SEBA Class 10 Maths Exercise 5.2 Solutions: Arithmetic Progressions | New Book

SEBA Class 10 Maths Exercise 5.2 Solutions: Arithmetic Progressions | New Book

Get free, step-by-step solutions to SEBA Class 10 Maths Exercise 5.2: Arithmetic Progressions from the new SCERT textbook. This article solves all textbook questions and MCQs in a simple way to help you prepare effectively for your upcoming matric examinations.

See More:

• SEBA Class 10 Maths Exercise 5.1 Solutions: Arithmetic Progressions

• SEBA Class 10 Maths Exercise 5.2 Solutions: Arithmetic Progressions

• SEBA Class 10 Maths Exercise 5.3 Solutions: Arithmetic Progressions

EXERCISE 5.2

1. Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n^{\text{th}}$ term of the AP:

$\mathbf{Sol^n.}$

(i) Given: $a = 7$, $d = 3$, $n = 8$

We know that:

$a_n = a + (n – 1)d$

$\implies a_8 = 7 + (8 – 1)3$

$= 7 + 7 \times 3$

$= 7 + 21$

$= 28$

$\therefore a_8 = \mathbf{28}$

(ii) Given: $a = -18$, $n = 10$, $a_n = 0$

We know that: $a_n = a + (n – 1)d$

$\implies 0 = -18 + (10 – 1)d$

$\implies 18 = 9d$

$\implies d = \frac{18}{9} = 2$

$\therefore d = \mathbf{2}$

(iii) Given: $d = -3$, $n = 18$, $a_n = -5$

We know that: $a_n = a + (n – 1)d$

$\implies -5 = a + (18 – 1)(-3)$

$\implies -5 = a + 17(-3)$

$\implies -5 = a – 51$

$\implies a = -5 + 51 = 46$

$\therefore a = \mathbf{46}$

(iv) Given: $a = -18.9$, $d = 2.5$, $a_n = 3.6$

We know that: $a_n = a + (n – 1)d$

$\implies 3.6 = -18.9 + (n – 1)(2.5)$

$\implies 3.6 + 18.9 = (n – 1)(2.5)$

$\implies 22.5 = (n – 1)(2.5)$

$\implies n – 1 = \frac{22.5}{2.5} = 9$

$\implies n = 9 + 1 = 10$

$\therefore n = \mathbf{10}$

(v) Given: $a = 3.5$, $d = 0$, $n = 105$

We know that: $a_n = a + (n – 1)d$

$\implies a_n = 3.5 + (105 – 1)(0)$

$\implies a_n = 3.5 + 0 = 3.5$

$\therefore a_n = \mathbf{3.5}$

2. Choose the correct choice in the following and justify:

(i) $30^{\text{th}}$ term of the AP: $10, 7, 4, \dots$, is

(A) 97

(B) 77

(C) -77

(D) -87

$\mathbf{Sol^n.}$

Here, $a = 10$, $d = 7 – 10 = -3$, and $n = 30$

We know that: $a_n = a + (n – 1)d$

$\implies a_{30} = 10 + (30 – 1)(-3)$

$\implies a_{30} = 10 + 29(-3)$

$\implies a_{30} = 10 – 87$

$\therefore a_{30} = -77$

So the correct option is (C) -77

(ii) $11^{\text{th}}$ term of the AP: $-3, -\frac{1}{2}, 2, \dots$, is

(A) 28

(B) 22

(C) -38

(D) $-48\frac{1}{2}$

$\mathbf{Sol^n.}$

Here, $a = -3$ and $n = 11$

$d = -\frac{1}{2} – (-3) = -\frac{1}{2} + 3 = \frac{5}{2}$

We know that: $a_n = a + (n – 1)d$

$\implies a_{11} = -3 + (11 – 1)\left(\frac{5}{2}\right)$

$\implies a_{11} = -3 + 10 \times \frac{5}{2}$

$\implies a_{11} = -3 + 25$

$\therefore a_{11} = 22$

So the correct option is (B) 22

3. In the following APs, find the missing terms in the boxes:

(i) $2, \; \square, \; 26$

$\mathbf{Sol^n.}$

Let the missing term be $a_2$

Here, $a_1 = a = 2$ and $a_3 = 26$

We know that $a_3 = a + 2d$

∴ $2 + 2d = 26$

$\implies 2d = 24$

$\implies d = 12$

Missing term:

$a_2 = a + d = 2 + 12 = 14$

$\therefore$ The missing term is 14.

