Q. 33.8( 237 Votes )

In the following APs, find the missing terms in the boxes :

(i) 2 , square , 26

(ii) square square , 3

(iii) 5 , square square , 9 1/2

(iv) -4 , square , square , square ,

(v) square , 38 , square , square , square ,-22

Answer :

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Let three terms a, b and c are in A.P

Therefore, b - a = c - b

Rearranging we get,  



And also nth term of an AP is given by 

an = a + (n - 1)d
 where, a = first term
n = number of terms
d = common difference



(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) We know: In AP, middle term is average of the other two terms
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
We know, nth term of an AP is

an = a + (n - 1)d

where a and d are first term and common difference respectively.


Now common difference:


a4 = a + 3 d

= 5 + 3d



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:


(iv) Here, a = - 4 and a6 = 6

We know, nth term of an AP is

an = a + (n - 1)d

where a and d are first term and common difference respectively.


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



(v) Given: Second term = 38 and sixth term = -22
So,

a + d = 38.......(1)
a = 38 - d
a + 5d = -22.....(2)
Putting the value of a in equation 2, we get,

38 - d + 5d = -22
38 + 4d = -22
4d = -22 - 38
4d = -60
d = -15

Putting the value of d in equation 1, we get,
a - 15 = 38
a = 38 + 15 = 53
Therefore, the series is
53, 38, 23, 8, -7, -22




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