(i) 30th term of the AP: 10, 7, 4, ..., is

Correct Choice:

Justification:

Given Arithmetic Progression is 10, 7, 4, ...

First term, a = 10

Common difference, d = 7 − 10 = −3

Number of terms, n = 30

Using the formula a_n = a + (n − 1)d:

a₃₀ = 10 + (30 − 1)(−3)

a₃₀ = 10 + 29(−3)

a₃₀ = 10 − 87

a₃₀ = −77

(ii) 11th term of the AP: −3, −1/2, 2, ..., is

Correct Choice:

Justification:

Given Arithmetic Progression is −3, −1/2, 2, ...

First term, a = −3

Common difference, d = −1/2 − (−3) = −1/2 + 3 = 5/2

Number of terms, n = 11

Using the formula a_n = a + (n − 1)d:

a₁₁ = −3 + (11 − 1)(5/2)

a₁₁ = −3 + 10(5/2)

a₁₁ = −3 + 25

a₁₁ = 22

(i) 2, , 26

Solution:

Here, the first term is a = 2 and the third term is a₃ = 26.

We know that a₃ = a + 2d

26 = 2 + 2d

24 = 2d

d = 12

The missing second term is a + d = 2 + 12 = 14.

(ii) , 13, , 3

Solution:

Here, the second term is a₂ = 13 and the fourth term is a₄ = 3.

a + d = 13     — (1)

a + 3d = 3     — (2)

Subtracting equation (1) from equation (2):

(a + 3d) − (a + d) = 3 − 13

2d = −10

d = −5

Substituting d = −5 into equation (1):

a + (−5) = 13

a = 18

The third term is a₃ = a + 2d = 18 + 2(−5) = 8.

(iii) 5, , , 9¹/₂

Solution:

Here, the first term is a = 5 and the fourth term is a₄ = 9¹/₂ = 19/2.

We know that a₄ = a + 3d

19/2 = 5 + 3d

19/2 − 5 = 3d

9/2 = 3d

d = 3/2 = 1.5

The second term is a + d = 5 + 1.5 = 6.5 (or 6¹/₂).

The third term is a + 2d = 5 + 2(1.5) = 8.

(iv) −4, , , , , 6

Solution:

Here, the first term is a = −4 and the sixth term is a₆ = 6.

We know that a₆ = a + 5d

6 = −4 + 5d

10 = 5d

d = 2

The missing terms are calculated sequentially by adding d = 2:

Second term = −4 + 2 = −2

Third term = −2 + 2 = 0

Fourth term = 0 + 2 = 2

Fifth term = 2 + 2 = 4

(v) , 38, , , , −22

Solution:

Here, the second term is a₂ = 38 and the sixth term is a₆ = −22.

a + d = 38     — (1)

a + 5d = −22     — (2)

Subtracting equation (1) from equation (2):

(a + 5d) − (a + d) = −22 − 38

4d = −60

d = −15

Substituting d = −15 into equation (1):

a + (−15) = 38

a = 53

The remaining missing terms are:

Third term = 38 + (−15) = 23

Fourth term = 23 + (−15) = 8

Fifth term = 8 + (−15) = −7

Solution:

Given Arithmetic Progression is 3, 8, 13, 18, ...

First term, a = 3

Common difference, d = 8 − 3 = 5

Let the nth term be 78, so a_n = 78.

Using the general term formula:

a_n = a + (n − 1)d

78 = 3 + (n − 1)5

78 − 3 = (n − 1)5

75 = (n − 1)5

n − 1 = 75 / 5

n − 1 = 15

n = 16

Hence, 78 is the term of the given AP.

(i) 7, 13, 19, . . . , 205

Solution:

Given AP is 7, 13, 19, ..., 205

First term, a = 7

Common difference, d = 13 − 7 = 6

Last term, a_n = 205

Using the formula:

205 = 7 + (n − 1)6

205 − 7 = (n − 1)6

198 = (n − 1)6

n − 1 = 198 / 6

n − 1 = 33

n = 34

Hence, the number of terms in this AP is .

(ii) 18, 15¹/₂, 13, . . . , −47

Solution:

Given AP is 18, 15¹/₂, 13, ..., −47

First term, a = 18

Common difference, d = 15.5 − 18 = −2.5 = −5/2

Last term, a_n = −47

Using the formula:

−47 = 18 + (n − 1)(−5/2)

−47 − 18 = (n − 1)(−5/2)

−65 = (n − 1)(−5/2)

Multiplying both sides by −2/5:

(−65) × (−2/5) = n − 1

13 × 2 = n − 1

26 = n − 1

n = 27

Hence, the number of terms in this AP is .

6. Check whether −150 is a term of the AP : 11, 8, 5, 2 . . .

Solution:

For the given AP 11, 8, 5, 2, ..., the first term is a = 11 and the common difference is d = 8 − 11 = −3.

