In the following AP’s find the missing terms:
(i) 2, __ , 26
(ii) __, 13, __, 3
(iii) 5, __, __, \({9}{\dfrac{1}{2}}\)
(iv) –4. __, __, __, __, 6
(v) __, 38, __, __, __, –22

(i) 2, __ , 26

We know that difference between consecutive terms is equal in any A.P.

Let the missing term be x.

\(\Rightarrow \) x – 2 = 26 – x
\(\Rightarrow \) 2x = 28
\(\Rightarrow \) x = 14

Therefore, missing term is 14.


(ii)__, 13, __, 3

Let missing terms be x and y.

The sequence becomes x, 13, y, 3

We know that difference between consecutive terms is constant in any A.P.

\(\Rightarrow \) y – 13 = 3 – y
\(\Rightarrow \) 2y = 16
\(\Rightarrow \) y = 8

And 13 – x = y – 13
\(\Rightarrow \) x + y = 26
But, we have y = 8,
\(\Rightarrow \) x + 8 = 26
\(\Rightarrow \) x = 18

Therefore, missing terms are 18 and 8.


(iii) 5, __, __, \({9}{\dfrac{1}{2}}\)

Here, first term = a = 5 And,
4th term = \(a_4 = {9}{\dfrac{1}{2}}\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_4\) = 5 + (4 - 1) d
\(\Rightarrow {{19} \over {2}}\) = 5 + 3d
\(\Rightarrow \) 19 = 2 (5 + 3d)
\(\Rightarrow \) 19 = 10 + 6d
\(\Rightarrow \) 6d = 19 – 10
\(\Rightarrow \) 6d = 9
\(\Rightarrow \) d = \({{3} \over {2}}\)

Therefore, we get common difference
= d = \({{3} \over {2}}\)

Second term
= a + d = \(5 + {{3} \over {2}}
= {{13} \over {2}}\)

Third term = second term + d
= \({{13} \over {2}} + {{3} \over {2}} = {{16} \over {2}} = 8\)

Therefore, missing terms are \({{13} \over {2}}\) and 8


(iv)–4. __, __, __, __, 6

Here, First term = a = –4 and 6th term = \(a_6 = 6\)

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(\Rightarrow a_6\) = -4 + (6 - 1) d
\(\Rightarrow \) 6 = -4 + 5d
\(\Rightarrow \) 5d = 10
\(\Rightarrow \) d = 2

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.


(v) __, 38, __, __, __, –22

We are given 2nd and 6th term.

Using formula \(a_n = a + (n - 1)d\) , to find nth term of arithmetic progression,

\(a_2\) = a + (2 - 1) d And \(a_6 \) = a + (6 - 1) d
38 = a + d And -22 = a + 5d

These are equations in two variables, we can solve them using any method.

Using equation (38 = a + d), we can say that a = 38 - d.

Putting value of a in equation (-22 = a + 5d),

\(\Rightarrow\) -22 = 38 – d + 5d
\(\Rightarrow \) 4d = -60
\(\Rightarrow \) d = -15

Using this value of d and putting this in equation 38 = a + d,

38 = a – 15\(\Rightarrow \) a = 53

Therefore, we get a = 53 and d = -15

First term = a = 53
Third term = second term + d
= 38 – 15 = 23
Fourth term = third term + d
= 23 – 15 = 8
Fifth term = fourth term + d
= 8 – 15 = –7

Therefore, missing terms are 53, 23, 8 and –7.