In the following APs, find the missing terms in the boxes :
(i)
(ii)
(iii)
(iv)
(v)
(i) We know: In AP, middle term is average of the other two terms
Hence, middle term = (2 + 26)/2 = 28/2 = 14
Thus, above AP can be written as 2, 14, 26
(ii) The middle term between 13 and 3 will be;
(13 + 3)/2 = 16/2 = 8
Now, a4 – a3 = 3 – 8 = - 5
a3 – a2 = 8 – 13 = - 5
Thus, a2 – a1 = - 5
Or, 13 – a1 = - 5
Or, a1 = 13 + 5 = 18
Thus, above AP can be written as 18, 13, 8, 3
(iii) We have, a = 5 and a4 = 91/2
Now common difference:
a4 = a + 3d
= 5 +
d =
Hence, using d, 2nd term and 3rd term can be calculated as:
a2 = a + d
= 5 +
=
a3 = a + 2d
= 5 +
= 8
Therefore, the A.P. can be written as:
5,
(iv) Here, a = - 4 and a6 = 6
Common difference:
a6 = a + 5d
6 = -4 + 5d
5d = 6 + 4 = 10
d = 2
The second, third, fourth and fifth terms of this AP are:
a2 = a + d = - 4 + 2 = - 2
a3 = a + 2d = - 4 + 4 = 0
a4 = a + 3d = - 4 + 6 = 2
a5 = a + 4d = - 4 + 8 = 4
Thus, the given AP can be written as: - 4, - 2, 0, 2, 4, 6
Let us take 38 as the first term and – 22 as the 5th term
Using this, common difference can be calculated as follows:
a5 = a + 4d
= 38 + 4d
4d = - 22 – 38 = - 60
d = - 15
If 38 is the second term, then first term:
a = a2 – d = 38 + 15 = 53
Third, fourth and fifth terms ar:
a3 = a + 2d = 53 + 2(- 15) = 53 – 30 = 23
a4 = a + 3d = 53 – 45 = 8
a5 = a + 4d = 53 – 60 = - 7
So, the AP can be written as: 53, 38, 23, 8, - 7, - 22