Let −150 be the nth term of this AP. Therefore, a_n = −150.

We know that a_n = a + (n − 1)d

−150 = 11 + (n − 1)(−3)

−150 − 11 = −3(n − 1)

−161 = −3(n − 1)

n − 1 = 161 / 3

n = 161 / 3 + 1

n = 164 / 3

Since n must be a positive integer, and 164 / 3 is a fraction, our assumption is wrong. Hence, −150 is not a term of the given AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

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

According to the given conditions:

a₁₁ = a + 10d = 38     — (1)

a₁₆ = a + 15d = 73     — (2)

Subtracting equation (1) from equation (2), we get:

(a + 15d) − (a + 10d) = 73 − 38

5d = 35

d = 7

Substituting the value of d into equation (1):

a + 10(7) = 38

a + 70 = 38

a = 38 − 70

a = −32

Now, finding the 31st term:

a₃₁ = a + 30d

a₃₁ = −32 + 30(7)

a₃₁ = −32 + 210

a₃₁ = 178

Hence, the 31st term of the AP is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Let the first term be a and the common difference be d.

Given that the number of terms is 50, the last term is the 50th term, so a₅₀ = 106.

The 3rd term is given as 12, so a₃ = 12.

Writing the linear representations:

a + 2d = 12     — (1)

a + 49d = 106     — (2)

Subtracting equation (1) from equation (2):

47d = 94

d = 2

Substituting the value of d into equation (1):

a + 2(2) = 12

a + 4 = 12

a = 8

Now, finding the 29th term:

a₂₉ = a + 28d

a₂₉ = 8 + 28(2)

a₂₉ = 8 + 56

a₂₉ = 64

Hence, the 29th term of the AP is 64.

9. If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?

Solution:

Let the first term be a and the common difference be d.

We are given a₃ = 4 and a₉ = −8.

a + 2d = 4     — (1)

a + 8d = −8     — (2)

Subtracting equation (1) from equation (2):

6d = −12

d = −2

Substituting the value of d into equation (1):

a + 2(−2) = 4

a − 4 = 4

a = 8

Let the nth term of this AP be zero, so a_n = 0.

a + (n − 1)d = 0

8 + (n − 1)(−2) = 0

−2(n − 1) = −8

n − 1 = 4

n = 5

Hence, the 5th term of the given AP is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

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

We know that a₁₇ = a + 16d and a₁₀ = a + 9d.

According to the given condition:

a₁₇ − a₁₀ = 7

(a + 16d) − (a + 9d) = 7

16d − 9d = 7

7d = 7

d = 1

Hence, the common difference of the AP is 1.

11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution:

For the given AP 3, 15, 27, 39, ..., the first term is a = 3 and the common difference is d = 15 − 3 = 12.

Let the required term be the nth term. Therefore, according to the question:

a_n = a₅₄ + 132

a + (n − 1)d = [a + 53d] + 132

Subtracting a from both sides:

(n − 1)12 = 53(12) + 132

(n − 1)12 = 636 + 132

(n − 1)12 = 768

n − 1 = 768 / 12

n − 1 = 64

n = 65

Hence, the 65th term will be 132 more than its 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Let the first term of the first AP be a₁ and that of the second AP be a₂. Let their identical common difference be d.

The 100th term of the first AP is a₁ + 99d and the 100th term of the second AP is a₂ + 99d.

The difference between their 100th terms is given as 100:

(a₁ + 99d) − (a₂ + 99d) = 100

a₁ − a₂ = 100     — (1)

Now, let us find the difference between their 1000th terms:

Difference = (a₁ + 999d) − (a₂ + 999d)

Difference = a₁ − a₂

From equation (1), we know that a₁ − a₂ = 100.

Hence, the difference between their 1000th terms is also 100.

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

Solution:

The smallest three-digit number divisible by 7 is 105, and the next numbers are 112, 119, and so on. The largest three-digit number divisible by 7 is 994.

This sequence forms an AP: 105, 112, 119, ..., 994.

Here, the first term is a = 105, the common difference is d = 7, and the last term is a_n = 994.

Using the general term formula:

a_n = a + (n − 1)d

994 = 105 + (n − 1)7

994 − 105 = 7(n − 1)

889 = 7(n − 1)

n − 1 = 889 / 7

n − 1 = 127

n = 128

Hence, there are 128 three-digit numbers divisible by 7.

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

Solution:

The smallest multiple of 4 greater than 10 is 12, and the largest multiple less than 250 is 248.

The sequence of multiples is given by the AP: 12, 16, 20, ..., 248.

Here, the first term is a = 12, the common difference is d = 4, and the last term is a_n = 248.