(ii) $\square, \; 13, \; \square, \; 3$

$\mathbf{Sol^n.}$

Here, $a_2 = 13$ and $a_4 = 3$

We know that:

$a_2 = a + d = 13 \quad \text{— (1)}$

$a_4 = a + 3d = 3 \quad \text{— (2)}$

Subtracting (1) from (2):

$(a + 3d) – (a + d) = 3 – 13$

$\implies 2d = -10$

$\implies d = -\frac{10}{2}$

$\implies d = -5$

Substituting $d = -5$ in equation (1):

$a + (-5) = 13$

$\implies a = 13 + 5$

$\implies a = 18$

Missing terms:

$a_1 = a = 18$

$a_3 = a_2 + d = 13 + (-5) = 8$

$\therefore$ The missing terms are 18 and 8.

(iii) $5, \; \square, \; \square, \; 9\frac{1}{2}$

$\mathbf{Sol^n.}$

Here, $a = 5$ and $a_4 = 9\frac{1}{2} = \frac{19}{2}$

We know that:

$a_4 = a + 3d$

$\implies \frac{19}{2} = 5 + 3d$

$\implies 3d = \frac{19}{2} – 5$

$\implies 3d = \frac{19 – 10}{2}$

$\implies 3d = \frac{9}{2}$

$\implies d = \frac{9}{2 \times 3}$

$\implies d = \frac{3}{2}$

Missing terms:

$a_2 = a + d = 5 + \frac{3}{2} = \frac{13}{2} = 6\frac{1}{2}$

$a_3 = a_2 + d = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8$

$\therefore$ The missing terms are $6\frac{1}{2}$ and 8.

(iv) $-4, \; \square, \; \square, \; \square, \; \square, \; 6$

$\mathbf{Sol^n.}$

Here, $a = -4$ and $a_6 = 6$

We know that:

$a_6 = a + 5d$

$\implies 6 = -4 + 5d$

$\implies 5d = 6 + 4$

$\implies 5d = 10$

$\implies d = \frac{10}{5}$

$\implies d = 2$

Missing terms:

$a_2 = a + d = -4 + 2 = -2$

$a_3 = a_2 + d = -2 + 2 = 0$

$a_4 = a_3 + d = 0 + 2 = 2$

$a_5 = a_4 + d = 2 + 2 = 4$

$\therefore$ The missing terms are -2, 0, 2, 4.

(v) $\square, \; 38, \; \square, \; \square, \; \square, \; -22$

$\mathbf{Sol^n.}$

Here, $a_2 = 38$ and $a_6 = -22$

We know that:

$a_2 = a + d = 38 \quad \text{— (1)}$

$a_6 = a + 5d = -22 \quad \text{— (2)}$

Subtracting (1) from (2):

$(a + 5d) – (a + d) = -22 – 38$

$\implies 4d = -60$

$\implies d = -\frac{60}{4}$

$\implies d = -15$

Substituting $d = -15$ in equation (1):

$a + (-15) = 38$

$\implies a = 38 + 15$

$\implies a = 53$

Missing terms:

$a_1 = a = 53$

$a_3 = a_2 + d = 38 + (-15) = 23$

$a_4 = a_3 + d = 23 + (-15) = 8$

$a_5 = a_4 + d = 8 + (-15) = -7$

$\therefore$ The missing terms are 53, 23, 8, -7.

4. Which term of the AP: $3, 8, 13, 18, \dots$, is 78?

$\mathbf{Sol^n.}$

Here, $a = 3$, $d = 8 – 3 = 5$, and $a_n = 78$

We know that:

$a_n = a + (n – 1)d$

$\implies 78 = 3 + (n – 1)5$

$\implies 78 – 3 = (n – 1)5$

$\implies 75 = (n – 1)5$

$\implies n – 1 = \frac{75}{5}$

$\implies n – 1 = 15$

$\implies n = 15 + 1$

$\implies n = 16$

$\therefore$ The $16^{\text{th}}$ term of the given AP is 78.

