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When doing some exercises on Khan Academy, the following question came up: "The sum of 3 consecutive odd numbers is 69. What is the third number in this sequence?" My answer was 27 because I thought -Odd- Numbers to show up as = 1, 3, 5, 7, etc. Apparently, it consists of x, 2, 4, 6, etc. Can someone explain this to me?
Consecutive odd numbers differs $2$.
Now suppose $a$ is a odd numbers, then $a-2, a+2$ are also odd numbers.
Now it is given that $$a-2+a+a+2=69\\\Rightarrow 3a=69\\\Rightarrow a=23$$
Then the third number in this sequence is $a+2=25$ (Ans.)
Odd numbers refer to the number that are not divisible by 2. E.g. 1, 3, 5, ...
As for the question, we need to assume 3 consecutive odd numbers: x, x+2 and x+4. (the difference between 2 consecutive-odds is 2)
$$ x+x+2+x+4=69$$ $$ 3x+6=69$$ $$ 3x= 69-6=63 $$ $$ x={63\over 3} = 21$$
Hence, the third number is 21+4= 25.
Using arithmetic progression formula
$$S_n=\frac{n}{2}(a+l)$$
where a is the first term and l is the last term.
$$S_n=\frac{n}{2}(2a+(n-1)d)$$
$$S_3=69$$
The difference between two odd numbers is two.
odd number sequences
$$1,3,5,7,9...$$
It is obvious that the sequence is arithmetic and have a common difference,
$$3-1=2=5-3$$
$$d=2$$
$$S_n=\frac{n}{2}(2a+(n-1)d)$$
where d is the common difference and a is the first term.
$$69=\frac{3}{2}(2a+(3-1)2)$$
$$a=21$$
By
$$S_n=\frac{n}{2}(a+l)$$
$$69=\frac{3}{2}(21+l)$$
$$l=25$$
