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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