5. Find the number of terms in each of the following APs:

(i) $7, 13, 19, \dots, 205$

$\mathbf{Sol^n.}$

Here, $a = 7$, $d = 13 – 7 = 6$, and $a_n = 205$

We know that:

$a_n = a + (n – 1)d$

$\implies 205 = 7 + (n – 1)6$

$\implies 205 – 7 = 6(n – 1)$

$\implies 198 = 6(n – 1)$

$\implies n – 1 = \frac{198}{6}$

$\implies n – 1 = 33$

$\implies n = 33 + 1$

$\implies n = 34$

$\therefore$ The number of terms in the given AP is 34.

(ii) $18, 15\frac{1}{2}, 13, \dots, -47$

$\mathbf{Sol^n.}$

Here, $a = 18$, $a_n = -47$

$d = 15\frac{1}{2} – 18 = \frac{31}{2} – 18 = \frac{31 – 36}{2} = -\frac{5}{2}$

We know that:

$a_n = a + (n – 1)d$

$\implies -47 = 18 + (n – 1)\left(-\frac{5}{2}\right)$

$\implies -47 – 18 = (n – 1)\left(-\frac{5}{2}\right)$

$\implies -65 = (n – 1)\left(-\frac{5}{2}\right)$

$\implies n – 1 = -65 \times \left(-\frac{2}{5}\right)$

$\implies n – 1 = 26$

$\implies n = 26 + 1$

$\implies n = 27$

$\therefore$ The number of terms in the given AP is 27.

6. Check whether -150 is a term of the AP: $11, 8, 5, 2, \dots$

$\mathbf{Sol^n.}$

Here, $a = 11$, $d = 8 – 11 = -3$

Let the $n^{\text{th}}$ term be $-150$:

We know that: $a_n = a + (n – 1)d$

$\implies -150 = 11 + (n – 1)(-3)$

$\implies -150 – 11 = -3(n – 1)$

$\implies -161 = -3(n – 1)$

$\implies n – 1 = \frac{161}{3}$

$\implies n = \frac{161}{3} + 1$

$\implies n = \frac{164}{3} = 54\frac{2}{3}$

Since $n$ must be a positive integer, -150 is not a term of the given AP.

7. Find the $31^{\text{st}}$ term of an AP whose $11^{\text{th}}$ term is 38 and the $16^{\text{th}}$ term is 73.

$\mathbf{Sol^n.}$

Let $a$ be the first term and $d$ be the common difference of the given AP.

Given:

$a_{11} = 38$
$\implies a + 10d = 38 \quad \text{— (1)}$

Also, $a_{16} = 73$
$ \implies a + 15d = 73 \quad \text{— (2)}$

Subtracting (1) from (2), we get

$(a + 15d) – (a + 10d) = 73 – 38$

$\implies 5d = 35$

$\implies d = \frac{35}{5}$

$\implies d = 7$

Substituting $d = 7$ in equation (1), we get

$a + 10(7) = 38$

$\implies a + 70 = 38$

$\implies a = 38 – 70$

$\implies a = -32$

Now, the $31^{\text{st}}$ term:

$a_{31} = a + (31 – 1)d$

$= a + 30d$

$= -32 + 30(7)$

$= -32 + 210$

$= 178$

$\therefore$ The $31^{\text{st}}$ term of the given AP is 178.

8. An AP consists of 50 terms of which $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term.

$\mathbf{Sol^n.}$

Given:

Total number of terms, $n = 50$

$a_3 = 12$
$\implies a + 2d = 12 \quad \text{— (1)}$

Last term, $a_{50} = 106$
$\implies a + 49d = 106 \quad \text{— (2)}$

Subtracting (1) from (2), we get

$(a + 49d) – (a + 2d) = 106 – 12$

$\implies 47d = 94$

$\implies d = \frac{94}{47}$

$\implies d = 2$

Substituting $d = 2$ in equation (1), we get

$a + 2(2) = 12$

$\implies a + 4 = 12$

$\implies a = 12 – 4$

$\implies a = 8$

Now, the $29^{\text{th}}$ term:

$a_{29} = a + 28d$

$= 8 + 28(2)$

$= 8 + 56$

$= 64$

$\therefore$ The $29^{\text{th}}$ term of the given AP is 64.

