1.Algebra Booster

1.72 Algebra Booster
fi
fi
Dividing Eq. (ii) by Eq. (i), we get
fi r = 3
Putting r = 3 in Eq. (i), we get
fi
fi a = 8
fi b = ar = 24; c = ar 2 = 72
Hence, the series is 8, 24, 72, 216, 648.
17. Given
and
It is also given that a, b, g and d are in HP.
fi 1 1 1 1 , , , AP
a b g d Œ
fi 1 + 1 = 1 +
1
a d b g
Also,
fi
fi
fi
fi
fi
Again,
2
Solving, we get,
ar ◊ ar
= 18
2
1 4
ar + ar
D = , a =
2 7
2
ar
18
1 + r =
…(ii)
b a
2
ar ar 18
g a
∏ =
1+ r 1+
r 6
3
d a
Thus,
3a = 6
1+
3
1 1 1 143
and B = = ◊ = gd g d 16
4 1
a + b = , ab =
A A
6 1
g + d = , gd =
B B
1 1 1 1
+ + + = 6+ 4=
10
a b d g
Ê 1 1ˆ
Ê 1 1ˆ
fi 12 12 12r
42
Á + ˜ + + = 10
Ëa d¯ Á
Ëb g ˜
r + + =
¯
Ê 1 1 ˆ 1 1
Á + ˜ = 5 = Ê +
ˆ
fi 1 1 r
42 7
r + + = 12 = 2
Ëa d¯ Á
Ëb g ˜
¯
fi 2(r 2 + r + 1) = 7r
Ê 1 1
fi 2r
ˆ
2 – 5r + 2 = 0
Á + ˜ = 5
fi 2r
Ëa
d¯
2 – 4r – r + 2 = 0
fi 2r(r – 2) – (r – 2) = 0
Ê 1 1 ˆ
fi (r – 2)(2r – 1) = 0
Á + + 3D˜
= 5
Ëa
a ¯
1
fi r = 2,
Ê 2 ˆ
2
Á + 3D˜
= 5
…(i)
Ëa
¯
When r = 2,
1 1
+ = 4
a b
+ + D =
a a
2 2
Ê 1ˆ Ê 1ˆ
a + = …(ii) 20. We have Áp
+ + q +
Ë p
˜
¯
Á
Ë q
˜
¯
fi 1 1 4
fi 2 D 4
1 1 7 1 9 4
= + D = + = fi b =
4 2 4 9
1 1 7 11 4
= + 2D
= + 1= fi g =
4 4 11
1 1 7 3 13 4
= + D = + = fi d =
4 2 4 13
1 1 1 63
A = = ◊ =
ab a b 16
18. Let three numbers be a, b and c
Given a + b + c = 42
…(i)
Also, a + 2, b + 2, c Р4 ΠAP
fi a + 2 + c – 4 = 2(b + 2)
fi a + c – 2 = 2b + 4
fi a + c = 2b + 6 …(ii)
From Eq. (i) and Eq. (ii), we get
2b + 6 + b = 42
fi 3b = 42 – 6 = 36
b = 12
fi ar = 12
From Eq. (i), we get
a + b + c = 42
fi a + ar + ar 2 = 42
fi a + 12 + 12r = 42
a = 6
Thus, the sequence is {6, 12, 24}.
When r = 1/2,
a = 24
Thus, the sequence is {24, 12, 6}.
2 2 Ê 1 1 ˆ
= ( p + q ) +
Á
+ + 4
2 2
Ë p q
˜
¯