The combinations of ordered triplets that are possible is equal to: A. 18.
How to find the possible number of ordered triplet?
Since the common difference (d) between the consecutive terms in an arithmetic progression (AP) is constant, we have:
d = y - x = z - y
Rearranging the terms, we have:
x = y - d
z = y + d
From the given equation:
x + y + z = 30
y - d + y + y + d = 30
3y = 30
y = 30/3
y = 10.
So the arithmetic progression (AP) is given by:
(10 - d), 10, (10 + d)....
Since the common difference (d) is a natural number which would be from 1 to 9, the combinations of ordered triplets that are possible is as follows:
Number of possibilities = 9 × 2
Number of possibilities = 18.
Read more on arithmetic progression here: https://brainly.com/question/18828482
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Complete Question:
If x, y, z are three natural numbers in A.P. such that x + y + z = 30 then the possible number of ordered triplet (x, y, z) is?
A. 18
B. 19
C. 20
D. 21
