### Problem 1: Show that the sequence defined by a_{n} = 5n – 7 is an A.P., find its common difference.

**Solution:**

Given:a

_{n}= 5n – 7Now putting n = 1, 2, 3, 4,5 we get,

a

_{1}= 5.1 – 7 = 5 – 7 = -2a

_{2}= 5.2 – 7 = 10 – 7 = 3a

_{3}= 5.3 – 7 = 15 – 7 = 8a

_{4}= 5.4 – 7 = 20 – 7 = 13We can see that,

a

_{2}– a_{1}= 3 – (-2) = 5a

_{3}– a_{2}= 8 – (3) = 5a

_{4}– a_{3}= 13 – (8) = 5Since, the successive difference of list is same i.e 5

∴ The given sequence is in A.P and have common difference of 5

### Problem 2: Show that the sequence defined by a_{n} = 3n^{2} – 5 is not an A.P.

**Solution:**

Given:a

_{n}= 3n^{2}– 5Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 3.1.1 – 5= 3 – 5 = -2a

_{2}= 3.2.2 – 5 = 12 – 5 = 7a

_{3}= 3.3.3 – 5 = 27 – 5 = 22a

_{4}= 3.4.4 – 5 = 48 – 5 = 43We can see that,

a

_{2}– a_{1}= 7 – (-2) = 9a

_{3}– a_{2}= 22 – 7 = 15a

_{4}– a_{5}= 43 – 22 = 21Since, the successive difference of list is not the same

∴ The given sequence is not in A.P

### Problem 3: The general term of a sequence is given by a_{n} = -4n + 15. Is the sequence an A.P.? If so, find its 15th term and the common difference.

**Solution:**

Given:a

_{n}= -4n + 15Now putting n = 1, 2, 3, 4 we get,

a

_{1}= -4.(1) + 15 = -4 + 15 = 11a

_{2}= -4.(2) + 15 = -8 + 15 = 7a

_{3}= -4.(3) + 15 = -12 + 15 = 3a

_{4}= -4.(4) + 15 = -16 + 15 = -1We can see that,

a

_{2}– a_{1}= 7 – (11) = -4a

_{3}– a_{2}= 3 – 7 = -4a

_{4}– a_{3}= -1 – 3 = -4Since, the successive difference of list is same i.e -4

∴ The given sequence is in A.P and have common difference of -4Hence, the 15th term will be

a

_{15}= -4(15) + 15 = -60 + 15 = -45

And, a_{15 }= -45

### Problem 4: Write the sequence with nth term :

### (i) a_{n} = 3 + 4n

**Solution:**

Given:a

_{n}= 3 + 4nNow putting n = 1, 2, 3, 4 we get,

a

_{1}= 3 + 4.1 = 7a

_{2}= 3 + 4.2 = 11a

_{3}= 3 + 4.3 = 15a

_{4}= 3 + 4.4 = 19

∴ The sequence is 7, 11, 15, 19We can see that,

a

_{2}– a_{1}= 11 – (7) = 4a

_{3}– a_{2}= 15 – (11) = 4a

_{4}– a_{3}= 19 – (15) = 4Since, the successive difference of list is same i.e 4

∴ The given sequence is in A.P

### (ii) a_{n} = 5 + 2n

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 5 + 2.1 = 7a

_{2}= 5 + 2.2 = 9a

_{3}= 5 + 2.3 = 11a

_{4}= 5 + 2.4 = 13

∴ The sequence is 7, 9, 11, 13We can see that,

a

_{2}– a_{1}= 9 – (7) = 2a

_{3}– a_{2}= 11 – (9) = 2a

_{4}– a_{3}= 13 – (11) = 2Since, the successive difference of list is same i.e 2

∴ The given sequence is in A.P

### (iii) a_{n} = 6 – n

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 6 – 1 = 5a

_{2}= 6 – 2 = 4a

_{3}= 6 – 3 = 3a

_{4}= 6 – 4 = 2

∴ The sequence is 5, 4, 3, 2We can see that,

a

_{2}– a_{1}= 4 – (5) = -1a

_{3}– a_{2}= 3 – (4) = -1a

_{4}– a_{3}= 2 – (3) = -1Since, the successive difference of list is same i.e -1

∴ The given sequence is in A.P

### (iv) a_{n} = 9 – 5n

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 9 – 5.1 = 4a

_{2}= 9 – 5.2 = -1a

_{3}= 9 – 5.3 = -6a

_{4}= 9 – 5.4 = -11

∴ The sequence is 4, -1, -6, -11We can see that,

a

_{2}– a_{1}= -1 – (4) = -5a

_{3}– a_{2}= -6 – (-1) = -5a

_{4}– a_{3}= -11 – (-6) = -5Since, the successive difference of list is same i.e -5

∴ The given sequence is in A.P

### Problem 5: The nth term of an A.P. is 6n + 2. Find the common difference.

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 6.1 + 2 = 8a

_{2}= 6.2 + 2 = 14a

_{3}= 6.3 + 2 = 20a

_{4}= 6.4 + 2 = 26We can see that,

a

_{2}– a_{1}= 14 – (8) = 6a

_{3}– a_{2}= 20 – (14) = 6a

_{4}– a_{3}= 26 – (20) = 6

Hence, the common difference is 6

### Problem 6: Justify whether it is true to say that the sequence, having following nth term is an A.P.

**(i) a**_{n} = 2n – 1

_{n}= 2n – 1

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 2.1 – 1 = 1a

_{2}= 2.2 – 1 = 3a

_{3}= 2.3 – 1 = 5a

_{4}= 2.4 – 1 = 7We can see that,

a

_{2}– a_{1}= 3 – (1) = 2a

_{3}– a_{2}= 5 – (3) = 2a

_{4}– a_{3}= 7 – (5) = 2Since, the successive difference of list is same i.e 2

Hence, the given sequence is in A.P

### (ii) a_{n} = 3n² + 5

**Solution:**

Now putting n = 1, 2, 3, 4 we get,

a

_{1}= 3.1.1 + 5 = 8a

_{2}= 3.2.2 + 5 = 17a

_{3}= 3.3.3 + 5 = 32a

_{4}= 3.4.4 + 5 = 53We can see that,

a

_{2}– a_{1}= 17 – (8) = 9a

_{3}– a_{2}= 32 – (17) = 15a

_{4}– a_{3}= 53 – (32) = 21Since, the successive difference of list is not the same

Hence, the given sequence is not in A.P

### (iii) a_{n} = 1 + n + n²

**Solution:**

Now putting n = 1, 2, 3 we get,

a

_{1}= 1 + 1 + 1.1 = 3a

_{2}= 1 + 2 + 2.2 = 7a

_{3}= 1 + 3 + 3.3 = 13We can see that,

a

_{2}– a_{1}= 7 – (3) = 4a

_{3}– a_{2}= 13 – (7) = 6Since, the successive difference of list is not the same

Hence, the given sequence is not in A.P