for the first term:
n= 1 there is 1 item
n=2 there are 2+1 = 3 items
n = 3 there are 3+2+1 = 3+2 = 6 items
n = 4 there are 4+3+2+1 = 4+6 = 10 items
We are adding the "n" term each time
n = 5 10+5 = 15
n= 6 15+6 = 21
n = 7 21+7 = 28
n = 8 28+8 = 36
There would be 36 items in the 8th step
If
is a number that is both divisible by 4 and 5, then

4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that
is a solution to the system, where
is any integer. Simply put, any multiple of 20 fits the bill.
Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have
when
, so the sum we want to compute is

Answer:
Step-by-step explanation:
You must “reverse” the inequality sign to make the statement true: When you multiply by a negative number, “reverse” the inequality sign. Whenever you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed in order to keep a true statement.
Step-by-step explanation:
Let, x^a =y......(1) and
y^b =z.....(2) and
z ^c =x......(3).
Now, using (1) in (2) we get,
x ^ab =z......(4).
Now, using (4) in (3) we get,
x ^abc =x
or, x ^abc =x ^1
or, abc=1.
Hope it will help :)