CHAPTER
: Binomial Theorem, Sequences And Series
1. The value of 2 + 4 + 6 +…….+ 2n is
(1) n(n-1)/2
(2) n(n+1)/2
(3) 2n(2n+1)/2
(4)
n(n + 1)
Solution
2. The coefficient of x6 in
(2 + 2x)10 is
(1) 10C6
(2) 26
(3) 10C6 26
(4)
10C6 210.
Solution
3. The coefficient of x8y12
in the expansion of (2x + 3y)20 is
(1) 0
(2) 28312
(3) 28312 + 21238
(4)
20C8 28312.
Solution
4. If nC10 > nCr
for all possible r, then a value of n is
(1) 10
(2) 21
(3) 19
(4)
20.
Solution
5. If a is the arithmetic mean and g is
the geometric mean of two numbers, then
(1) a ≤ g
(2)
a ≥ g
(3) a = g
(4) a > g.
Solution
6. If (1 + x2)2 (1
+ x)n = a0 + a1x + a2x2
+……..+ xn+4 and if a0, a1, a2 are
in AP, then n is
(1) 1
(2) 2
(3)
3
(4) 4.
Solution
7. If a, 8, b are in AP, a, 4, b are in
GP, and if a, x, b are in HP then x is
(1)
2
(2) 1
(3) 4
(4) 16.
Solution
8. The sequence form an
(1) AP
(2) GP
(3)
HP
(4) AGP.
Solution
9. The HM of two positive numbers whose
AM and GM are 16, 8 respectively is
(1) 10
(2) 6
(3) 5
(4)
4.
Solution
10. If Sn denotes the sum of
n terms of an AP whose common difference is d, the value of Sn - 2Sn-1 + Sn-2 is
(1)
0
(2) 2d
(3) 4d
(4) d2.
11. The remainder when 3815
is divided by 13 is
(1)
12
(2) 1
(3) 11
(4) 5.
Solution
12. The nth term of the
sequence 1, 2, 4, 7, 11,…….is
(1) n3 + 3n2 + 2n
(2) n3 - 3n2
+ 3n
(3) n(n+1)(n+2) / 3
(4)
n2-n+2 / 2 .
13. The sum up to n terms of the series is
Ans:
4
Solution
14. The nth term of the
sequence 1/2 , 3/4 , 7/8 , 15/16 ,……….is
(1) 2n - n - 1
(2)
1 - 2-n
(3) 2-n + n - 1
(4) 2n-1.
Solution
15. The sum up to n terms of the series √2
+ √8 + √18 + √32 +……….is
(1) n(n+1) / 2
(2) 2n(n + 1)
(3)
n(n+1) / 2
(4) 1.
Solution
16. The value of the series 1/2 + 7/4 +
13/8 + 19/16 +…………. is
(1) 14
(2)
7
(3) 4
(4) 6.
Solution
17. The sum of an infinite GP is 18. If
the first term is 6, the common ratio is
(1) 1/3
(2)
2/3
(3) 1/6
(4) 3/4 .
Solution
18. The coefficient of x5 in
the series e-2x is
(1) 2/3
(2) 3/2
(3)
-4/15
(4) 4/15 .
Solution
19. The value of 1/2! + 1/4! + 1/6! +………
is
(1) e2+1 / 2e
(2) (e+1)2 / 2e
(3)
(e-1)2
/ 2e
(4) e2+1 / 2e .
Solution
20. The value of is
(1) log (5/3)
(2)
3/2 log(5/3)
(3) 5/3 log(5/3)
(4) 2/3 log(2/3)
Solution
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