Class 10 Arithmetic Progressions Practice Questions and Solutions

Ques­tion 2#

Is 1, $-1$, 1, $-1$, … an AP?

The dif­fer­ences are $-1 - 1 = -2$ and $1 - (-1) = 2$. These are not equal.

Answer: No. The list alter­nates, so it has no com­mon dif­fer­ence.

Ques­tion 3#

Find $p$ if $2p$, $p + 10$ and $3p + 2$ are con­sec­u­tive terms of an AP. Write the three terms.

For three num­bers in AP, twice the mid­dle one equals the sum of the other two:

$$\begin{aligned} 2(p + 10) &= 2p + (3p + 2) \\ 2p + 20 &= 5p + 2 \\ 18 &= 3p \\ p &= 6 \end{aligned}$$

The terms are $2 \times 6 = 12$, $6 + 10 = 16$ and $3 \times 6 + 2 = 20$, which have a com­mon dif­fer­ence of 4.

Answer: $p = 6$; the terms are 12, 16, 20.

Why this works: $b - a = c - b$ rearranges to $2b = a + c$.

The nth term#

Ques­tion 4#

Find the 25th term of the AP $-8, -5, -2, \ldots$

Here $a = -8$ and $d = -5 - (-8) = 3$.

$$a_{25} = -8 + 24 \times 3 = -8 + 72 = 64$$

Answer: 64.

Ques­tion 5#

Which term of the AP 4, 9, 14, … is 249?

Here $a = 4$ and $d = 5$.

$$\begin{aligned} 4 + (n - 1) \times 5 &= 249 \\ (n - 1) \times 5 &= 245 \\ n - 1 &= 49 \\ n &= 50 \end{aligned}$$

Answer: The 50th term.

Ques­tion 6#

Is 302 a term of the AP 6, 13, 20, … ? Give a rea­son.

Here $a = 6$ and $d = 7$. If $a_n = 302$, then $(n - 1) \times 7 = 296$, so $\displaystyle n - 1 = \frac{296}{7}$, which is not a whole num­ber.

You can also see this directly: $a_{43} = 6 + 42 \times 7 = 300$ and $a_{44} = 307$, and 302 lies strictly between them.

Answer: No, 302 is not a term, because $n$ would not be a whole num­ber.

Ques­tion 7#

The 7th term of an AP is $-4$ and the 13th term is $-16$. Find the AP and its 20th term.

Write the two facts as equa­tions:

$$a + 6d = -4 \qquad \text{and} \qquad a + 12d = -16$$

Sub­tract­ing the first from the sec­ond gives $6d = -12$, so $d = -2$. Then $a = -4 - 6 \times (-2) = 8$.

$$a_{20} = 8 + 19 \times (-2) = 8 - 38 = -30$$

Answer: The AP is 8, 6, 4, 2, …, and its 20th term is $-30$.

Ques­tion 8#

How many terms are there in the AP $20, 17.5, 15, \ldots, -45$?

Here $a = 20$, $d = -2.5$ and $l = -45$.

$$\begin{aligned} 20 + (n - 1)(-2.5) &= -45 \\ (n - 1)(-2.5) &= -65 \\ n - 1 &= 26 \\ n &= 27 \end{aligned}$$

Answer: 27 terms.

Sums of APs#

Ques­tion 9#

Find the sum of the first 22 terms of the AP $\displaystyle \frac{1}{2}, 1, \frac{3}{2}, \ldots$

Here $\displaystyle a = \frac{1}{2}$, $\displaystyle d = \frac{1}{2}$ and $n = 22$.

$$\displaystyle S_{22} = \frac{22}{2}\left[2 \times \frac{1}{2} + 21 \times \frac{1}{2}\right] = 11\left[1 + \frac{21}{2}\right] = 11 \times \frac{23}{2} = \frac{253}{2}$$

Answer: $\displaystyle \frac{253}{2}$, that is, 126.5.

Ques­tion 10#

Find the sum $46 + 43 + 40 + \cdots + 10$.

Here $a = 46$, $d = -3$ and $l = 10$. The num­ber of terms comes from $46 - 3(n - 1) = 10$, so $n - 1 = 12$ and $n = 13$.

$$\displaystyle S_{13} = \frac{13}{2}(46 + 10) = \frac{13}{2} \times 56 = 364$$

Answer: 364.

Ques­tion 11#

Find the sum of all two-digit num­bers that are divis­i­ble by 7.

These num­bers are 14, 21, …, 98, an AP with $a = 14$, $d = 7$ and $l = 98$. From $14 + 7(n - 1) = 98$ we get $n - 1 = 12$, so $n = 13$.

$$\displaystyle S_{13} = \frac{13}{2}(14 + 98) = 13 \times 56 = 728$$

Answer: 728.

Ques­tion 12#

The sum of the first $n$ terms of an AP is $S_n = 6n - n^2$. Find the first term, the sec­ond term, the $n$th term and the 10th term.

The first term is $S_1 = 6 - 1 = 5$. For $n \ge 2$:

$$\begin{aligned} a_n &= S_n - S_{n-1} \\ &= (6n - n^2) - \left[6(n - 1) - (n - 1)^2\right] \\ &= 6n - n^2 - 6n + 6 + n^2 - 2n + 1 \\ &= 7 - 2n \end{aligned}$$

This also gives $a_1 = 5$, so the for­mula works for every $n$. Then $a_2 = 3$ and $a_{10} = 7 - 20 = -13$.

