Given 2, 4, 6, 8, . . .
Here, the first term, a1 = 2
Second term, a2 = 4
And a3 = 6
Now, common difference = a2 – a1 = 4 – 2 = 2
Also, a3 – a2 = 6 – 4 = 2
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 2.
(2) 2, 5/2, 3, 7/3, . . .
Solution:
(3) – 10, – 6, – 2, 2, . . .
Solution:
Given – 10, – 6, – 2,2, . . .
Here, the first term, a1 = – 10
Second term, a2 = – 6
a3 = – 2
Now, common difference = a2 – a1 = – 6 – (– 10) = – 6 + 10 = 4
Also, a3 – a2 = – 2 – (– 6) = – 2 + 6 = 4
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 4.
(4) 0.3, 0.33, .0333, . . .
Solution:
Given 0.3, 0.33, 0.333, . . .
Here, the first term, a1 = 0.3
Second term, a2 = 0.33
a3 = 0.333
Now, common difference = a2 – a1 = 0.33 – 0.3 = 0.03
Also, a3 – a2 = 0.333 – 0.33 = 0.003
Since, the common difference is not same.
Hence the terms are not in Arithmetic progression
(5) 0, – 4, – 8, – 12, . . .
Solution:
Given 0, – 4, – 8, – 12, . . .
Here, the first term, a1 = 0
Second term, a2 = – 4
a3 = – 8
Now, common difference = a2 – a1 = – 4 – 0 = – 4
Also, a3 – a2 = – 8 – (– 4) = – 8 + 4 = – 4
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = – 4.
(6) -1/5, -1/5, – 1/5, . . .
Solution:
(7) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ….
Solution:
Given
3, 3 + √2, 3 + 2√2, 3 + 3√2, ….
Here, the first term, a1 = 3
Second term, a2 = 3 + √2
a3 = 3 + 2√2
Now, common difference = a2 – a1 = 3 + √2 – 3 = √2
Also, a3 – a2 = 3 + 2√2 – (3 + √2) = 3 + 2√2 – 3 – √2 = √2
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = √2.
(8) 127, 132, 137, . . .
Solution:
Given 127, 132, 137, . . .
Here, the first term, a1 = 127
Second term, a2 = 132
a3 = 137
Now, common difference = a2 – a1 = 132 – 127 = 5
Also, a3 – a2 = 137 – 132 = 5
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 5.