9. If the $3^{\text{rd}}$ and the $9^{\text{th}}$ terms of an AP are 4 and -8 respectively, which term of this AP is zero?

$\mathbf{Sol^n.}$

Given:

$a_3 = 4$
$\implies a + 2d = 4 \quad \text{— (1)}$

$a_9 = -8$
$\implies a + 8d = -8 \quad \text{— (2)}$

Subtracting (1) from (2), we get

$(a + 8d) – (a + 2d) = -8 – 4$

$\implies 6d = -12$

$\implies d = -\frac{12}{6}$

$\implies d = -2$

Substituting $d = -2$ in equation (1), we get

$a + 2(-2) = 4$

$\implies a – 4 = 4$

$\implies a = 4 + 4$

$\implies a = 8$

Let the $n^{\text{th}}$ term be zero:

$a_n = 0$

$\implies a + (n – 1)d = 0$

$\implies 8 + (n – 1)(-2) = 0$

$\implies -2(n – 1) = -8$

$\implies n – 1 = \frac{-8}{-2}$

$\implies n – 1 = 4$

$\implies n = 4 + 1$

$\implies n = 5$

$\therefore$ The $5^{\text{th}}$ term of this AP is zero.

10. The $17^{\text{th}}$ term of an AP exceeds its $10^{\text{th}}$ term by 7. Find the common difference.

$\mathbf{Sol^n.}$

We know that:

$a_{17} = a + 16d$

$a_{10} = a + 9d$

According to the question:

$a_{17} – a_{10} = 7$

$\implies (a + 16d) – (a + 9d) = 7$

$\implies a + 16d – a – 9d = 7$

$\implies 7d = 7$

$\implies d = \frac{7}{7}$

$\implies d = 1$

$\therefore$ The common difference is 1.

11. Which term of the AP: $3, 15, 27, 39, \dots$ will be 132 more than its $54^{\text{th}}$ term?

$\mathbf{Sol^n.}$

Here, $a = 3$, $d = 15 – 3 = 12$

Let the $54^{\text{th}}$ term be $a_{54}$.

We know that $a_{54} = a + 53d$

$= 3 + 53(12)$

$= 3 + 636$

$= 639$

Let the required term be $a_n$.

According to the question:

$a_n = a_{54} + 132$

$\implies a_n = 639 + 132$

$\implies a_n = 771$

Now:

$a + (n – 1)d = 771$

$\implies 3 + (n – 1)12 = 771$

$\implies 12(n – 1) = 771 – 3$

$\implies 12(n – 1) = 768$

$\implies n – 1 = \frac{768}{12}$

$\implies n – 1 = 64$

$\implies n = 64 + 1$

$\implies n = 65$

$\therefore$ The $65^{\text{th}}$ term will be 132 more than its $54^{\text{th}}$ term.

12. Two APs have the same common difference. The difference between their $100^{\text{th}}$ terms is 100, what is the difference between their $1000^{\text{th}}$ terms?

$\mathbf{Sol^n.}$

Let the first term of the first AP be $a$ and
The first term of the second AP be $A$.
And two APs have the same common difference $d$.

Then:

$a_{100} = a + 99d$

$A_{100} = A + 99d$

According to the question:

$a_{100} – A_{100} = 100$

$\implies (a + 99d) – (A + 99d) = 100$

$\implies a – A = 100 \quad \text{— (1)}$

Now, the difference between their $1000^{\text{th}}$ terms is:

$a_{1000} – A_{1000}$

$= (a + 999d) – (A + 999d)$

$= a – A$

$a – A = 100$ [From equation (1)]

$\therefore$ The difference between their $1000^{\text{th}}$ terms of the two AP is 100.