Answer: $a_1 = 5$, $a_2 = 3$, $a_n = 7 - 2n$, $a_{10} = -13$.

Ques­tion 13#

The first and last terms of an AP are 4 and 52, and the sum of all its terms is 364. Find the num­ber of terms and the com­mon dif­fer­ence.

Using $\displaystyle S_n = \frac{n}{2}(a + l)$:

$$\displaystyle 364 = \frac{n}{2}(4 + 52) = 28n \quad \Rightarrow \quad n = 13$$

Then $52 = 4 + 12d$, so $12d = 48$ and $d = 4$.

Answer: 13 terms; $d = 4$.

Word prob­lems and exam-level ques­tions#

Ques­tion 14#

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 in the last row, and how many seats are there in the hall?

The seats per row form an AP with $a = 20$, $d = 2$ and $n = 25$.

$$a_{25} = 20 + 24 \times 2 = 68$$

$$\displaystyle S_{25} = \frac{25}{2}(20 + 68) = \frac{25}{2} \times 88 = 1100$$

Answer: 68 seats in the last row; 1100 seats in all.

Ques­tion 15#

How many terms of the AP 17, 15, 13, … must be taken so that their sum is 72? Explain the dou­ble answer.

Here $a = 17$ and $d = -2$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[34 - 2(n - 1)\right] &= 72 \\ \frac{n}{2}(36 - 2n) &= 72 \\ n(18 - n) &= 72 \\ n^2 - 18n + 72 &= 0 \\ (n - 6)(n - 12) &= 0 \end{aligned}$$

So $n = 6$ or $n = 12$. Both are pos­i­tive whole num­bers. The 7th to 12th terms are $5, 3, 1, -1, -3, -5$, which add up to 0, so the sum of 12 terms equals the sum of 6 terms.

Answer: 6 terms or 12 terms.

Ques­tion 16#

For the AP 7, 13, 19, …, 241, find (a) the num­ber of terms, (b) the two mid­dle terms, and (c) the sum of the last 10 terms.

(a) With $a = 7$ and $d = 6$: $7 + 6(n - 1) = 241$ gives $n - 1 = 39$, so $n = 40$.

(b) Since 40 is even, the mid­dle terms are the 20th and 21st:

$$a_{20} = 7 + 19 \times 6 = 121, \qquad a_{21} = 7 + 20 \times 6 = 127$$

(c) The last 10 terms are the 31st to the 40th. Here $a_{31} = 7 + 30 \times 6 = 187$ and $a_{40} = 241$, so

$$\displaystyle \text{Sum} = \frac{10}{2}(187 + 241) = 5 \times 428 = 2140$$

Answer: (a) 40; (b) 121 and 127; (c) 2140.

Why this works: the last 10 terms are them­selves an AP, so the first-and-last form applies to them directly.

Ques­tion 17#

A club gives 6 prizes with a total value of ₹1200. Each prize is ₹30 less than the prize just above it. Find the value of each prize.

Let the top prize be ₹$a$. The prizes form an AP with $d = -30$ and $n = 6$:

$$\displaystyle \begin{aligned} \frac{6}{2}\left[2a + 5 \times (-30)\right] &= 1200 \\ 2a - 150 &= 400 \\ a &= 275 \end{aligned}$$

The prizes are ₹275, ₹245, ₹215, ₹185, ₹155 and ₹125. Check: these add up to ₹1200.

Answer: ₹275, ₹245, ₹215, ₹185, ₹155 and ₹125.

Ques­tion 18#

The sec­ond and third terms of an AP are 11 and 17. Find the sum of its first 30 terms.

The com­mon dif­fer­ence is $d = 17 - 11 = 6$, so the first term is $a = 11 - 6 = 5$.

$$\displaystyle S_{30} = \frac{30}{2}\left[2 \times 5 + 29 \times 6\right] = 15 \times 184 = 2760$$

Answer: 2760.

Key terms#

Arith­metic pro­gres­sion

A sequence in which every term after the first is the pre­vi­ous term plus a fixed num­ber $d$.

Com­mon dif­fer­ence

The fixed num­ber $d$; any term minus the term before it.

nth term

$a_n = a + (n - 1)d$.

Last term

The final term $l$ of a finite AP.

Sum of n terms

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

Mid­dle term

For an odd num­ber of terms $n$, the $\displaystyle \frac{n + 1}{2}$th term; for even $n$ there are two mid­dle terms.

Com­mon ques­tions#

How much work­ing should I show in a 3-mark AP ques­tion?#

Write $a$ and $d$, the for­mula you are using, the sub­sti­tu­tion, and the final answer with units. Exam­in­ers give marks for each of these steps.

What if solv­ing for n gives a frac­tion?#

If the ques­tion asks which term has a value, a frac­tional $n$ means the value is not a term. If it asks how many terms give a sum, reject the frac­tional root.

When are both roots of the qua­dratic in n accept­able?#

When both are pos­i­tive whole num­bers. This hap­pens when the terms change sign and a block of terms adds up to zero, as in Ques­tion 15.

Can I use either sum for­mula?#

Yes. Use $\displaystyle \frac{n}{2}(a + l)$ when the last term is known, and $\displaystyle \frac{n}{2}[2a + (n - 1)d]$ oth­er­wise. Both give the same value.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.