1.Algebra Booster

1.28 Algebra Booster
8.
3 - 1
2
9. 2 2( 2 - 1)
10. a = 3
1 1
11. b Œ Ê -a, ˘ and r Œ È - , a
ˆ
Ë
Á
˜
3˚
˙
Î
Í 27 ¯
1/3 1/4 1/3 1/4 1/3 1/4
12.
(2x y - x - y + 1) x y
1/3 1/4 2
( x + y -1)
13. n = 2, 3, 4
14. ± 2
15. 1/201
16. 8, 24, 72, 216, 648
17. A = 3, B = 8
18. 24, 12, 6
22. –5050
23. –(2) 99 ¥ 10000
2 2
24.
n(4n - 1) c
3
25. 14
26. 3/2
27. 7
28. 2 n–1 – n – 2
nn ( + 1)( n+
2)
29.
6
30. 200
4
p
31.
96
n+ 1 3 nn ( + 1)( n+
2)
32. 2(2 -n
- 2) +
6
p
33.
4
Ê
n
x ˆ
34. Á1
-
( x+ 1)( x+ 2) º ( x+
n)
˜
Ë
¯
35. (2 –n + 2n – 1)
36. 0
37. 51
nn+ ( 1)
38.
4
INTEGER TYPE QUESTIONS
1. 2 2. 4 3. 7 4. 3 5. 3
6. 6 7. 7 8. 5 9. 6 10. 5
11. 6 12. 6 13. 4 14. 3 15. 8
COMPREHENSIVE LINK PASSAGES
Passage I: (i) Æ b, (ii) Æ c, (iii) Æ d;
Passage II: (i) Æ b, (ii) Æ a, (iii) Æ c;
Passage III: (i) Æ d, (ii) Æ a, (iii) Æ c;
Passage IV: (i) Æ d, (ii) Æ b, (iii) Æ c;
Passage V: (i) Æ a, (ii) Æ a, (iii) Æ d;
Passage VI: (i) Æ b, (ii) Æ b, (iii) Æ c;
Passage VII: (i) Æ a, (ii) Æ b, (iii) Æ a, (iv) Æ a, (v) Æ a;
Passage VIII: (i) Æ c, (ii) Æ b, (iii) Æ c;
Passage IX: (i) Æ c, (ii) Æ b, (iii) Æ c;
MATRIX MATCH
1. (A) Æ R; (B) Æ P ;(C) Æ P, Q, R; (D) Æ P
2. (A) Æ R; (B) Æ P; (C) Æ Q;
3. (A) Æ R; (B) Æ P; (C) Æ Q; (D) Æ S
4. (A) Æ Q; (B) Æ P; (C) Æ S; (D) Æ R
MATCHING LISTS
5. (C)
6. (B)
7. (D)
8. (B)
HINTS AND SOLUTIONS
LEVEL I
1. We have t n+1
– t n
= (3(n + 1) + 5) – (3n + 5)
= 3.
Hence, its common difference is 3.
2. Let the first term = a and the common difference = d
t p
= q fi a + (p – 1)d = q
t q
= p fi a + (q – 1)d = p
On subtracting, we get
(p – q)d = (p – q)
( p - q)
fi d = = –1
( q - p)
Also, a + (p – 1) ◊ (–1) = q
fi a – (p – 1) = q
fi a = q + p – 1
Thus, t n
= a + (n – 1)d
= q + p – 1 – (n – 1)
= p + q + n
3. Let the first term = a and the common difference = d
We have
mt m
= nt n
fi m(a + (m – 1)d) = n(a + (n – 1)d)
fi (m – n)a = –(m 2 – m – n 2 + n)d
fi (m – n)a = –[m 2 – n 2 – (m – n)]d
fi a = –(m + n – 1)d
Now, t m+n
= a + (m + n – 1)d
= –(m + n – 1)d + (m + n – 1)d
= 0.
Hence, the result.