1.Algebra Booster

1.44 Algebra Booster
115. It is given that,
a 1
, a 2
, a 3
ΠAP fi 2a 2
= a 1
+ a 3
Also
and,
Now,
fi
fi
fi
fi
fi
2
2, 3, 4ŒGPfi 3 = 2◊
4
a a a a a a
2aa
3 5
a3, a4, a5ŒHPfi a4=
a + a
a
2
3 = 2◊
4
a a a
Êa + a ˆ Ê 2a a ˆ
Á ˜ Á ˜
¯
2 1 3 3 5
3 = ¥
Ë 2 ¯ Ëa3+
a5
aa
2 3 5
3 = ( 1+ 3)
¥ Á
Ëa3+
a ˜
5¯
2
3 3+ 5 = 3 5 1+
3
a a a
a ( a a ) a a ( a a )
3 2 2
3 + 3 5= 1 3 5+
3 5
a a a a a a a a
a
3
3 = a1a3a5
2
Ê
3 5
fi a3 = a1a5
fi a 1
, a 3
, a 5
ΠGP
116. a x = b y = c z = d w = k (say)
It is given that a, b, c and d are in GP.
a b c
\ = =
b c d
fi
fi
1 1 1
x y z
k k
=
k
=
k k k
1 1 1
y z w
1- 1 1-1 1 1
x y y z -
z w
k = k = k
fi 1 – 1 = 1 – 1 =
1 –
1
x y y z z w
fi 1 , 1 , 1 , 1 AP
x y z w Œ
fi x, y, z and w ΠHP
117. It is given that x, y and z are in GP.
\ y 2 = xz
fi log(y 2 ) = log(xz)
fi 2log(y) = log(x) + log(z)
fi log(x), log(y), log(z) ΠAP
fi 1 + log(x), 1 + log(y), 1 + log(z) ΠAP
fi
1 1 1
, , HP
1+ log( x) 1+ log( y) 1+
log( z)
Œ
1 1
118. In AP, tp
= fi A+ ( p - 1) d =
a
a
1 1
tq
= fi A+ ( q - 1) d =
b
b
(i)
(ii)
and
1 1
tr
A ( r 1) d
c
c
(iii)
ˆ
Solving Eqs (i), (ii) and (iii), we get
b - a
( p - q)
=
abd
( b - c)
( q – r)
=
bcd
( c – a)
and ( r – p)
=
acd
Thus,
q -r
r - p p -q
+ +
a b c
( b -c) ( c -a) ( a -b)
= + +
bcd acd abd
1Ê1 1 1 1 1 1ˆ
= Á - + - + -
d Ë
˜
c b a c b a¯
1
= ¥ 0
d
= 0.
119. It is given that
x 1
, x 2
, x 3
, …, x n
are in HP.
1 1 1 1
\ , , ,…, AP
x x x x Œ
fi
fi
1 2 3
n
1 1 1 1 1 1
- = - =º= - = d
x x x x x x
2 1 3 2 n n–1
1-
x2
x2-
x x
3
n-1
xn
= =º= -
=
xx 1 2 x2x3 xn–1xn
x
Now,
x 1
x 2
+ x 2
x 3
+ … + x n–1
x n
x1-
x2
x2-
x x
3
n-1
- x
= + +º+
d d d
1
= ( x1- x2+ x2- x3+ xn-1-
xn
)
d
1
= ( x1
- xn
)
d
= ( n-1)
x1
xn
120. It is given that a, b and c are in AP.
\ 2b = a + c
It is given that
fi
fi
fi
  Â
n n n
x = a , y = b , z = c
n= 0 n= 0 n=
0
1 1 1
x = , y = , z =
1-a 1-b 1-c
1 1 1
(1 - a) = , (1 - b) = , (1 - c)
=
x y z
1 1 1
a = Ê Á1 - ˆ , b = Ê 1 - ˆ , c = Ê Á1-
ˆ ˜
Ë
˜
x¯ Á
Ë y
˜
¯ Ë z¯
d
n