1.Algebra Booster

1.86 Algebra Booster
37. Let a and b be the roots of the given equation
Thus, HM of the roots of a and b
2ab
=
a + b
Ê8+
2 5ˆ
2Á
5 2
˜
Ë + ¯
= Ê 4+
5 ˆ
Á
5 2
˜
Ë + ¯
2(8 + 2 5) 4(4 + 5)
= = = 4
(4 + 5) (4 + 5)
a 3
38. Given 4, ar
1- r = = 4
fi
3
4r = 4
1-
r
fi 3 4(1 – r)
4r fi 16r(1 – r) = 3
fi 16r 2 – 16r + 3 = 0
fi 16r 2 – 12r – 4r + 3 = 0
fi 4r(4r – 3) – 1(4r – 3) = 0
fi (4r – 3)(4r – 1) = 0
fi
3 1
r = or
4 4
1
When r = , a = 3.
4
3
When r = , a = 1.
4
40. Given T n+1
– T n
= 21
fi n+1 C 3
– n C 3
= 21
fi
( n + 1)! ( n)!
- = 21
3!( n + 1 -3)! 3!( n -3)!
fi ( n + 1)! ( n
- )! = 126
( n – 2)! ( n - 3)!
fi (n + 1)n(n – 1) – n(n – 1)(n – 2) = 126
fi n(n 2 – 1) – n(n 2 – 3n + 2) = 126
fi (n 3 – n – n 3 + 3n 2 – 2n) = 126
fi 3n 2 – 3n = 126
fi n 2 – n – 42 = 0
fi (n – 7)(n + 6) = 0
fi n = 7 or –6
fi Since n should be a natural number, so n = 7
41. Given a, b, c, d ΠAP
fi 1 , 1 , 1 , 1 ŒHP
a b c d
fi
abcd abcd abcd abcd
, , , ŒHP
a b c d
fi bcd, acd, abd, abc ΠHP
42.
43.
44. Given c = a 1
◊ a 2
◊ a 3
… a n
As we know that, AM ≥ GM
a + a +º+ 2a + a
fi
1 2 n-1
n
a + a +º+ 2a + a
1 2 n-1
n n
n
≥
≥
n
2( a ◊a … a )
2c
1 2
n
fi (a 1
+ a 2
+ … + 2a n – 1
+ a n
) ≥ n(2c) 1/n
Hence, the minimum value of
a 1
+ a 2
+ … + a n – 1
+ 2a n
is n(2c) 1/n .
45. Given a, b, c are in AP
fi 2b = a + c …(i)
Also, a 2 , b 2 , c 2 are in GP.
fi b 4 = a 2 c 2 …(ii)
fi
fi
3
a + b + c =
2
3
3b =
2
1
fi b =
2
From Eq. (ii), we get,
2 2 Ê1ˆ
1
ac = Á =
Ë
˜
2¯
16
1
fi ac =±
4
Also a + c = 2b = 1
Thus, a and c are the roots of
fi
fi
fi
4
2 1
x - x ± = 0
4
2 1 2 1
x - x + = 0, x - x - = 0
4 4
2 2
Ê 1 1 1
Áx
– ˆ = 0, Ê x –
ˆ =
Ë
˜
2¯ Á
Ë
˜
2¯
2
1 1 1
x = , x = ±
2 2 2
1 1
fi x = ± , since a < b < c
2 2
1 1
Thus, a = - , as a < b < c
2 2
47. Given a, b, c ΠAP
fi 2b = a + c
Also, a 2 , b 2 , c 2 ΠHP
1 1 1
fi , , AP
2 2 2
a b c Œ
2 = 1 +
1
b a c
fi
2 2 2
n