The document examines arithmetic progressions (APs) and determines which number sequences provided are APs. It justifies the classifications by showing the constant difference between consecutive terms. It then provides the first five terms for six given APs where the first term (a) and common difference (d) are specified.

1. EX. NO. 2

2. 1. Which of the following lists of numbers
are Arithmetic Progressions? Justify.
(i) 1, 3, 6, 10, .....
Let,
t1 = 1, t2 = 3, t3 = 6, t4 = 10
t2 – t1 = 3 – 1 = 2
t3 – t2 = 6 – 3 = 3
t4 – t3 = 10 – 6 = 4
As, The difference between two consecutive
terms is not constant.
So, The sequence is not an A.P.

3. ii. 3, 5, 7, 9, 11, .....
Sol. Let, t1 = 3, t2 = 5, t3 = 7, t4 = 9, t5 =
11
t2 – t1 = 5 – 3 = 2
t3 – t2 = 7 – 5 = 2
t4 – t3 = 9 – 7 = 2
t5 – t4 = 11 – 9 = 2
As, The difference between two
consecutive terms is 2 which is constant.
So, The sequence is an A.P.

4. (iii) 1, 4, 7, 10, ....
Sol. Let, t1 = 1, t2 = 4, t3 = 7, t4 = 10
t2 – t1 = 4 – 1 = 3
t3 – t2 = 7 – 4 = 3
t4 – t3 = 10 – 7 = 3
As, The difference between two
consecutive terms 3 which is
constant.
So, The sequence is an A.P.

5. (iv) 3, 6, 12, 24, ....
Sol. t1 = 3, t2 = 6, t3 = 12, t4 = 24
t2 – t1 = 6 – 3 = 3
t3 – t2 = 12 – 6 = 6
t4 – t3 = 24 – 12 = 12
Since, The difference between
two consecutive terms is not
constant.
So, The sequence is not an A.P.

6. (v) 22, 26, 28, 31, ...
Sol. t1 = 22, t2 = 26, t3 = 28, t4 = 31
t2 – t1 = 26 – 22 = 4
t3 – t2 = 28 – 26 = 2
t4 – t3 = 21 – 28 = 3
Since, The difference between two
consecutive terms is not constant.
So, The sequence is not an A.P.

7. (vi) 0.5, 2, 3.5, 5, ...
Sol. t1 = 0.5, t2 = 2, t3 = 3.5, t4 = 5
t2 – t1 = 2 – 0.5 = 1.5
t3 – t2 = 3.5 – 2 = 1.5
t4 – t3 = 5 – 3.5 = 1.5
As, The difference between two
consecutive terms is 1.5 which is
constant.
So, The sequence is an A.P.

8. (vii) 4, 3, 2, 1, ....
Sol. t1 = 4, t2 = 3, t3 = 2, t4 = 1,
t2 – t1 = 3 – 4 = – 1
t3 – t2 = 2 – 3 = – 1
t4 – t3 = 1 – 2 = – 1
As, The difference between two
consecutive terms is –1 which is
constant.
So, The sequence is an A.P.

9. (viii) – 10, – 13, – 16, – 19, .....
Sol. t1 = – 10, t2 = – 13, t3 = – 16, t4 = –
19
t2 – t1 = – 13 – (– 10) = – 13 + 10 = – 3
t3 – t2 = – 16 – (– 13) = – 16 + 13 = – 3
t4 – t3 = – 19 – (– 16) = – 19 + 16 = – 3
As, The difference between two
consecutive terms is – 3 which is
constant. So, The sequence is an A.P.

10. 2. Write the first five terms of the following
Arithmetic Progressions where,
the common difference ‘d’ and the first term ‘a’ are
given :
(i) a = 2, d = 2.5
Sol. a = 2, d = 2.5
Here, t1 = a = 2
t2 = t1 + d = 2 + 2.5 = 4.5
t3 = t2 + d = 4.5 + 2.5 = 7
t4 = t3 + d = 7 + 2.5 = 9.5
t5 = t4 + d = 9.5 + 2.5 = 12
Thus, The first five terms of the A.P. are 2, 4.5, 7,
9.5 and 12.

11. (ii) a = 10, d = – 3
Sol. a = 10, d = – 3
Here, t1 = a = 10
t2 = t1 + d = 10 + (– 3) = 10 – 3 = 7
t3 = t2 + d = 7 + (– 3) = 7 – 3 = 4
t4 = t3 + d = 4 + (– 3) = 4 – 3 = 1
t5 = t4 + d = 1 + (– 3) = 1 – 3 = – 2
Thus, The first five terms of the
A.P. are 10, 7, 4, 1 and – 2.

12. (iii) a = 4, d = 0
Sol. a = 4, d = 0
Here, t1 = a = 4
t2 = t1 + d = 4 + 0 = 4
t3 = t2 + d = 4 + 0 = 4
t4 = t3 + d = 4 + 0 = 4
t5 = t4 + d = 4 + 0 = 4
Thus, The first five terms of the
A.P. are 4, 4, 4, 4 and 4.

13. (iv) a = 5, d = 2
Sol. a = 5, d = 2
Here, t1 = a = 5
t2 = t1 + d = 5 + 2 = 7
t3 = t2 + d = 7 + 2 = 9
t4 = t3 + d = 9 + 2 = 11
t5 = t4 + d = 11 + 2 = 13
Thus, The first five terms of the
A.P. are 5, 7, 9, 11 and 13.

14. (v) a = 3, d = 4
Sol. a = 3, d = 4
Here, t1 = a = 3
t2 = t1 + d = 3 + 4 = 7
t3 = t2 + d = 7 + 4 = 11
t4 = t3 + d = 11 + 4 = 15
t5 = t4 + d = 15 + 4 = 19
Thus, The first five terms of the
A.P. are 3, 7, 11, 15 and 19.

15. (vi) a = 6, d = 6
Sol. a = 6, d = 6
Here, t1 = a = 6
t2 = t1 + d = 6 + 6 = 12
t3 = t2 + d = 12 + 6 = 18
t4 = t3 + d = 18 + 6 = 24
t5 = t4 + d = 24 + 6 = 30
Thus,The first five terms of A.P. are
6, 12, 18, 24 and 30.