1.Algebra Booster

Sequence and Series 1.79
Also, a, x, y, z, b are in HP.
fi 1 , 1 , 1 , 1 ,
1 are in HP.
a x y z b
fi 1 1 1 3 Ê 1 1 ˆ
+ + = Á +
x y z 2 Ë
˜
a b¯
3Êa
+ bˆ 3Ê10ˆ
= Á =
2Ë ˜
ab ¯
Á ˜
2Ëab¯
fi 3 Ê 10 ˆ
Á
5 =
2Ë
˜
ab¯
3
fi ab = 9
Clearly a, b are the roots of
t 2 – (a + b)t + ab = 0
fi t 2 – 10t + 9 = 0
fi (t – 9)(t – 1) = 0
fi t = 9 or 1
fi a = 9, b = 1
Hence, the value of (a – b – 2) is 6
10. We have
6
1 - p
1 - p
2 3 5
= (1 + 3 x + (3 x) + (3 x) + … + (3 x ))
6 6
1- p 1 -(3 x)
fi =
1- p 1 -(3 x)
Comparing, we get
p = 3x
p
fi 2 3 2 5
x + = + =
11. We have
n+
5
 4( x - 3) = 4 ( x - 3)
x=
5
n+
5
Â
x=
5
= 4[2 + 3 + 4 + … + (n + 2)]
n
= 4 ◊ (2+ n + 2)
2
= 2n(n + 4)
= 2n 2 + 8n
Comparing with An 2 + B n
+ C we get
A = 2, B = 8, C = 0
Hence, the value of (A + B – C – 4) is 6.
12. We have
n
Â
n
Â
r = 1
rr ( + 1)(2r+
3)
3 2
= (2r + 5r + 3 r)
r = 1
n
3
n
2
n
  Â
= 2 r + 5 r + 3 r
r= 1 r= 1 r=
1
2
Ênn ( + 1) ˆ Ênn ( + 1)(2n+ 1) ˆ Ênn
( + 1) ˆ
= 2Á + 5Á
˜ + 3
Ë
˜
2 ¯ Ë
Á ˜
6 ¯ Ë 2 ¯
Ênn ( + 1) ˆ È Ênn ( + 1) ˆ Ê(2n+ 1) ˆ ˘
= Á 2 + 5 + 3
Ë
˜
2 ¯ Í Á ˜ Á ˜
2 3
˙
Î Ë ¯ Ë ¯ ˚
1 ( 1)(3
2
= nn+ n + 13 n+
14)
6
1 ( 1)(3 3 13 2
= n + n + n + 14 n )
6
1 (3 4 16 3 27 2
= n + n + n + 14 n )
6
Ê3+
27 ˆ
Now, (a + c + 1) = Á + 1
Ë
˜
6 ¯
= 5+ 1=
6
13. As we know that
fi
Ê
2 2
x + y ˆ Êx + yˆ
Á ≥
Ë
˜
2 ¯
Á
Ë
˜
2 ¯
2
Ê8ˆ Êx
+ yˆ
Á ≥ Á ˜
Ë
˜
2¯
Ë 2 ¯
2
Êx
+ yˆ
fi 4 ≥ Á
Ë
˜
2 ¯
fi (x + y) 2 £ 16
fi (x + y) £ 4
Hence, the maximum value of (x + y) is 4.
14. We have
n
Ê
k
2ˆ
ÂÂ Á m ˜
k= 1Ëm=
1 ¯
n
Êkk
( + 1)(2k+
1) ˆ
= Â Á
Ë
˜
6 ¯
k = 1
n
1
2
= Â ( k(2k + 3k
+ 1))
6
k = 1
n
2
1 3 2
= Â (2k + 3 k + k)
6
k = 1
2
1 È Ênn ( + 1) ˆ Ênn ( + 1)(2n+ 1) ˆ Ênn
( + 1) ˆ˘
= Í2Á ˜ + 3Á ˜ + Á ˜˙
6Í
Ë 2 ¯ Ë 6 ¯ Ë 2 ¯
Î
˙˚
1 Ênn
( + 1) ˆÈ Ênn
( + 1) ˆ ˘
= Á 2 + 2n
+ 1+
1
6Ë ˜
2 ¯Í
Á ˜
2
˙
Î Ë ¯ ˚
1 Ênn
( + 1) ˆ 2
= Á ( n + 3n
+ 2)
6Ë
˜
2 ¯
1 (
2 2
= n + n)( n + 3n
+ 2)
12
1 (
4 4
3 5
2
= n + n + n + 2 n )
24
Comparing with an 4 + bn 3 + cn 2 + d n
+ e,
We have
1 1
a = and d =
12 6