Arithmetic sequence question

x, y and z form an arithmetic sequence.
Let a and d are the first term and the common difference respectively.
x = a cm
y = (a + d) cm
z = (a + 2d) cm

Perimeter:
a + (a + d) + (a + 2d) = 42 ...... (1)

Pythagoras theorem:
a2 + (a + d)2 = (a + 2d)2 ...... (2)

(1):
a + a + d + a + 2d = 42
3a + 3d = 42
a + d = 14
d = 14 - a ...... (3)

(2):
a2 + (a2 + 2ad + d2) = a2 + 4ad + 4d2
a2 - 4ad - 3d2 = 0 ...... (4)

Put (3) into (4):
a2 - 4a(14 - a) + 3(14 - a)2 = 0
a2 - 4a(14 - a) + 3(196 - 28a + a2) = 0
a2 - 56a + 4a2 + 588 - 84a + 3a2 = 0
8a2 - 140a + 588 = 0
2a2 - 35a + 147 = 0
(2a - 21)(a - 14) = 0
a = 21/2 or a = 14 (rejected)
a = 10.5

Put a = 10.5 into (3)
d = 14 - 10.5
d = 3.5

x = a cm
x = 10.5 cm

y = (a + d) cm
y = 14 cm

z = (a + 2d) cm
z = 17.5 cm

Ans: The sides of the triangle are 10.5 cm, 14 cm and 17.5 cm respectively.
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