13. How many three-digit numbers are divisible by 7?

$\mathbf{Sol^n.}$

The list of three-digit numbers divisible by 7 are:

$105, 112, 119, \dots, 994$

The above sequence is an AP with:

$a = 105$, $d = 7$, and $a_n = 994$

We know that:

$a_n = a + (n – 1)d$

$\implies 994 = 105 + (n – 1)7$

$\implies 994 – 105 = 7(n – 1)$

$\implies 889 = 7(n – 1)$

$\implies n – 1 = \frac{889}{7}$

$\implies n – 1 = 127$

$\implies n = 127 + 1$

$\implies n = 128$

$\therefore$ There are 128 three-digit numbers divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

$\mathbf{Sol^n.}$

The multiples of 4 lying between 10 and 250 are:

$12, 16, 20, \dots, 248$

The above sequence forms an AP with:

$a = 12$, $d = 4$, and $a_n = 248$

We know that:

$a_n = a + (n – 1)d$

$\implies 248 = 12 + (n – 1)4$

$\implies 248 – 12 = 4(n – 1)$

$\implies 236 = 4(n – 1)$

$\implies n – 1 = \frac{236}{4}$

$\implies n – 1 = 59$

$\implies n = 59 + 1$

$\implies n = 60$

$\therefore$ There are 60 multiples of 4 lying between 10 and 250.

15. For what value of $n$, are the $n^{\text{th}}$ terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

$\mathbf{Sol^n.}$

The first AP is $63, 65, 67, \dots$

Here, $a_1 = 63$, $d_1 = 65 – 63 = 2$

Let the $n^{\text{th}}$ term be $a_{n}$

∴ $a_n = 63 + (n – 1)2$

$= 63 + 2n – 2$

$= 61 + 2n$

The second AP is: $3, 10, 17, \dots$

Here, $a_2 = 3$, $d_2 = 10 – 3 = 7$

Let the $n^{\text{th}}$ term be $a’_{n}$

∴$a’_n = 3 + (n – 1)7$

$= 3 + 7n – 7$

$= 7n – 4$

According to the question:

$a_n = a’_n$

$\implies 61 + 2n = 7n – 4$

$\implies 61 + 4 = 7n – 2n$

$\implies 65 = 5n$

$\implies n = \frac{65}{5}$

$\implies n = 13$

$\therefore$ the $13^{\text{th}}$ terms of the two APs are equal.

16. Determine the AP whose third term is 16 and the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12.

$\mathbf{Sol^n.}$

Given:

Third term, $a_3 = 16$
$\implies a + 2d = 16 \quad \text{— (1)}$

Also, the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by 12.

∴ $a_7 – a_5 = 12$

$\implies (a + 6d) – (a + 4d) = 12$

$\implies 2d = 12$

$\implies d = \frac{12}{2}$

$\implies d = 6$

Substituting $d = 6$ in equation (1), we get

$a + 2(6) = 16$

$\implies a + 12 = 16$

$\implies a = 16 – 12$

$\implies a = 4$

The terms of the AP are:

$a_1 = 4$

$a_2 = 4 + 6 = 10$

$a_3 = 10 + 6 = 16$

$a_4 = 16 + 6 = 22$

$\therefore$ The required AP is $4, 10, 16, 22, \dots$

17. Find the $20^{\text{th}}$ term from the last term of the AP: $3, 8, 13, \dots, 253$.

$\mathbf{Sol^n.}$

Given AP: $3, 8, 13, \dots, 253$

Writing the given AP in reverse order, then we have

$253, 248, 243, \dots, 13, 8, 3$

Here:

First term ($a$) = 253

Common difference ($d$) = $248 – 253 = -5$

$n = 20$

We know that:

$a_n = a + (n – 1)d$

$\implies a_{20} = 253 + (20 – 1)(-5)$

$= 253 + 19(-5)$

$= 253 – 95$

$= 158$

$\therefore$ The $20^{\text{th}}$ term from the last term is 158.

18. The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is 24 and the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44. Find the first three terms of the AP.

$\mathbf{Sol^n.}$

We know that: $a_n = a + (n – 1)d$

So, $a_4 = a + 3d$

And, $a_8 = a + 7d$

Given, the sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is 24

∴$a_4 + a_8 = 24$

$\implies (a + 3d) + (a + 7d) = 24$

$\implies 2a + 10d = 24$

∴ $a + 5d = 12 \quad \text{— (1)}$

Also given, the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44.

