Let $S$ be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in $S$ with perimeter less than $30$ is

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Solution

The correct option is C $2$
Let the sides of the triangle required be $a, a + d, a + 2 d such that d > 0 \Rightarrow 3 a + 3 d < 30 \dots (1) & a^{2} + (a + d)^{2} = (a + 2 d)^{2} \Rightarrow a^{2} - 2 a d - 3 d^{2} = 0 \Rightarrow a^{2} - 3 a d + a d - 3 d^{2} = 0 \Rightarrow a (a - 3 d) + d (a - 3 d)) = 0 \Rightarrow a = 3 d 12 d < 30 (using (1)) \Rightarrow d = 1, 2 \Rightarrow a = 3, 6$
Therefore, $2$ triangles are possible.

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An arithmetic progression is defined as a sequence of numbers with a constant difference between consecutive terms.

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Let S be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in S with perimeter less than 30 is

The correct option is C 2Let the sides of the triangle required be a,a+d,a+2d such that d>0⇒3a+3d<30⋯(1)& a2+(a+d)2=(a+2d)2⇒a2−2ad−3d2=0⇒a2−3ad+ad−3d2=0⇒a(a−3d)+d(a−3d))=0⇒a=3d12d<30 (using (1))⇒d=1,2 ⇒a=3,6 Therefore, 2 triangles are possible.

Let $S$ be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in $S$ with perimeter less than $30$ is