1.Algebra Booster

1.46 Algebra Booster
Now,
Ê a + b ˆ Ê c + b ˆ
Á +
Ë2a -b˜ ¯
Á
Ë2c -b˜
¯
Ê b ˆ Ê b ˆ
1+ 1+
Á a ˜ Á c ˜
= Á b˜ + Á b˜
Á2- ˜ Á2-
˜
Ë a¯ Ë c¯
Ê 2c
ˆ Ê 2a
ˆ
1+ 1+
Á a + c ˜ Á a + c ˜
= Á ˜ + Á ˜
2c
2a
2 2
Á - -
Ë a + c ˜
¯
Á
Ë a + c˜
¯
Ê a + c + 2c ˆ Ê a + c + 2a
ˆ
= Á +
Ë2a + 2c - 2c˜ ¯
Á
Ë2a + 2c - 2a˜
¯
Êa + 3cˆ Ê3a + cˆ
= Á +
Ë
˜
2a
¯
Á
Ë
˜
2c
¯
2 2
ac + 3c + 3a + ac
=
2ac
2 2
3c + 3a + 2ac
=
2ac
2 2
3( a + c )
= + 1
2ac
3 Êa
cˆ
= Á + + 1
2 Ë
˜
c a¯
3
≥ ¥ 2+ 1=
4
2
Hence, the value of l is 4.
127. As we know that,
AM ≥ GM
a b c
+ +
fi
b c a a b c
≥ 3 ◊ ◊ = 1
3 b a a
Êa b cˆ
fi Á + + ≥3
Ë
˜
b c a¯
Hence, the result.
128. It is given that a, b and c are in HP, so we can write,
1 1 1
, , are in AP.
a b c
2 1 1
Thus, = +
b a c
Ê1 1 1ˆÊ1 1 1ˆ
Now, Á + - + -
Ë
˜Á
b c a¯Ë ˜
c a b¯
Ê3 1ˆÊ2 1ˆ
= Á - -
Ë
˜Á
b a¯Ë ˜
b b¯
1Ê3 1ˆ
3 1
= Á - = -
2 .
bË
˜
b a¯
b ab
129. Given a, b, c and d are in HP.
fi 1 , 1 , 1 ,
1 are in AP
a b c d
Let D be the common difference.
Thus, 1 - 1 = D
b a
fi
1 ab = ( a - b)
D
1
Similarly, bc = ( b - c)
D
and
1 cd = ( c - d)
D
Now, ab + bc + cd
1 1 1
= ( a - b) + ( b - c) + ( c - d)
D D D
1
= ( a - b + b – c + c – d )
D
1
= ( a – d )
D
Ê 3ad
ˆ
= Á ¥ ( a – d)
Ëa
- d˜
¯
= 3ad
130. Since a, b and c are in AP
fi 2b = a + c …(i)
Also, x, y and z are in HP
2xz
fi y = x + z
…(ii)
And, ax, by and cz are in GP.
\ b 2 y 2 = axcz …(iii)
fi
2 2 2
Êa + cˆ 4x z
Á ¥
Ë
˜
2
2 ¯ ( x + z)
= acxz [from Eqs. (i) and (ii)]
fi
fi
2 2
( a + c) ( x + z)
=
ac xz
2 2 2 2
a + c + 2ac x + z + 2xz
=
ac
xz
fi a + c + 2= x + z + 2
c a z x
fi a + c = x +
z
c a z x
Hence, the result.
131. We have,
a - x a - y a - z
= =
px qy rz