∴ $a_6 + a_{10} = 44$

$\implies (a + 5d) + (a + 9d) = 44$

$\implies 2a + 14d = 44$

$a + 7d = 22 \quad \text{— (2)}$

Subtracting (1) from (2), we get

$(a + 7d) – (a + 5d) = 22 – 12$

$\implies 2d = 10$

$\implies d = \frac{10}{2}$

$\implies d = 5$

Substituting $d = 5$ in equation (1), we have

$a + 5(5) = 12$

$\implies a + 25 = 12$

$\implies a = 12 – 25$

$\implies a = -13$

Therefore, the first three terms are:

$a_1 = a = -13$

$a_2 = a + d = -13 + 5 = -8$

$a_3 = a_2 + d = -8 + 5 = -3$

$\therefore$ The first three terms of the AP are -13, -8, -3.

19. Chandana started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did her income reach Rs 7000?

$\mathbf{Sol^n.}$

Let in the $n^{\text{th}}$ year, Chandana’s income reach Rs 7000.

∴ $a_n = 7000$

Chandana’s annual salaries forms an AP:

$5000, 5200, 5400, \dots, 7000$

Here, $a = 5000$, $d = 200$, and $a_n = 7000$

We know that:

$a_n = a + (n – 1)d$

$\implies 7000 = 5000 + (n – 1)200$

$\implies 7000 – 5000 = 200(n – 1)$

$\implies 2000 = 200(n – 1)$

$\implies n – 1 = \frac{2000}{200}$

$\implies n – 1 = 10$

$\implies n = 10 + 1$

$\implies n = 11$

Year $= 1995 + (11 – 1) = 1995 + 10 = 2005$

$\therefore$ In the $11^{\text{th}}$ year, i.e., in the year 2005, her income reached Rs 7000.

20. Ramcharan saved Rs 5 in the first week of a year and then increased his weekly savings by Rs 1.75. If in the $n^{\text{th}}$ week, his weekly savings become Rs 20.75, find $n$.

$\mathbf{Sol^n.}$

Ramcharan’s weekly savings forms an AP with:

$a = 5$, $d = 1.75$, and $a_n = 20.75$

We know that:

$a_n = a + (n – 1)d$

$\implies 20.75 = 5 + (n – 1)(1.75)$

$\implies 20.75 – 5 = (n – 1)(1.75)$

$\implies 15.75 = (n – 1)(1.75)$

$\implies n – 1 = \frac{15.75}{1.75}$

$\implies n – 1 = \frac{1575}{175}$

$\implies n – 1 = 9$

$\implies n = 9 + 1$

$\implies n = 10$

$\therefore$ The value of $n$ is 10.

21. Match the column

Choose the correct match:

A) 1, 4, 3, 2

B) 3, 1, 4, 2

C) 2, 1, 4, 3

D) 4, 3, 2, 1

$\mathbf{Sol^n.}$

• For P: AP is $2, 4, 6, 8, \dots$

  $a_5 = a + 4d $
  $= 2 + 4(2) $
  $= 2 + 8 $
  $= 10 \rightarrow \mathbf{3}$

• For Q: AP is $-1.2, -3.2, -5.2, -7.2, \dots$

  $d = -3.2 – (-1.2) $
  $= -3.2 + 1.2 = -2 \rightarrow \mathbf{1}$

• For R: AP is $3, 6, 9, 12, \dots$

  $a = 3 \rightarrow \mathbf{4}$

• For S: AP is $0, -4, -8, -12, \dots$

  $a_4 – a_3 = -12 – (-8)$
  $= -12 + 8 = -4 \rightarrow \mathbf{2}$

Therefore, the matching order is P-3, Q-1, R-4, S-2.

So the correct option is B) 3, 1, 4, 2

22. Which of the following statements is true or false?

P) A sequence $a_1, a_2, a_3, \dots, a_n, \dots$ is an AP if $a_{n+1} – a_n$ is independent of $n$.

Q) If $a_1, a_2, a_3, a_4, \dots, a_n$ is an AP with common difference ‘$d$‘, then $a_p – a_q = (p – q)d$.

