Progression (Sequence and Series)Medium
Question
With usual notations in ᐃABC, it is given that r1, r2, r3 are in harmonic progression. The perimeter of triangle is 24 units and area is 24 square units, then -
Options
A.a & c are roots of equation x2 - 16x + 63 = 0
B.a & c are roots of equation x2 - 16x + 60 = 0
C.r2 + r = 8
D.r2 + r = 10
Solution
r1, r2, r3 in H.P. ⇒ a, b, c are in A.P.
2s = a + b + c = 24 ⇒ b = 8
⇒ a + c = 16, s = 12
ᐃ =
⇒ ac = 60
⇒ x2 - 16x + 60 = 0 has roots a & c
r2 + r = r1, r2, r3 in H.P. ⇒ a, b, c are in A.P.
2s = a + b + c = 24 ⇒ b = 8
⇒ a + c = 16, s = 12
ᐃ = ⇒ ac = 60
⇒ x2 - 16x + 60 = 0 has roots a & c
r2 + r =
2s = a + b + c = 24 ⇒ b = 8
⇒ a + c = 16, s = 12
ᐃ =
⇒ ac = 60⇒ x2 - 16x + 60 = 0 has roots a & c
r2 + r = r1, r2, r3 in H.P. ⇒ a, b, c are in A.P.
2s = a + b + c = 24 ⇒ b = 8
⇒ a + c = 16, s = 12
ᐃ =
⇒ ac = 60⇒ x2 - 16x + 60 = 0 has roots a & c
r2 + r =
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