Using the general term formula:

a_n = a + (n − 1)d

248 = 12 + (n − 1)4

248 − 12 = 4(n − 1)

236 = 4(n − 1)n − 1 = 236 / 4n − 1 = 59n = 60Hence, there are 60 multiples of 4 that lie between 10 and 250.

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution:

For the first arithmetic progression 63, 65, 67, ..., the first term is a = 63 and the common difference is d = 65 − 63 = 2.

The nth term of this first progression is given by the formula:

a_n = 63 + (n − 1)2

For the second arithmetic progression 3, 10, 17, ..., the first term is A = 3 and the common difference is D = 10 − 3 = 7.

The nth term of this second progression is given by the formula:

A_n = 3 + (n − 1)7

According to the given condition, their respective nth terms are equal to each other:

63 + (n − 1)2 = 3 + (n − 1)7

63 − 3 = 7(n − 1) − 2(n − 1)

60 = 5(n − 1)

n − 1 = 60 / 5

n − 1 = 12

n = 13

Hence, the 13th terms of both the given arithmetic progressions are equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

Let the first term of the required progression be a and its common difference be d.

Given that the third term is 16, we can write our first linear relationship:

a₃ = a + 2d = 16     — (1)

We are also told that the 7th term exceeds the 5th term by a value of 12:

a₇ − a₅ = 12

(a + 6d) − (a + 4d) = 12

6d − 4d = 12

2d = 12

d = 6

Substituting this value of the common difference d = 6 back into equation (1):

a + 2(6) = 16

a + 12 = 16

a = 16 − 12

a = 4

The sequence terms are determined sequentially by adding the common difference to the first term:

First term = 4

Second term = 4 + 6 = 10

Third term = 10 + 6 = 16

Fourth term = 16 + 6 = 22

Hence, the required Arithmetic Progression is 4, 10, 16, 22, ...

17. Find the 20th term from the last term of the AP : 3, 8, 13, . . . , 253.

Solution:

For the given AP 3, 8, 13, ..., 253, the first term is a = 3, the common difference is d = 8 − 3 = 5, and the last term is l = 253.

To find the nth term from the last term, we use the formula: l − (n − 1)d.

Here, we need to find the 20th term from the end, so n = 20.

Substituting the values into our formula:

20th term from the last term = 253 − (20 − 1) × 5

20th term from the last term = 253 − 19 × 5

20th term from the last term = 253 − 95

20th term from the last term = 158

Hence, the 20th term from the last term of the given AP is 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

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

According to the first condition, the sum of the 4th and 8th terms is 24:

a₄ + a₈ = 24

(a + 3d) + (a + 7d) = 24

2a + 10d = 24

Dividing the equation by 2, we get:

a + 5d = 12     — (1)

According to the second condition, the sum of the 6th and 10th terms is 44:

a₆ + a₁₀ = 44

(a + 5d) + (a + 9d) = 44

2a + 14d = 44

Dividing the equation by 2, we get:

a + 7d = 22     — (2)

Subtracting equation (1) from equation (2):

(a + 7d) − (a + 5d) = 22 − 12

2d = 10

d = 5

Substituting d = 5 into equation (1):

a + 5(5) = 12

a + 25 = 12

a = 12 − 25

a = −13

The first three terms are determined sequentially by adding the common difference:

First term, a₁ = −13

Second term, a₂ = −13 + 5 = −8

Third term, a₃ = −8 + 5 = −3

Hence, the first three terms of the AP are −13, −8, and −3.

19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution:

The annual salary received by Subba Rao in subsequent years forms an Arithmetic Progression: 5000, 5200, 5400, ..., 7000.

Here, the first term is a = 5000, the common difference is d = 200, and the nth term representing the target salary is a_n = 7000.

Using the general term formula a_n = a + (n − 1)d:

7000 = 5000 + (n − 1) × 200

7000 − 5000 = 200(n − 1)

2000 = 200(n − 1)

n − 1 = 2000 / 200

n − 1 = 10

n = 11

Therefore, it takes 11 years for his salary to reach ₹ 7000.

The corresponding calendar year is calculated as: 1995 + (11 − 1) = 1995 + 10 = 2005.

Hence, Subba Rao's income reached ₹ 7000 in the year 2005.

20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Solution:

The weekly savings accumulated by Ramkali form an Arithmetic Progression where the first term is a = 5 and the common weekly increment is d = 1.75.

Given that her savings in the nth week become ₹ 20.75, we have a_n = 20.75.

Using the general term formula a_n = a + (n − 1)d:

20.75 = 5 + (n − 1) × 1.75

20.75 − 5 = (n − 1) × 1.75

15.75 = (n − 1) × 1.75

n − 1 = 15.75 / 1.75

n − 1 = 1575 / 175

n − 1 = 9

n = 9 + 1

n = 10

Hence, the value of n is 10.