1.Algebra Booster

Sequence and Series 1.89
fi 11a 2 + 2(1 + 2 + 3 + … + 10)ad
+ d 2 (1 2 + 2 2 + …
+ 10 2 ) = 990
fi 11a 2 + 2.55ad + 55.7 ◊ d 2 = 990
fi a 2 + 10ad + 35d 2 = 90
fi 225 + 150d + 35d 2 = 90
fi 35d 2 + 150d + 135 = 0
fi 7d 2 + 30d + 27 = 0
fi 7d 2 + 21d + 9d + 27 = 0
fi 7d(d + 3) + 9(d + 3) = 0
fi (d + 3)(7d + 9) = 0
fi d = –3 or – 9/7
27
Since a 2 < , so d = –3.
2
Now, a1+ a2+º+
a11
11
a + ( a + d) + ( a + 2 d) + … + ( a + 10 d)
=
11
11 a + d(1 + 2 + 3 +º+ 10)
=
11
11a
+ 55d
=
11
= a + 5d
= 15 -15
= 0.
57. Given a 1
, a 2
, …, a 100
are in AP.
Here, a 1
= 3 and S = ( a ),1 £ p£
100
p
p
Â
i = 1
5 n
S
(6 (5 1) )
m S
+ n-
d
5n
Now, = = 2
Sn
S n
n (6 + ( n-1) d)
2
(6 – d) + 5nd
= 5 ¥
(6 - d)
+ nd
It is independent of n only when 6 – d = 0
Thus, d = 6
Now, a 2
= a 1
+ d = 3 + 6 = 9
58. Given a 1
, a 2
, …, a n
ΠHP
fi
Given
fi
1 1 1
, ,…, AP
a a a Œ
1 2
20 1
n
1 1
= + 19d
a a
1 1
19d = -
a a
20 1
fi
1 1 4
19d = - = -
25 5 25
fi
4
d =-
25 ¥ 19
Also,
1
a < 0 , so
n
i
1
( n 1) d 0
a + - <
1
fi 1 Ê
( 1) 4 ˆ
+ n - 0
5 Á- <
Ë 25¥
19˜
¯
fi
fi
fi
59. We have S
4( n - 1) 1
>
25 ¥ 19 5
19 ¥ 5
( n - 1) >
4
19 ¥ 5
n > 1+ ≥25
4
n
4n
Â
k = 1
k( k+
1)
2
= (-1)
◊k
= –1 2 – 2 2 + 3 2 + 4 2 – 5 2 – 6 2 + 7 2 + 8 2 – 9 2 – 10 2 + …
– (4n – 1) 2 + (4n) 2
= (3 2 – 1 2 ) + (4 2 – 2 2 ) + (7 2 – 5 2 ) + (8 2 – 6 2 ) + …
+ ((4n) 2 – (4n – 2) 2 )
= 2(3 + 1) + 2(4 + 2) + 2(7 + 5) + …
+ 2(4n + 4n – 2)
= 2(3 + 1) + 2(4 + 2) + 2(7 + 5) + …
+ 2(4n – 1 + 4n – 3) + 2(4n + 4n – 2)
= 2(1 + 2 + 3 + 4 + … + (4n – 1) + 4n)
2.4 n(4n + 1)
=
2
= 4n(4n + 1)
When 4n(4n + 1) = 1056
fi 4n 2 + n – 264 = 0
fi n = 8
When 4n(4n + 1) = 1132
fi n(4n + 1) = 283
fi 4n 2 + n – 283 = 0
fi n = 9
60. Given a, b, c are in GP.
fi b 2 = ac
2
b
fi c =
a
a + b + c
Now, = b + 2
3
fi a + b + c = 3b + 6
fi a – 2b + c = 6
fi
fi
fi
2
b
a - 2b
+ = 6
a
2
6
1 - 2 b b
2
a
+ a
= a
2
Êb
ˆ 6
Á - 1 =
Ë
˜
a ¯ a
fi Since b is an integer, so a = 6
a
Ê
2
a + a - 14ˆ
36 + 6 -14 28
Now, Á
= = = 4
Ë a + 1
˜
¯ 7 7
2