Question 1 |

Find the Arithmetic Mean of series: 2, 6, 10, 14, 18, 22, 26, 30.

16 | |

8 | |

64 | |

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

=(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

32 | |

88 | |

128 | |

110 |

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Question 2 Explanation:

Sum of AP = (n/2)[2a+(n-1)d]

= 4*[4+7*4]

=128

= 4*[4+7*4]

=128

Question 3 |

Find the AM of series: 10, 7, 4, 1, -2

13/2 | |

14/3 | |

4 | |

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

40 | |

21 | |

20 | |

18 |

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Question 4 Explanation:

Sum of series = (n/2)[2a+(n-1)d]

= 20

= 20

Question 5 |

Find sum of series: 2, 2.5, 3, 3. 5, 4, 4. 5..........11

120 | |

123.5 | |

126.5 | |

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

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

13/2 | |

25/8 | |

19 | |

22/9 |

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Question 6 Explanation:

AM of Series: (2+11)/2

=13/2

=13/2

Question 7 |

Find the sum of series: 1, 3, 9, 27, 81, ..............3

^{9}[(1-3 ^{10})]/(1-3) | |

18 | |

10 | |

20 |

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Question 7 Explanation:

Sol: Sn=[a(1-r

^{n})]/(1-r) =[1(1-3^{10})]/(1-3) =[(1-3^{10})]/(1-3)Question 8 |

Find the sum of series: 1/3, 1/9, 1/27, 1/81.................

1/2 | |

1/3 | |

1/4 | |

1/6 |

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Question 8 Explanation:

Sn=a/(1-r)

= (1/3)/(1-1/3)

=1/2

= (1/3)/(1-1/3)

=1/2

Question 9 |

If the product of n positive integers is n

^{n}, then their sum is:A negative integer | |

Equal to n | |

Equal to n+(1/n) | |

Never less than n ^{2} |

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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 = 3

^{3}. This can also be written as n^{n}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:540 m | |

600 m | |

720 m | |

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:

Remain unchanged | |

Increase by 1 | |

Increase by 1.5 | |

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 |

log

_{10}10 + log_{10}10^{2}+ ... + log_{10}10^{n}n ^{2} + 1 | |

n ^{2} - 1 | |

[(n ^{2} + n)/2] * [(n(n+1))/3] | |

(n ^{2} + 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 = (n

^{2}+n)/2Question 13 |

x, 17, 3x - y

^{2}- 2 and 3x + y^{2}- 30 are four consecutive terms of an increasing arithmetic sequence. The sum of the four numbers is divisible by:2 | |

3 | |

5 | |

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

^{13}∑_{n=1}1/n can also be written as x/13! What would be the remainder if x is divided by 11?2 | |

4 | |

7 | |

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

^{100}∑_{i=2}1/log_{i}100! is:0.01 | |

0.1 | |

1 | |

10 |

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Question 15 Explanation:

The expression

^{100}∑_{i=2}1/log_{i}100! can be rewritten as: 1/log_{2}100! + 1/log_{3}100! + ... + 1/log_{100}100! = log_{100!}2 + log_{100!}3 + ... + log_{100!}100 = log_{100!}2*3*...*100 = log_{100!}100! = 1.
There are 15 questions to complete.