CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Question 3
In the following AP’s find the missing terms:
(i) 2, __ , 26
(ii) __, 13, __, 3
(iii) 5, __, __,
(iv) -4. __, __, __, __, 6
(v) __, 38, __, __, __, -22
Solution (i)
2, __ , 26
We know that difference between consecutive terms is constant in any Arithmetic progression. (1)
Let the missing term be
.
Using property (1), we can say that
Therefore, missing term is
.
Solution (ii)
__, 13, __, 3
Let missing terms be
and
.
The sequence becomes:
We know that difference between consecutive terms is constant in any Arithmetic progression. (1)
Using property (1), we can say that
Also, using property (1), we can say that
But, we have
, Putting value of
in the above equation, we get
Therefore, missing terms are 18 and 8.
Solution (iii)
We can also use the following method to solve similar problems. It is also easy. The first two questions can also be solved by the same method.
5, __, __,
Here, first term =
And, 4th term =
Using formula
, to find nth term of arithmetic progression, we get
Therefore, we get common difference = d =
Second term = a+ d =
Third term = second term + d =
Therefore, missing terms are
and
Solution (iv)
-4. __, __, __, __, 6
Here, First term = a = -4
6th term =
= 6
Using formula
, to find nth term of arithmetic progression, we get
Therefore, common difference = d = 2
Second term = first term + d = a+ d = -4 + 2 = -2
Third term = second term + d = -2 + 2 = 0
Fourth term = third term + d = 0 + 2 = 2
Fifth term = fourth term + d = 2 + 2 = 4
Therefore, missing terms are -2, 0, 2 and 4.
Solution (v)
__, 38, __, __, __, -22
In this problem, we are given 2nd and 6th term.
Using formula
, to find nth term of arithmetic progression, we get
And
And
These are equations in two variables, we can solve them using any method. Lets solve them using substitution method.
Using equation (
), we can say that
. Putting value of
in equation (
), we get
Using this value of
and putting this in equation
, we get
Therefore, we get
and
First term = a = 53
Third term = second term + d = 38 – 15 = 23
Fourth term = third term + d = 23 – 15 = 8
Fiftht term = fourth term + d = 8 – 15 = -7
Therefore, missing terms are 53, 23, 8 and -7.
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