Solve the following arithmetic progression problems:

1. The 7th term of an AP is -39/12 and the 15th term is -103/12. What is the 27th term?

A. -187/12

B. -191/12

C. -199/12

D. -205/12

Sol : Option C
The 7th term is a + 6d = -39/12
The 15th term is a + 14d = -103/12
Equating these terms with their values and solving as simultaneous equations, we get a = 3/4 and d = – 2/3.So the 27th term is (a + 26d) = 3/4 + (-52/3) = -199/12.

2. In an AP of 21 terms, the sum of the first 3 terms is – 33 and that of the middle 3 is 75. What is the sum of the AP?

Sol : Option C
The AP can be expressed as a, (a + d), ---, (a +20d). The sum of the first 3 terms is (3a + 3d) = -33 and the sum of the middle 3 terms is (3a + 30d) = 75. Solving these two equations, we get a = -15 and d = 4. The sum of the 1st 21 terms is (21/2) (2 * -15 + 20 * 4) = 525.

Q.3. The 6th term of an AP is 6 and the 16th term is 14. What is the 27th term?

A. 106/5

B. 22/5

C. 118/5

D. 114/5

Sol : Option D
The 6th term is a + 5d = 6
The 16th term is a + 15d = 14
Equating these terms with their values and solving as simultaneous equations, we get a = 2 and d = 4/5. So the 27th term is (a + 26d) = 2 + (104/5) = 114/5.

Q.4. In an AP, the ratio of the 2nd term to the 7th term is 1/3. If the 5th term is 11, what is the 15th term?

A. 28

B. 31

C. 33

D. 36

Sol : Option B
The 2nd and the 7th terms are (a + d) and (a + 6d) respectively. The ratio of these terms is 1/3. Solving this ratio, we get 2a = 3d. The 5th term is (a + 4d) = 11. Substituting for a, we get a = 3 and d = 2. Therefore, the 15th term is (a + 14d) = 31.

Q.5. The sum of the first 3 terms in an AP is 6 and that of the last 3 is 16. If the AP has 13 terms, what is the sum of the middle three terms?

A. 7

B. 9

C. 11

D. 13

Sol : Option C
The AP can be expressed as a, (a + d), ---, (a +12d). The sum of the first 3 terms is (3a + 3d) = 6 and that of the last 3 is (3a + 33d) = 16. Solving these equations, we get a = 5/3 and d = 1/3. The sum of the middle 3 terms is (3a + 18d) = 11.

Q.6. In an AP, the ratio of the 7th term to the 10th term is -1. If the 16th term is -15, what is the 3rd term?

A. 11

B. 13

C. -11

D. -13

Sol : Option A
The 7th and the 10th terms are (a + 6d) and (a + 9d) respectively. The ratio of these terms is -1. Solving this ratio, we get -2a = 15d. The 16th term is (a + 15d) = -15. Substituting for a, we get a = 15 and d = -2. Therefore, the 3rd term is (a + 2d) = 11.

Q.7. In an AP, the sum of the first 3 terms is -36 and that of the last 3 is 27. If there are 10 terms, what are the 1 st term and the common difference respectively?

A. 15, 3

B. -15, 3

C. 15, -3

D. -15, -3

Sol : Option B
The AP can be expressed as a, (a + d), ---, (a + 9d). The sum of the first 3 terms is (3a + 3d) = -36 and the sum of the last 3 terms is (3a + 24d) = 27. Solving these two equations, we get a = -15 and d = 3.

Q8. In an AP, the sum of the first 3 terms is – 60 and that of the last 3 are 84. If there are 15 terms, what is the sum of the middle 3 terms?

A. 8

B. 12

C. 16

D. 24

Sol : Option B
Since the number of terms is odd, the AP can be expressed as (a – 7d), (a – 6d), ---, a, ---, (a + 6d), (a + 7d). The sum of the first 3 terms is (3a – 18d) = -60 and the sum of the last 3 terms is (3a + 18d) = 84. Solving these two equations, we get a = 4 and d = 4. Since the middle term is 4, the sum of the middle 3 terms is 12.

Q9. In an AP, the ratio of the 2nd term to the 6th term is 2/5. If the 8th term is 26, what is the 10th term?

A. 28

B. 29

C. 32

D. 33

Sol : Option C
The 2nd and the 6th terms are (a + d) and (a + 5d) respectively. The ratio of these terms is 2/5. Solving this ratio, we get 3a = 5d. The 8th term is (a + 7d) = 26. Substituting for a, we get a = 5 and d = 3. Therefore, the 10th term is (a + 9d) = 32

Q10. The sum of the first 3 terms in an AP is 51 and that of the last 3 is 99. If the AP has 11 terms, what is the arithmetic mean of the middle 3 terms?

A. 17

B. 19

C. 21

D. 25

Sol : Option D
The AP can be expressed as a, (a + d), ---, (a +10d). The sum of the first 3 terms is (3a + 3d) = 51 and that of the last 3 is (3a + 27d) = 99. Solving these equations, we get a = 15 and d = 2. The arithmetic mean of the 3 terms is the value of the middle term = 25.