1.Algebra Booster

Sequence and Series 1.43
Ê
fi c ˆ Êcˆ
dÁ - 2e + f = 0
Ë
˜ Á ˜
a¯ Ëb¯
fi Ê d e f
Á ˆ - 2Á
Ê ˜
ˆ + = 0
Ë
˜
a¯
Ëb¯
c
fi
Êdˆ f Êeˆ
Á + = 2Á ˜
Ë
˜
a¯
c Ëb¯
d e f
fi , , are in AP
a b c
111. Let two numbers be a and b respectively. It is given
that,
a + b = 6 ab
fi (a + b) 2 = 36ab
fi (a – b) 2 = (a + b) 2 – 4ab = 32ab
Now,
2
( a + b) 36ab
=
2
( a - b)
32ab
fi
2
( a + b) 9
=
2
( a - b)
8
fi ( a + b ) =
3
( a - b) 2 2
fi ( a + b ) + ( a - b ) 3 +
=
2 2
( a + b) -( a -b) 3-
2 2
fi 2 a 3 +
= 2 2
2b
3 - 2 2
fi
a 3+
2 2
=
b 3 - 2 2
1 1
112. In AP, tm
= fi a + ( m- 1) d =
n
n
1 1
and tn
= fi a + ( n - 1) d =
m
m
On subtraction, we get,
1 1 ( m-
n)
( m- n)
d = - =
n m mn
1
fi d =
mn
1
Put d = in (i), we get,
mn
fi
1 1
a + ( m- 1) =
mn n
1 1 1
a + - =
n mn n
1
fi a =
mn
Thus, t mn
= a + (mn – 1)d
= 1 + ( mn -1)
1
mn mn
= 1
Hence, the mnth term of HP is
= 1 = 1 1
1
=
t mn
113. In an AP,
1 1
tm
= fi a + ( m- 1) d =
n
n
1 1
and tn
= fi a + ( n - 1) d =
m
m
On subtraction, we get,
1 1 ( m-
n)
( m- n)
d = - =
n m mn
1
fi d =
mn
1
Put d = in Eg. (i), we get
mn
1 1
a + ( m- 1) =
mn n
1 1 1
fi a + - =
n mn n
1
fi a =
mn
Thus, t m+n
= a + (m + n – 1)d
1 ( m+ n -1)
= +
mn mn
1 ( m+
n) 1
= + -
mn mn mn
Êm+
nˆ
= Á
Ë
˜
mn ¯
Ê mn ˆ
Hence, the (m + n)th term is Á
Ëm+
n˜
¯ .
114. In an AP,
t p
= q fi a + (p – 1)d = q
and t q
= p fi a + (q – 1)d = p
On subtraction, we get
(p – q)d = (q – p)
fi d = –1
Put d = –1, in Eg. (i), we get
fi a – (p – 1) = q
fi a = p + q – 1
Now,
t n
= a + (n – 1)d
= p + q – 1 – (n – 1)
= p + q – n
Hence, the nth term of the HP is
1 1
=
t p + q -n
n
(i)