In the following APs find the missing term in the boxes.
(i) For this A.P.,
a = 2
a3 = 26
We know that, 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) For this A.P.,
a2 = 13 and
a4 = 3
We know that, 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), we get
- 10 = 2d
d = - 5
From equation (i), we get
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) For this A.P.,
a = 5 and
a4 = 19/2
We know that, an = a + (n − 1) d
a4 = a + (4 - 1) d
19/2 = 5 + 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2
a2 = a + (2 - 1) d
a2 = 5 + 3/2
a2 = 13/2
a3 = a + (3 - 1) d
a3 = 5 + 2×3/2
a3 = 8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For this A.P.,
a = −4 and
a6 = 6
We know that,
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.
(v)
For this A.P.,
a2 = 38
a6 = −22
We know that
an = a + (n − 1) d
a2 = a + (2 − 1) d
38 = a + d ... (i)
a6 = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 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.
