Question 1
Find the Arithmetic Mean of series: 2, 6, 10, 14, 18, 22, 26, 30.
A
16
B
8
C
64
D
36
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Question 1 Explanation: 
AM = (a1+a2+a3+......+an)/n
=(n(a1+an)/2)/n
=(a1+an)/2 = (2 + 30)/2 = 16
Question 2
Find the Sum of series: 2, 6, 10, 14, 18, 22, 26, 30
A
32
B
88
C
128
D
110
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Question 2 Explanation: 
Sum of AP = (n/2)[2a+(n-1)d]
= 4*[4+7*4]
=128
Question 3
Find the AM of series: 10, 7, 4, 1, -2
A
13/2
B
14/3
C
4
D
16/5
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Question 3 Explanation: 
AM = (a1+a2+a3+......+an)/n =(n(a1+an)/2)/n =a1+a2/2 =10-2/2 = 4
Question 4
Find the Sum of series: 10, 7, 4, 1, -2
A
40
B
21
C
20
D
18
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Question 4 Explanation: 
Sum of series = (n/2)[2a+(n-1)d]
= 20
Question 5
Find sum of series: 2, 2.5, 3, 3. 5, 4, 4. 5..........11
A
120
B
123.5
C
126.5
D
118.5
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Question 5 Explanation: 
sum of AP = (n/2)[2a+(n-1)d]
n=19, a=2, d=1/2
S = (19/2)[2*2+(19-1)1/2]
=(19/2)[4+9]
=9.5*13 = 123.5
Question 6
Find Arithmetic Mean of series: 2, 2.5, 3, 3. 5, 4, 4. 5..........11
A
13/2
B
25/8
C
19
D
22/9
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Question 6 Explanation: 
AM of Series: (2+11)/2
=13/2
Question 7
Find the sum of series: 1, 3, 9, 27, 81, ..............39
A
[(1-310)]/(1-3)
B
18
C
10
D
20
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Question 7 Explanation: 
Sol: Sn=[a(1-rn)]/(1-r) =[1(1-310)]/(1-3) =[(1-310)]/(1-3)
Question 8
Find the sum of series: 1/3, 1/9, 1/27, 1/81.................
A
1/2
B
1/3
C
1/4
D
1/6
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Question 8 Explanation: 
Sn=a/(1-r)
= (1/3)/(1-1/3)
=1/2
Question 9
If the product of n positive integers is nn, then their sum is:
A
A negative integer
B
Equal to n
C
Equal to n+(1/n)
D
Never less than n2
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Question 9 Explanation: 
Clearly, since the given integers are positive, their sum can't be negative. Also, since the numbers are all integers their sum can't be a fraction. Let's take 1, 3 and 9. The product of these three integers is 27 = 33. This can also be written as nn where n=3. As we can see, the sum of these 3 integers is not equal to 3. Therefore, we are left with the fourth option.
Question 10
A tennis ball is initially dropped from a height of 180 m. After striking the ground, it rebounds (3/5)th of the height from which it has fallen. The total distance that the ball travels before it comes to rest is:
A
540 m
B
600 m
C
720 m
D
900 m
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Question 10 Explanation: 
The total distance traveled by the ball is the sum of two infinite series: a. Series 1: the distance traveled by the ball when it's falling down b. Series 2: the distance traveled by the ball when it's bouncing up S1 = a1 / (1 - r1) and S2 = a2 / (1 - r2) S1 = 180 / (1 - 3/5) and S2 = (180 * 3/5) / (1 - 3/5) S1 = 180 / (2/5) and S2 = 108 / (2/5) S1 = 180 * 5/2 and S2 = 108 * 5/2 S1 = 450 and S2 = 270 Therefore, S = S1+S2 = 720 m.
Question 11
The average of 7 consecutive number is P. If the next three numbers are also added, the average shall:
A
Remain unchanged
B
Increase by 1
C
Increase by 1.5
D
Increase by 2
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Question 11 Explanation: 
We are given P = [(a-3)+(a-2)+(a-1)+a+(a+1)+(a+2)+(a+3)]/7 assuming that the middle number is a. Then, P = 7a/7 = a. If we add, the next three integers then the average becomes Q = {[(a-3)+(a-2)+(a-1)+a+(a+1)+(a+2)+(a+3)] + [(a+4)+(a+5)+(a+6)]}/10 = {7a + 3a + 15}/10 = {10a + 15}/10 = a + 1.5. So, the average is increased by Q-P = 1.5
Question 12
log1010 + log10102 + ... + log1010n
A
n2 + 1
B
n2 - 1
C
[(n2 + n)/2] * [(n(n+1))/3]
D
(n2 + n)/2
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Question 12 Explanation: 
Rewriting the expression as 1 + 2 + ... + n We know, sum of n natural numbers is given by n*(n+1)/2 = (n2+n)/2
Question 13
x, 17, 3x - y2 - 2 and 3x + y2 - 30 are four consecutive terms of an increasing arithmetic sequence. The sum of the four numbers is divisible by:
A
2
B
3
C
5
D
7
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Question 13 Explanation: 
Let's rewrite the 4 terms as (17-d), 17, (17+d), (17+2d) where d is the common difference. Then, Sum = (17-d) + 17 + (17+d) + (17+2d) = 17*4 + 2d = 2 * (17*2 + d). So, their sum is divisible by 2.
Question 14
Expression 13n=11/n can also be written as x/13! What would be the remainder if x is divided by 11?
A
2
B
4
C
7
D
9
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Question 14 Explanation: 
We have, x/13! = 1/1 + 1/2 + 1/3 ... + 1/13 x = 13!/1 + 13!/2 + 13!/3 ... + 13!/13 Clearly, all the terms in the summation are divisible by 11 except 13!/11. We can rewrite it as 13*12*11*...*2*1 / 11 = 13*12*10! We have to find the remainder for 13*12*10! / 11. But, we know, remainder for ((n-1)!/n) = -1. Then, Remainder for 13*12*10! / 11 = 2*1*(-1) = -2. Converting the negative remainder to positive, -2+11 = 9.
Question 15
The value of the expression 100i=2 1/logi100! is:
A
0.01
B
0.1
C
1
D
10
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Question 15 Explanation: 
The expression 100i=2 1/logi100! can be rewritten as: 1/log2100! + 1/log3100! + ... + 1/log100100! = log100!2 + log100!3 + ... + log100!100 = log100!2*3*...*100 = log100!100! = 1.
There are 15 questions to complete.

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