In the following APs, find the missing terms :

(i) 2, ……, 26.

(ii) ……, 13, ……, 3.

(iii) 5, ……, ……., $9\dfrac{1}{2}$

(iv) -4, ……, ….., ….., ……, 6

(v) ……, 38, ……, ….., ……, -22

(i) Let x be the missing term.

A.P. : 2, x, 26

Since, above list is in A.P. so common difference will be equal among consecutive numbers.

∴ x - 2 = 26 - x

⇒ x + x = 26 + 2

⇒ 2x = 28

⇒ x = $\dfrac{28}{2}$ = 14.

Hence, missing term = 14.

(ii) Let x and y be the missing term.

A.P. : x, 13, y, 3.

Since, above list is in A.P. so common difference will be equal among consecutive numbers.

∴ 3 - y = y - 13

⇒ y + y = 13 + 3

⇒ 2y = 16

⇒ y = $\dfrac{16}{2}$

⇒ y = 8.

Common difference (d) : y - 13 = 8 - 13 = -5.

⇒ 13 - x = d

⇒ 13 - x = -5

⇒ x = 13 + 5 = 18

Hence, missing terms are 18 and 8.

(iii) Let x and y be the missing term.

A.P. : 5, x, y, $9\dfrac{1}{2}$

In above A.P.,

First term (a) = 5

Fourth term (a₄) = $9\dfrac{1}{2}$

By formula,

a_n = a + (n - 1)d

$\Rightarrow a_4 = 5 + (4 - 1) \times d \\[1em] \Rightarrow 9\dfrac{1}{2} = 5 + 3d \\[1em] \Rightarrow \dfrac{19}{2} - 5 = 3d \\[1em] \Rightarrow 3d = \dfrac{19- 10}{2} \\[1em] \Rightarrow 3d = \dfrac{9}{2} \\[1em] \Rightarrow d = \dfrac{9}{2} \times \dfrac{1}{3} \\[1em] \Rightarrow d = \dfrac{3}{2}.$

In above A.P.

x = 2nd term = a + d = $5 + \dfrac{3}{2} = \dfrac{10 + 3}{2} = \dfrac{13}{2} = 6\dfrac{1}{2}$.

y = 3rd term = a + 2d = $5 + 2 \times \dfrac{3}{2}$ = 5 + 3 = 8.

Hence, missing terms are $6\dfrac{1}{2}, 8$.

(iv) By formula,

a_n = a + (n - 1)d

First term (a) = -4

Sixth term (a₆) = 6

⇒ a + (6 - 1)d = 6

⇒ -4 + 5d = 6

⇒ 5d = 6 + 4

⇒ 5d = 10

⇒ d = $\dfrac{10}{5}$ = 2.

Second term = a + d = -4 + 2 = -2

Third term = a + 2d = -4 + 4 = 0

Fourth term = a + 3d = -4 + 6 = 2

Fifth term = a + 4d = -4 + 8 = 4

Hence, the missing terms are -2, 0, 2, 4.

(v) By formula,

a_n = a + (n - 1)d

Let the first term be a and common difference be d.

Given,

Second term = 38

⇒ a + (2 - 1)d = 38

⇒ a + d = 38 ….(1)

Given,

Sixth term = -22

⇒ a + (6 - 1)d = -22

⇒ a + 5d = -22 ….(2)

On Subtracting equation (1) from equation (2), we get :

⇒ a + 5d - (a + d) = -22 - 38

⇒ a - a + 5d - d = -60

⇒ 4d = -60

⇒ d = $\dfrac{-60}{4}$ = -15.

Substituting value of d in equation (1), we get :

⇒ a + (-15) = 38

⇒ a = 38 + 15 = 53.

Third term = a₃

= a + (3 - 1)d

= a + 2d

= 53 + 2(-15)

= 53 - 30 = 23.

Fourth term = a₄

= a + (4 - 1)d

= a + 3d

= 53 + 3(-15)

= 53 - 45

= 8.

Fifth term = a₅

= a + (5 - 1)d

= a + 4d

= 53 + 4(-15)

= 53 - 60

= -7.

Hence, the missing terms are 53, 23, 8, -7.