Goto Page 1,Ex-5.2
Question 3:
I. 2,
, 26
For this A.P.,
a = 2, a3 = 26
Since, an = a + (n − 1) d
a3 = 2 + (3 − 1) d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2 − 1) 12
= 14
Therefore, 14 is the missing term.
II.
Given
a2 = 13 and, a4 = 3
Since, an = a + (n − 1) d
a2 = a + (2 − 1) d
13 = a + d (I)
a4 = a + (4 − 1) d
3 = a + 3d (II)
On subtracting (I) from (II),
−10 = 2d
d = −5
From equation (I),
13 = a + (−5)
a = 18
a3 = 18 + (3 − 1) (−5)
= 18 + 2 (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
III.
Given
a = 5
Since, an = a + (n − 1) d
Therefore, the missing terms are 13/2 and 8 respectively.
IV.
Given
a = −4 and a6 = 6
Since, an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
Given
a2 = 38
a6 = −22
Since, an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d …………………………..(1)
a6 = a + (6 − 1) d
−22 = a + 5d ………………………(2)
On subtracting equation (1) from (2),
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
NCERT MATHS CLASS - X
Arithmetic Progressions
Exercise 5.2
I. 2,
, 26For this A.P.,
a = 2, a3 = 26
Since, an = a + (n − 1) d
a3 = 2 + (3 − 1) d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2 − 1) 12
= 14
Therefore, 14 is the missing term.
II.
Given
a2 = 13 and, a4 = 3
Since, an = a + (n − 1) d
a2 = a + (2 − 1) d
13 = a + d (I)
a4 = a + (4 − 1) d
3 = a + 3d (II)
On subtracting (I) from (II),
−10 = 2d
d = −5
From equation (I),
13 = a + (−5)
a = 18
a3 = 18 + (3 − 1) (−5)
= 18 + 2 (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
III.
Given
a = 5
Since, an = a + (n − 1) d
Therefore, the missing terms are 13/2 and 8 respectively.
IV.
Given
a = −4 and a6 = 6
Since, an = a + (n − 1) d
a6 = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 (2) = 0
a4 = a + 3d = − 4 + 3 (2) = 2
a5 = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
Given
a2 = 38
a6 = −22
Since, an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d …………………………..(1)
a6 = a + (6 − 1) d
−22 = a + 5d ………………………(2)
On subtracting equation (1) from (2),
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.