1.Algebra Booster

Sequence and Series 1.83
fi 1 2 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + …
2
… + (2m
-1)
2
2 m(2m+
1)
2
= - 2 ◊(2 m)
2
= m((2m + 1) 2 – 4m 2 )
= m(2m – 1) 2
Put 2m – 1 = n, we get
n 2 ( n + 1)
1 2 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + … + n 2 = .
2
21. Let a 1
, a 2
, …, a n
ΠAP with the common difference d
and a 1
= A and a 2n – 1
= B.
Then a 2n – 1
– a 1
= B – A
(2n – 1 – 1)d = B – A
2(n – 1)d = B – A
B - A
d =
2( n - 1)
Thus, a = a n
= a 1
+ (n – 1)d
( B - A)
Ê A+
Bˆ
= A + =Á
2 Ë
˜
2 ¯
Next, if b 1
, b 2
, …, b 2n – 1
ΠGP with the common ratio r
and b 1
= A, b 2n – 1
= B
b2n
-1
B
Thus, =
b1
A
2n-2
B
r =
A
n-1
B
r =
A
n-1
Now, b = b = br = AB
n
1
2ab
Similarly, c = a + b
It is easy to show that ac – b 2 = 0.
1
Now, a - b = ( A+ B ) - AB
2
1 ( ) 2
= A - B ≥ 0
2
Similarly, 1 - 1 ≥0 fib≥c
c b
Thus, a ≥ b ≥ c.
23. Given
x Ê x 7ˆ
log32, log 3(2 -5), log3Á2
- ˜ are in AP.
Ë 2¯ x
Ê x 7ˆ
fi 2log 3(2 - 5) = log32 + log3Á2
-
Ë
˜
2¯
Ï Ê x 7ˆ¸
log (2 - 5) = log Ì2 ◊Á2
- ˜˝
Ó Ë 2¯˛
fi
x 2
3 3
fi
x 2 Ï Ê x 7ˆ¸
(2 - 5) = Ì2 ◊Á2
- ˜˝
Ó Ë 2¯˛
2 Ï Ê 7ˆ¸
x
fi ( a - 5) = Ì2 ◊Áa - ˜˝, a = 2
Ó Ë 2¯˛
fi a 2 – 10a + 25 = 2a – 7
fi a 2 – 12a + 32 = 0
fi (a – 8)(a – 4) = 0
fi a = 4 or 8
When a = 4
fi 2 x = 4 = 2 2 fi x = 2
When a = 8
fi 2 x = 2 3
fi x = 3
But x = 2 does not satisfy the given equation.
Thus, the solution is x = 3.
24. Let a and b be two numbers respectively.
b - a na + b
Then p = a + =
n + 1 n+
1
and 1 1 Ê n 1 ˆ
= Á + ˜
q n + 1Ëa b¯
Ê 1 1
Thus, ( n 1) p na b and n + n ˆ
+ - = Á - =
Ë q a
˜
¯ b
Eliminating b, we get
Ên+
1 nˆ
(( n+ 1) p -na) Á - = 1
Ë q a
˜
¯
fi ((n + 1)p – na)(a(n + 1) – nq) = qa
fi (n + 1) 2 ap – npq(n + 1) – na 2 (n + 1) + (n 2 – 1)aq
= 0
fi (n + 1)ap – npq – na 2 + (n – 1)aq = 0
fi na 2 – ((n – 1)q + (n + 1)p)a + npq = 0
As a is real, so
((n – 1)q + (n + 1)p) 2 – 4n 2 pq ≥ 0
fi (n – 1) 2 q 2 – 2(n 2 + 1)pq + (n + 1) 2 p 2 ≥ 0
Thus q cannot lie between the roots of
(n – 1) 2 x 2 – 2(n 2 + 1)xp + (n + 1) 2 p 2 = 0
One of the roots of this equation is p and the other is
2
Ên
+ 1ˆ
Á p.
Ën
- 1˜
¯
2
Ên
+ 1ˆ
2
Also product of the roots = Á p
Ën
- 1˜
¯
2
Ên
+ 1ˆ
Thus q cannot lie between the roots of p and Á p.
Ën
- 1˜
¯
1
25. We have S 1 = = 2
1
1 -
2