A) 5 |
B) 3 |
C) 1 |
D) 7 |
A) 5 |
Let the 5 consecutive odd positive integersbex+1,x+ 3,x+ 5,x+ 7,x+ 9
Average = Sum of all terms / Number of terms
\( \Large 9 = \frac{x+1+x+3+x+5+x+7+x+9}{5} = 9 : 5 = 5 \)
\( \Large 9 = \frac{5x +25}{5} \)
=> 45 = 5x + 25
5x = 45 - 25 = 20
5x = 20 => x = 4
The least one is x+ 1 = 4 + 1 = 5