Answer:
Let's choose the three odd consecutive numbers 1, 3, and 5.
3^2=9
1•5=5
9-5=4
Let's try the same thing, but with the numbers 101, 103, and 105.
103^2=10609
101•105=10605
10609-10605=4
So, yes, the square of the second number out of three consecutive odd numbers is four greater than the product of the first and the third numbers.
Answer:
1042
Step-by-step explanation:
-38,-8,22,52
The common difference is 52 - 22 = 30
a_1 = -38
a_2 = -38 + 30
a_3 = -38 + 2 * 30
a_4 = -38 + 3 * 30 (notice that the common difference is multiplied by 1 less than the term number)
a_n = -38 + (n - 1)30 (since the term number is n, we multiply the common difference by n - 1)
a_n = -38 + 30(n - 1)
a_37 = -38 + 30(37 - 1)
a_37 = 1042
Using the change in base property, we have
approximately.
