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) isA. 18B. 19C. 20D. 21

Hint: Take the common difference as â€˜dâ€™. Substitute the values of x, y and z in the equation $x+y+z=30$. From the simplified equation, find the AP and substitute different values of d, which gives the total number of possibilities of ordered triplet.

It is said that x, y and z represent 3 natural numbers in AP.
AP represents Arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant.
Difference here means the second minus the first term.
Let us consider â€˜dâ€™ as the common difference of the Arithmetic progression and â€˜dâ€™ is a natural number.
Here â€˜xâ€™ is the first term of AP.
y is the 2nd term of AP and â€˜zâ€™ the 3rd term.
Now, we need to find the difference between consecutive terms.
So, 2nd term â€“ 1st term = d
$\Rightarrow y-x=d$
Rearrange the terms $x=y-d$.
Similarly, 3rd term â€“ 2nd term = d.
$z-y=d\Rightarrow z=y+d$
Now, substitute the value of x and z in the equation $x+y+z=30$.
\begin{align} & \Rightarrow \left( y-d \right)+y+\left( y+d \right)=30 \\ & \Rightarrow y-d+y+y+d=30 \\ & \Rightarrow y+y+y=30 \\ & \Rightarrow 3y=30 \\ & \therefore y=\dfrac{30}{3}=10 \\ & \therefore x=y-d=10-d \\ & z=y+d=10+d \\ \end{align}
$\therefore$ The AP of x, y, z is now equal to (10 â€“ d), 10, (10 + d).
Now, letâ€™s consider d =1, then AP becomes 9, 10, 11.
Similarly,
d = 2, AP = 8, 10, 12
d = 3, AP = 7, 10, 13
d = 4, AP = 6, 10, 14
d = 5, AP = 5, 10, 15
d = 6, AP = 4, 10, 16
d = 7, AP = 3, 10, 17
d = 8, AP = 2, 10, 18
d = 9, AP = 1, 10, 19
So these combinations of ordered triplets are possible.
i.e. now there are 9 possibilities which are the same for (19, 1), (18, 2)â€¦..
$\therefore$ Number of possibilities $=9\times 2=18$ possibilities
Hence option A is the correct answer.

Note: The possibilities available can be represented as