R) The $n^{\text{th}}$ term $a_n$ of an AP with first term ‘$a$‘ and common difference ‘$d$‘ is given by $a_n = a + (n – 1)d$.

A) P is false, Q and R are true

B) P and Q are true, R is false

C) All of P, Q and R are true

D) All of P, Q and R are false

$\mathbf{Sol^n.}$

• Statement P is true: If $a_{n+1} – a_n$ is independent of $n$, the difference between consecutive terms is constant ($d$).

• Statement Q is true:

  $a_p – a_q = [a + (p – 1)d] – [a + (q – 1)d] $
  $= a + pd – d – a – qd + d $
  $= (p – q)d$.

• Statement R is true: The standard formula for the general term of an AP is $a_n = a + (n – 1)d$.

So the correct option is C) All of P, Q and R are true

23. The steps used to find the total numbers of terms in the AP $-1, 3, 7, 11, \dots, 95$ are jumbled below:

(i) $-1 + 4n – 4 = 95$

(ii) $n = 25$

(iii) $a_n = 95$

(iv) $-1 + (n – 1)4 = 95$

Choose the correct option:

(A) (i) $\rightarrow$ (iii) $\rightarrow$ (ii) $\rightarrow$ (iv)

(B) (iii) $\rightarrow$ (i) $\rightarrow$ (ii) $\rightarrow$ (iv)

(C) (iii) $\rightarrow$ (iv) $\rightarrow$ (i) $\rightarrow$ (ii)

(D) (iii) $\rightarrow$ (iv) $\rightarrow$ (ii) $\rightarrow$ (i)

$\mathbf{Sol^n.}$

Here, $a = -1$, $d = 3 – (-1) = 4$

1. Setting the last term: $a_n = 95 \rightarrow \mathbf{(iii)}$

2. Substituting values: $-1 + (n – 1)4 = 95 \rightarrow \mathbf{(iv)}$

3. Expanding brackets: $-1 + 4n – 4 = 95 \rightarrow \mathbf{(i)}$

4. Solving for $n$: $4n – 5 = 95 $
   $\implies 4n = 100 $
   $\implies n = 25 \rightarrow \mathbf{(ii)}$

The correct sequence is (iii) $\rightarrow$ (iv) $\rightarrow$ (i) $\rightarrow$ (ii).

So the correct option is (C) (iii) $\rightarrow$ (iv) $\rightarrow$ (i) $\rightarrow$ (ii)

24. If the first three terms of an AP are $b, c$ and $2b$, then find the ratio of $b$ and $c$.[cite: 3]

$\mathbf{Sol^n.}$

Given that $b, c, 2b$ are the consecutive terms of an AP[cite: 3].

Therefore, the common differences are equal:

$c – b = 2b – c$

$\implies c + c = 2b + b$

$\implies 2c = 3b$

$\implies \frac{b}{c} = \frac{2}{3}$

$\therefore$ The ratio of $b$ and $c$ is $2 : 3$.

25. The digits of a positive number of three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

$\mathbf{Sol^n.}$

Let the digits of the three-digit number at the hundreds, tens, and units places be $(a + d)$, $a$, and $(a – d)$ respectively.

According to the first condition:

$(a + d) + a + (a – d) = 15$

$\implies 3a = 15$

$\implies a = \frac{15}{3}$

$\implies a = 5$

The original number is:

$= 100(a + d) + 10a + (a – d)$

$= 100a + 100d + 10a + a – d$

$= 111a + 99d$

The number obtained by reversing the digits is:

$= 100(a – d) + 10a + (a + d)$

$= 100a – 100d + 10a + a + d$

$= 111a – 99d$

According to the second condition:

$\text{Reversed number} = \text{Original number} – 594$

$\implies 111a – 99d = (111a + 99d) – 594$

$\implies -99d – 99d = -594$

$\implies -198d = -594$

$\implies d = \frac{-594}{-198}$

$\implies d = 3$

Now, determining each digit:

• Hundreds digit $= a + d = 5 + 3 = 8$

• Tens digit $= a = 5$

• Units digit $= a – d = 5 – 3 = 2$

$\therefore$ The required number is 852.