Correct option is B
Given:
The smallest side of a triangle is 5 units
The three sides are in Arithmetic Progression (A.P.) with a positive integer difference
We are to find how many such valid triangles are possible
Solution:
Let the three sides of the triangle in A.P. be:
a − d, a, a + d
Or to simplify (since smallest side is 5):
Let the sides be:
5, 5 + d, 5 + 2d
Where d is a positive integer.
Triangle Inequality Rule:
Sum of any two sides must be greater than the third side.
We apply this to:
5 + (5 + d) > (5 + 2d)
=> 10 + d > 5 + 2d
=> 5 > d
So d < 5
And since d is a positive integer,
d = 1, 2, 3, 4 → 4 values
There are 4 possible values of d, hence 4 valid triangles
Final Answer:
S. Ans. (B) 4
