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Veseljchak [2.6K]
3 years ago
10

An arithmetic progression is a sequence of numbers in which the distance (or difference) between any two successive numbers if t

he same. This in the sequence 1, 3, 5, 7, ..., the distance is 2 while in the sequence 6, 12, 18, 24, ..., the distance is 6. Given the positive integer distance and the integers m and n, create a list consisting of the arithmetic progression between (and including) m and n with a distance of distance (if m > n, the list should be empty.) For example, if distance is 2, m is 5, and n is 12, the list would be [5, 7, 9, 11]. Associate the list with the variable arith_prog.
Mathematics
1 answer:
notsponge [240]3 years ago
3 0

Answer:[m, m+d, m+2d, - - - - -, n]

Step-by-step explanation:

We know the formula for arithmetic progression is a_(n) = a_(1) + (n-1)d

Where a_(n) is the nth term of the sequence

a_(1) is the first term of the sequence

n is the number of the term like if we are talking about 7th term so the n is 7.

d is the difference between two successive terms.

For this problem we know our first term that is m, our last term that is n and our difference that is d.

For second term we will use the formula

a_(2) = m + (2-1)d

a_(2) = m + (1)d

a_(2) = m + d

Similarly,

a_(3) = m + (3-1)d

a_(3) = m + (2)d

a_(3) = m + 2d

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Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

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For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

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