Represent these consecutive numbers (assuming that they are all integers):
x
x+1
x+2
x+3
x+4
x+5
and so on
x+8
x+9 is the tenth number. x+9 = 10, so x = 9.
Think of it this way: there are 10 consecutive numbers, and the last one is 10.
Working backwards, we get the sequence 10, 9, ... 3, 2, 1.
The sum of such an arith sequence is equal to the count of the numbers times the average of the first and last terms:
sum here = 10(1+10)/2 = 5(11) = 55 (answer)
Answer:
5;5 and 65; they are congruent
H-5 = -30
Add 5 to both sides to get variable h by itself
h = -25
