Question

In each of the arithmetic sequences below, some numbers are missing. Their positions are marked with a. Find these numbers.

Answer 1

(1)

First number = 24

Second number = 42

Let the third number be x and the fourth number be y.

Thus, the arithmetic sequence is 24, 42, x, y, …

⇒ 42 − 24 = x − 42

⇒ x = 42 − 24 + 42

⇒ x = 60

Also, x − 42 = y − x

⇒ 60 − 42 = y − 60

⇒ y = 60 + 60 − 42

⇒ y = 78

Therefore, the third and the fourth numbers are 60 and 78 respectively.

Thus, the arithmetic sequence is:

(2)

Second number = 24

Third number = 42

Let the first number be x and the fourth number be y.

Thus, the arithmetic sequence is x, 24, 42, y, …

⇒ 24 − x = 42 − 24

⇒ x = 24 − 42 + 24

⇒ x = 6

Also, 42 − 24 = y − 42

⇒ y = 42 − 24 + 42

⇒ y = 60

Therefore, the first and the fourth numbers are 6 and 60 respectively.

Thus, the arithmetic sequence is:

(3)

Third number = 24

Fourth number = 42

Let the first number be x and the second number be y.

Thus, the arithmetic sequence is x, y, 24, 42, …

⇒ 24 − y = 42 − 24

⇒ y = 24 − 42 + 24

⇒ y = 6

Also, y − x = 24 − y

⇒ 6 − x = 24 − 6

⇒ x = 6 − 24 + 6

⇒ x = −12

Therefore, the first and the second numbers are −12 and 6 respectively.

Thus, the arithmetic sequence is:

(4)

First number = 24

Third number = 42

Let the second number be x and the fourth number be y.

Thus, the arithmetic sequence is 24, x, 42, y, …

⇒ x − 24 = 42 − x

⇒ x + x = 42 + 24

⇒ 2x = 66

⇒ x = 33

Also, 42 − x = y − 42

⇒ 42 − 33 = y − 42

⇒ y = 42 − 33 + 42

⇒ y = 51

Therefore, the second and the fourth numbers are 33 and 51 respectively.

Thus, the arithmetic sequence is:

(5)

Second number = 24

Fourth number = 42

Let the first number be x and the third number be y.

Thus, the arithmetic sequence is x, 24, y, 42, …

⇒ y − 24 = 42 − y

⇒ y + y = 42 + 24

⇒ 2y = 66

⇒ y = 33

Also, 24 − x = y − 24

⇒ 24 − x = 33 − 24

⇒ x = 24 − 33 + 24

⇒ x = 15

Therefore, the first and the third numbers are 15 and 33 respectively.

Thus, the arithmetic sequence is:

(6)

First number = 24

Fourth number = 42

Let the second number be x and the third number be y.

Thus, the arithmetic sequence is 24, x, y, 42, …

⇒ x − 24 = y − x

⇒ 2x − y − 24 = 0 … (1)

Also, y − x = 42 − y

⇒ 2y − x − 42 = 0 … (2)

Multiplying equation (2) by 2:

4y − 2x − 84 = 0 … (3)

On solving equations (1) and (3):

3y − 108 = 0

⇒ y = 36

From equation (1), x = 30

Therefore, the second and the third numbers are 30 and 36 respectively.

Thus, the arithmetic sequence is: