FINDING ARITHMETIC MEAN BETWEEN NUMBERS

Problem 1 :

Find the arithmetic mean between 4 and 10.

Solution :

Let a = 4 and b = 10

Arithmetic mean = (4 + 10)/2

= 14/2

= 7

Problem 2 :

Insert 4 arithmetic mean between 4 and 324.

Solution :

Let first term a = 4, and last term = 324

4, __, __, __, __, 324

a = 4, a + d, a + 2d, a + 3d, a + 4d are unknowns

a + 5d = 324

Applying the vale of a, we get

4 + 5d = 324

5d = 324 - 4

5d = 320

d = 320/5

d = 64

So, the four arithmetic means are 68, 132, 196 and 260.

Problem 3 :

The two arithmetic means between -6 and 14 is

a)  2/3, 1/3      b)  2/3, 7 1/3       c)  -2/3, -7  1/3    d)  none

Solution :

Let first term a = -6, and last term = 14

-6, __, __, 14

a = -6, a + d, a + 2d are unknowns

a + 3d = 14

Applying the vale of a, we get

-6 + 3d = 14

3d = 14 + 6

3d = 20

d = 20/3

a + d ==> -6 + (20/3)

= (-18 + 20)/3

= 2/3

a + 2d ==> -6 + 2(20/3)

= (-18 + 40)/3

= 22/3 or 7  1/3

So, the answer is option b.

Problem 4 :

The arithmetic mean between 33 and 77 is 

a)  50   b)  45   c)  55   d) none

Solution :

Let a = 33 and b = 77

Arithmetic mean = (33 + 77)/2

= 110/2

= 55

So, the answer is c.

Problem 5 :

The 4 arithmetic means between -2 and 23 are

a) 3, 13, 8, 18     b)  18, 3, 8, 13

c)  3, 8, 13, 18       d) none 

Solution :

Let first term a = -2, and last term = 23

-2, __, __, __, __, 23

a = -2, a + d, a + 2d, a + 3d, a + 4d are unknowns

a + 5d = 23

Applying the value of a, we get

-2 + 5d = 23

5d = 23 + 2

5d = 25

d = 5

So, the solution is 3, 8, 13, 18, option c.

Problem 6 :

The arithmetic mean of two positive numbers is 40 and their geometric mean is 24. The numbers are 

a)  72, 8     b)  70, 10     c)  60, 20

Solution :

Let a and b be two numbers.

(a + b)/2 = 40

a + b = 40(2)

a + b = 80

b = 80 - a-----(1)

√ab = 24

ab = (24)²

ab = 576-----(2)

Applying (1) in (2), we get

a(80 - a) = 576

80a - a² = 576

a² - 80a + 576 = 0

(a - 8)(a - 72) = 0

a = 8 and a = 72

Applying the value a in b = 576/a

So, the answer is 72 and 8, option a.

Problem 7 :

The numbers x, 8 and y are in G.P and the numbers x, y, -8 are in A.P The values of x, y are ___

a) 16, 4    b)  4, 16     c)  both    d) none

Solution :

x, 8 and y are in G.P

√xy = 8

xy = 64

y = 64/x  -------(1)

x, y, -8 are in A.P

y = (x + (-8))/2

2y = x - 8

y = (x - 8)/2 ---(2)

Applying the value of (1) in (2)

64/x = (x - 8)/2

128 = x² - 8x

x² - 8x - 128 = 0

(x - 16)(x + 8) = 0

x = 16 and x = -8

So, the answer is x = 16 and y = 4.