1. a square has a side of 2 decimeters. a new square is formed by joining the midpoints of the sides. If this procedure is reoeated 6 times, what is the perimetet and area of the seventh square
2. find the sum of those numbers between 15 and 810 which have 8 as their units digit
3. in an arithmetic sequence, the sixth term is twice the 3rd term and the 30th term is 60. find the common difference?
1. To find the perimeter and area of the seventh square, we first need to understand how the new square is formed. In each step, we are joining the midpoints of the sides of the previous square. This means that each side of the new square is half the length of the previous square's side.
Given that the side length of the first square is 2 decimeters, we can calculate the side length of the seventh square by continuously halving the side length.
To calculate the side length of the seventh square:
1st square: 2 decimeters
2nd square: (1/2) * 2 = 1 decimeter
3rd square: (1/2) * 1 = 0.5 decimeters
4th square: (1/2) * 0.5 = 0.25 decimeters
5th square: (1/2) * 0.25 = 0.125 decimeters
6th square: (1/2) * 0.125 = 0.0625 decimeters
Now that we know the side length of the seventh square is 0.0625 decimeters, we can find the perimeter and area.
The perimeter of a square is calculated by multiplying the side length by 4, since all sides are equal.
Perimeter of the seventh square = 4 * 0.0625 decimeters = 0.25 decimeters.
The area of a square is calculated by multiplying the side length by itself.
Area of the seventh square = 0.0625 decimeters * 0.0625 decimeters = 0.00390625 square decimeters.
Therefore, the perimeter of the seventh square is 0.25 decimeters and the area is 0.00390625 square decimeters.
2. To find the sum of numbers between 15 and 810 which have 8 as their units digit, we need to find all the numbers that satisfy this condition and then sum them up.
The units digit 8 repeats every 10 numbers (18, 28, 38, ..., 808), so we can calculate how many 10-number sequences are between 15 and 810 by subtracting 15 from 810 and dividing by 10:
(810 - 15) / 10 = 79.5
Since we can't have half a sequence, we discard the decimal part and have 79 complete 10-number sequences.
Within each sequence, the units digit 8 appears once. So within 79 sequences, the units digit 8 appears 79 times.
To calculate the sum of these 79 numbers, we can use the arithmetic series formula: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.
Plugging in the values, we get:
Sn = 79/2 * (18 + 808) = 79/2 * 826 = 32687
Therefore, the sum of numbers between 15 and 810 which have 8 as their units digit is 32687.
3. In an arithmetic sequence, the difference between consecutive terms is constant. Let's say the common difference is represented by 'd'.
According to the problem, the 6th term is twice the 3rd term. We can set up the equation as follows:
3rd term + 2d = 6th term
Let's represent the 3rd term as 'a + 2d'. Then the 6th term would be 'a + 2(2d)'.
Substituting these values into the equation:
a + 2d + 2d = a + 4d
Simplifying the equation:
4d = 2d
2d = 0
From this equation, we get d = 0.
However, this implies that the common difference is zero, which contradicts the definition of an arithmetic sequence where the difference between consecutive terms is constant.
Therefore, no common difference can satisfy the given conditions of the problem.