1.Algebra Booster

1.78 Algebra Booster
5. We know that
Ê1
+ aˆ Á ≥
Ë
˜
2 ¯
fi (1 + a) ≥2
a
a
Similarly, (1 + b) ≥2
b
(1 + c) ≥2
c
(1 + d) ≥2
d
Multiplying, we get
(1 + a)(1 + b)(1 + c)(1 + d) ≥16
abcd
Thus, l = 16
4
Hence, the value of ( l + 1) is 3.
6. Let the common ratio be r.
Given a 1
+ a n
= 66
fi a + ar n – 1 = 66 …(i)
Also, a 2
◊ a n – 1
= 128
fi ar ◊ ar n – 2 = 128
fi a 2 r n – 1 = 128 …(ii)
From Eqs (i) and (ii), we get
Ê 128ˆ
aÁ1+ = 66
Ë 2 ˜
a ¯
128
fi a + = 66
a
fi a 2 – 66a + 128 = 0
fi (a – 2)(a – 64) = 0
fi a = 2 or 64
From Eq. (i), we get,
r n – 1 = 32
Again,
n
 i
i = 1
a = 126
fi (a 1
+ a 2
+ a 3
+ … + a n
) = 126
fi (a + ar + ar 2 + … + ar n – 1 ) = 126
fi a(1 + r + r 2 + … + r n – 1 ) = 126
fi
n
1 r
a Ê - ˆ
Á = 126
Ë 1-
r
˜
¯
fi
Ê
n
1 - r ˆ
2Á
= 126
Ë 1 - r
˜
¯
fi
Ê
n
1 - r ˆ
Á = 63
Ë 1 - r
˜
¯
fi
Ê1-
32r
ˆ
Á = 63
Ë 1 - r ˜
¯
fi 1 – 32r = 63 – 63r
fi 31r = 62
fi r = 2
Now, from Eq. (iii), we get,
2 n – 1 = 32 = 2 5
fi n – 1 = 5
fi n = 6
7. Given
1 1 1 10
+ + =
a + b b+ c c + a 3
fi
1 1 1 10
+ + =
a + b b+ c c + a 3
fi
3 3 3
+ + = 10
b + c c + a a + b
fi
a + b + c a + b + c a + b + c
+ + = 10
b + c c + a a + b
fi
a + ( b + c) b + ( a + c) c + ( a + b)
+ + = 10
b + c c + a a + b
fi a + 1+ + 1+ + 1=
10
b + c c + a a + b
fi a + b + c = 10 - 3 = 7
b + c c + a a + b
Hence, the value of
a b c
+ + is 7.
b + c c + a a + b
8. Let
Now,
1 1 1 1
4 8 16 4.2 n
s = + + + … +
-1
1Ê
1 1 1 ˆ
= Á1
+ + +º+
2 1
4Ë
n-
˜
2 2 2 ¯
Ê
n
Ê1ˆ
ˆ
1
n
1
Á - Á
Ë
˜
2¯ ˜
1Ê Ê1ˆ
ˆ
= Á ˜ = 1
4 1 Á -Á
˜ ˜
1
2Ë
Ë2¯
¯
Á -
Ë
˜
2 ¯
b n
Ê1ˆ
= Á
Ë ˜
5¯
Ê1ˆ
= Á
Ë ˜
5¯
= (5)
1Ê
n
Ê1ˆ
ˆ
log 1
5 Á - ˜
2Ë
Á
Ë
˜
2¯
¯
1Ê
n
Ê1ˆ
ˆ
2log 5 Á1-
˜
2Ë
Á
Ë
˜
2¯
¯
Ê n
1 Ê1ˆ
ˆ
-2log5 1
2Á
- ˜
Ë
Á
Ë
˜
2¯
¯
-2log
1
5 2
log 5 (4)
(5) , when n
= (5) = 4
Thus, lim ( b ) + 3 = 7
n
n
9. Given a, x, y, z, b are in AP.
Clearly x, y, z are the AM between a and b
Êa
+ bˆ
Thus, x + y + z = 3Á Ë
˜
2 ¯
Êa
+ bˆ fi 3Á
= 15
Ë
˜
2 ¯
fi (a + b) = 10