If a sequence is arithmetic, the difference between any two consecutive terms will be same along the sequence.
Let t_{1}, t_{2}, t_{3}, t_{4}, ............ be a sequence.
In the above sequence, if the difference between any two consecutive terms is 'd' along the sequence, then the sequence is arithmetic.
d = t_{2 }-ta_{1}
d = t_{3 }- t_{2}
d = t_{4 }- t_{3}
The difference 'd' is called common difference.
Question 1 :
Check whether the following sequences are in A.P.
(i) a - 3, a - 5, a -7,...
Solution :
d = t_{2} - t_{1} d = (a - 5) - (a - 3) = a - 5 - a + 3 d = -2 |
d = t_{3} - t_{2} d = (a - 7) - (a - 5) = a - 7 - a + 5 d = -2 |
Since the common difference area same, the given sequence is arithmetic progression.
(ii) 1/2, 1/3, 1/4, 1/5........
Solution :
d = t_{2} - t_{1} d = (1/3) - (1/2) = (2 - 3)/6 d = -1/6 |
d = t_{3} - t_{2} d = (1/4) - (1/3) = (3 - 4)/12 d = -1/12 |
The common differences are not equal. Hence the given sequence is not A.P.
(iii) 9, 13, 17, 21, 25,...
Solution :
d = t_{2} - t_{1} d = 13 - 9 d = 4 |
d = t_{3} - t_{2} d = 17 - 13 d = 4 |
The given sequence is arithmetic progression.
(iv) -1/3, 0, 1/3, 2/3.........
Solution :
d = t_{2} - t_{1} d = 0 - (-1/3) d = 1/3 |
d = t_{3} - t_{2} d = (1/3) - 0 d = 1/3 |
(v) 1, -1, 1, -1, 1, -1,............
Solution :
d = t_{2} - t_{1} d = -1 - 1 d = -2 |
d = t_{3} - t_{2} d = 1 - (-1) = 1 + 1 = 2 |
The given sequence is not arithmetic progression.
Question 2 :
First term a and common difference d are given below. Find the corresponding A.P.
(i) a = 5 , d = 6
Solution :
General form of A.P
a, a + d, a + 2d,...........
5, (5+6), (5, + 2(6)), ......................
5, 11, 17, ...................
(ii) a = 7 , d = -5
Solution :
a, a + d, a + 2d,...........
a = 7
a + d = 7 + (-5) = 2
a + 2d = 7 + 2(-5) = 7 - 10 = -3
The required sequence is 7, 2, -3, .................
(iii) a = 3/4, d = 1/2
Solution :
a = 3/4
a + d = (3/4) + (1/2) = (3+2)/4 = 5/4
a + d = (3/4) + 2(1/2) = (3/4) + 1 = 7/4
Hence the required sequence is 3/4, 5/4, 7/4,...............
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