind, together with a few that test the for­mu­las in alge­braic form. In every prob­lem the first job is the same: decide what $a$, $d$ and $n$ stand for, and whether the ques­tion asks for a term $t_n$ or a sum $S_n$.

The two for­mu­las#

$$\displaystyle t_n = a + (n - 1)d \qquad S_n = \frac{n}{2}\left[2a + (n - 1)d\right] = \frac{n}{2}(a + l)$$

Ask your­self: "How much in the $n$th step?" is a term. "How much alto­gether in $n$ steps?" is a sum.

Exam­ple 1#

Ravi saves ₹50 in the first week of the year and increases his sav­ing by ₹20 each week. How much does he save in the 12th week, and how much in the first 12 weeks alto­gether?

The weekly sav­ings form an A.P. with $a = 50$ and $d = 20$.

In the 12th week: $t_{12} = 50 + 11 \times 20 = 270$, so ₹270.

Alto­gether: $\displaystyle S_{12} = \frac{12}{2}(50 + 270) = 6 \times 320 = 1920$, so ₹1920.

Exam­ple 2#

A hall has 25 rows of seats. The first row has 20 seats and each row has 2 more seats than the row in front of it. How many seats are there in the hall?

$a = 20$, $d = 2$, $n = 25$.

$$\displaystyle S_{25} = \frac{25}{2}\left[40 + 24 \times 2\right] = \frac{25}{2} \times 88 = 25 \times 44 = 1100$$

There are 1100 seats.

Exam­ple 3#

Sita repays a loan of ₹3250 in monthly instal­ments. The first instal­ment is ₹20 and each instal­ment is ₹15 more than the one before. In how many months is the loan cleared, and what is the last instal­ment?

$a = 20$, $d = 15$, and we need $S_n = 3250$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[40 + 15(n - 1)\right] &= 3250 \\ n(15n + 25) &= 6500 \\ 15n^2 + 25n - 6500 &= 0 \\ 3n^2 + 5n - 1300 &= 0 \end{aligned}$$

By the qua­dratic for­mula, $\displaystyle n = \frac{-5 \pm \sqrt{25 + 15600}}{6} = \frac{-5 \pm 125}{6}$. The pos­i­tive value is $\displaystyle n = \frac{120}{6} = 20$. So the loan is cleared in 20 months. The last instal­ment is $t_{20} = 20 + 19 \times 15 = 305$, that is ₹305.

Exam­ple 4#

165 logs are stacked so that the bot­tom row has 21 logs, the next row 20, the next 19, and so on. How many rows are there, and how many logs are in the top row?

$a = 21$, $d = -1$, $S_n = 165$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[42 - (n - 1)\right] &= 165 \\ n(43 - n) &= 330 \\ n^2 - 43n + 330 &= 0 \\ (n - 10)(n - 33) &= 0 \end{aligned}$$

So $n = 10$ or $n = 33$. But with 33 rows, the 22nd row would have $21 - 21 = 0$ logs and later rows a neg­a­tive num­ber, which is impos­si­ble. So there are 10 rows, and the top row has $t_{10} = 21 - 9 = 12$ logs.

Always ask whether each solu­tion of the equa­tion makes sense in the sit­u­a­tion.

Exam­ple 5#

Find the sum of all two-digit num­bers that are divis­i­ble by 3.

These are $12, 15, 18, \ldots, 99$. Count them: $12 + 3(n - 1) = 99$ gives $n - 1 = 29$, so $n = 30$. Then

$$\displaystyle S_{30} = \frac{30}{2}(12 + 99) = 15 \times 111 = 1665$$

Exam­ple 6#

The angles of a quadri­lat­eral are in A.P. and the largest angle is $120^\circ$. Find all the angles.

Let the angles be $a, a + d, a + 2d, a + 3d$. Their sum is $360^\circ$, so $4a + 6d = 360$. The largest is $a + 3d = 120$. From the sec­ond, $a = 120 - 3d$. Sub­sti­tut­ing:

$$4(120 - 3d) + 6d = 360 \implies 480 - 6d = 360 \implies d = 20$$

So $a = 60$, and the angles are $60^\circ, 80^\circ, 100^\circ, 120^\circ$.

Exam­ple 7#

The $p$th term of an A.P. is $q$ and the $q$th term is $p$, where $p \neq q$. Show that its $r$th term is $p + q - r$.

From the gen­eral term:

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

Sub­tract­ing, $(p - q)d = q - p$, so $d = -1$ (we may divide by $p - q$ because $p \neq q$). Then $a = q + (p - 1) = p + q - 1$. So

$$t_r = (p + q - 1) + (r - 1)(-1) = p + q - r$$

In par­tic­u­lar the $(p + q)$th term is 0.

Exam­ple 8#

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the A.P.

The first 14 terms add up to $63 + 161 = 224$. So

$$\displaystyle S_7 = \frac{7}{2}(2a + 6d) = 63 \implies a + 3d = 9$$

$$\displaystyle S_{14} = \frac{14}{2}(2a + 13d) = 224 \implies 2a + 13d = 32$$

From the first, $2a + 6d = 18$. Sub­tract from the sec­ond: $7d = 14$, so $d = 2$ and $a = 3$. The A.P. is $3, 5, 7, 9, \ldots$

Prac­tice#

1. Anil earns ₹12000 in the first year of a job, with an increase of ₹1500 every year. What is his salary in the 10th year, and what does he earn in total in 10 years?
2. A the­atre has 30 rows. The first row has 16 seats and each row has 3 more than the one before. How many seats are there in all?
3. A clock strikes once at 1 o'clock, twice at 2 o'clock, and so on. How many times does it strike in 12 hours?
4. Find the sum of all three-digit num­bers that leave remain­der 1 when divided by 5.
5. Find the sum of all two-digit odd num­bers.
6. The angles of a pen­ta­gon are in A.P. and the small­est angle is $84^\circ$. Find the com­mon dif­fer­ence and the largest angle.
7. A spi­ral is made of semi­cir­cles whose lengths increase by 2 cm each time, start­ing at 5 cm. What is the total length of the first 15 semi­cir­cles?
8. 120 logs are stacked with 15 in the bot­tom row, 14 in the next, and so on. How many rows are there, and how many logs are in the top row?
9. The sum of the first 5 terms of an A.P. is 40 and the sum of the first 10 terms is 155. Find the A.P.
10. The 5th term of an A.P. is 3 times the 2nd term, and the 9th term is 34. Find the A.P.

Answers#