Question

The average of 41 consecutive odd numbers is 49. What is the largest number.

Answer 1

Let the numbers be a, a + 2, a + 4,......a(n)

This forms an Arithmetic Progression AP

General term: a + (n-1)d

here d = 2

and n = 49

Now,

Sum of n terms in AP is given by:

Sum = n/2 [ 2a + (n-1)d ]

Average of 41 numbers is 49

Average = Sum of numbers / Total numbers

41 = Sum / 49

Sum = 41 x 49 = 2009

Putting this sum in our formula, we get

2009 = 49/2 [ 2a + (49 - 1)2 ]

4018 = 49 [ 2a + 96 ]

82 = 2a + 96

2a = -14

a = -7

Largest number is 49th number

49th number is = a + (n-1)d

-7 + 48(2) = 89

Largest number is 89

Answer 2

Let x is first odd number
Then, x , x + 2, x + 4, x + 6, x + 8, ........... x + 2 × (41 - 1)
e.g., 41 consecutive odd numbers are : x , x + 2, x + 4, x + 6, ........... x + 80
now, sum of 41 consecutive odd numbers = (x + x + 2 + x + 4 + x + 6 + ..... + x + 80)
= 41x + (2 + 4 + 6 + 8 + 10 + 12 + ..... + 80)
= 41x + 40/2(2 + 80)
= 41x + 20 × 82
= 41x + 1640

Now, average of 41 consecutive odd numbers = sum of observations/total number of observations
49 = (41x + 1640)/41
49 = x + 40
x = 9

Hence, largest odd number = x + 80 = 89