1.Algebra Booster

1.84 Algebra Booster
2
S2
= = 3
1
1 -
3
3
S3
= = 4
1
1 -
4
o
n
Sn
= = n + 1
1
1 - n + 1
2 2 2
1 + 2 +º+ 2n
1
Thus, S S S -
= 2 2 + 3 3 + 4 2 + … + (2n) 2
= [1 2 + 2 2 + 3 3 + 4 2 + … + (2n) 2 ] – 1
2 n(2n+ 1)(4n+
1)
= -1
6
26. Let two positive numbers be a and b respectively.
2ab
a + b 4
Given =
ab 5
fi
fi
ab 2
a + b = 5
a + b =
ab
5
2
fi 2 25ab
( a + b)
= …(i)
4
Now, (a – b) 2 = (a + b) 2
25ab
9ab
= - 4ab
= …(ii)
4 4
Dividing Eq. (i) and Eq. (ii), we get
fi
2
( a + b) 25
=
2
( a - b)
9
( a + b) 5
=
( a - b) 3
fi ( a + b ) + ( a - b ) 5 +
=
3
( a + b) -( a -b) 5–3
fi
27. Given
fi
fi
a 8 4
b = 2 = 1
2
cos n
n=
0
x = Â
j
x = 1 + cos 2 j + cos 4 j + …
1 1
2
x = = = cosec j
2 2
1 - cos j sin j
Similarly,
y = sec 2 j
1
and z =
2 2
1 - cos j sin j
Thus,
1 xy
z = =
1 1
1 - ◊
xy - 1
y x
fi xyz – z = xy
fi xyz = xy + z
Again,
1 1 2 2
+ = sin j + cos j = 1
x y
fi x + y = xy = xyz – z
fi x + y + z = xyz
28. Given
ln(a + c), ln(a Рc), ln(a Р2b + c) ΠAP
fi (a + c), (a Рc), (a Р2b + c) ΠGP
fi (a – c) 2 = (a + c)(a + c – 2b)
fi a 2 + c 2 – 2ac = a 2 + ac – 2ab + ac + c 2 – 2bc
fi 2ab + 2bc = 4ac
fi ab + bc = 2ac
fi b(a + c) = 2ac
fi
2ac
b a c
Thus, a, b, c are in HP
31. Now, x 1
+ x 2
+ x 3
= 1 …(i)
x 1
x 2
+ x 2
x 3
+ x 3
x 1
= b
…(ii)
x 1
x 2
x 3
= –g
…(iii)
From Eq. (i), we get,
1
2x2+ x2= 1fi x2=
3
From Eq. (ii), we get
b = x 2
(x 1
+ x 3
) + x 1
x 3
fi
fi
fi
2
2 x1x3
= 2x
+
2 Ê1 ˆÊ1
ˆ
= + Á - d + d
9 Ë
˜Á
3 ¯Ë
˜
3 ¯
2 1 2
= + -d
9 9
1 2
= -d
3
Ê 1ˆ
2
Áb
- = - d £ 0
Ë
˜
3¯
1
b £
3
b Ê 1
Á , ˘
Ë 3 ˙˚
From Eq. (iii), we get
x 1
x 2
x 3
= –g
fi
Ê1 ˆ1 Ê1
ˆ
Á -d ◊ + d = -g
Ë
˜
3 ¯
Á ˜
3 Ë3
¯