The average of five consecutive odd integers is -21. What is the least of these integers?

Answer 1

#-25#

Take #x#. This is the smallest integer. Since these are consecutive odd integers, the second must be #2# greater than the first. The third number must be #2# greater than the second. And so forth.
For example, #1, 3, 5, 7, and 9# are five consecutive odd integers, and they are all two more than the last. So, our five numbers are
#x, x+2, (x+2)+2, ((x+2)+2)+2, and (((x+2)+2)+2)+2#

which means

#x, x+2, x+4, x+6, and x+8#
According to the question, their average is #-21#. So,
#(x+(x+2)+(x+4)+(x+6)+(x+8))/5 = -21#

Therefore, by simplifying,

#(5x+20)/5 = -21#

So

#5x+20 = -105#

Then

#5x = -125#

and

#x=-25#
Shortcut: Since these are odd integers that are consecutive, you can take #-21# as the middle number, #-23# as the second, #-19# to even out the #-23# and maintain the average of #-21#, then #-25# as the first, then #-17# as the last. This is a little hard to explain but makes sense if you really think about it.
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Answer 2

# "The answer is:" \qquad \qquad \qquad -25. #

# "[And the 5 integers in question are:" #

# \qquad \qquad \qquad \qquad \qquad \quad -25, -23, -21, -19, -17.] #

# "Let the smallest of these odd integers be:" \qquad \qquad 2 n - 1. #
# "The remaining 4 odd integers are:" #
# \qquad \qquad \qquad \qquad \qquad \quad \ 2 n + 1, \quad 2 n + 3, \quad 2 n + 5, \quad 2 n + 7. \quad #
# "The average of all 5 odd integers is:" #
# { ( 2 n - 1 ) + ( 2 n + 1 ) + ( 2 n + 3 ) + ( 2 n + 5 ) + ( 2 n + 7 ) } / 5. #
# "The average of all 5 odd integers is given to be -21. Thus:" #
# { ( 2 n - 1 ) + ( 2 n + 1 ) + ( 2 n + 3 ) + ( 2 n + 5 ) + ( 2 n + 7 ) } / 5 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad = -21. #
# \qquad \qquad \qquad \qquad \qquad { 5 ( 2 n ) - 1 + 1 + 3 + 5 + 7 } / 5 \ = \ -21 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad { 10 n + 15 } / 5 \ = \ -21 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ { color{red}cancel{ 5 } ( 2 n + 3 ) } / color{red}cancel{ 5 } \ = \ -21 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 2 n + 3 \ = \ -21 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 2 n \ = \ -21 - 3 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 2 n \ = \ -24 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ \ n \ = \ -12. #
# "At the start here, we had:" #
# \qquad \qquad \quad \ "the smallest of these odd integers is:" \qquad \quad 2 n - 1. #
# "As we found" \quad n \ = -12, \ "we have:" #
# \qquad \qquad \quad \ "the smallest of these odd integers is:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 2 ( -12 ) - 1 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ -24 - 1 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ -25. #
# "This is our answer:" \qquad \qquad \qquad -25. \qquad \qquad \qquad \qquad \qquad \qquad !! #
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Answer 3

The least of these integers is -25.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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