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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 1 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 1
The value of $$\displaystyle \sum_{r = 16}^{30}{ (r + 2)(r - 3) } $$ is equal to :
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$$7782$$
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$$7787$$
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$$7790$$
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$$7780$$
Explanation
Let $$S= \displaystyle \sum_{r = 16}^{30}{ (r + 2)(r - 3) } =\displaystyle \sum_{r = 1}^{30}{ (r + 2)(r - 3) } -\sum_{r = 1}^{15}{ (r + 2)(r - 3) }$$
$$ = \displaystyle \sum_{r = 1}^{30}{ (r^2-r -6) } -\sum_{r = 1}^{15}{ (r^2 -r - 6) }$$
[By using $$ \sum_{i=1}^{n} k^2 =1^2 + 2^2 + 3^2 +… + n^2=\dfrac{n(n+1)(2n+1)}{6}$$]
$$ = \displaystyle \left( \frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2} -6r\right)_{r=30} -\left( \frac{r(r+1)(2r+1)}{6}-\frac{r(r+1)}{2} -6r\right)_{r=15}$$
$$ = \displaystyle \left( \frac{30(30+1)(60+1)}{6}-\frac{30(30+1)}{2} -180\right) -\left( \frac{15(15+1)(30+1)}{6}-\frac{15(15+1)}{2} -90\right)$$
$$=7780$$
Let $$\mathrm{a}_{1},\ \mathrm{a}_{2},\ \mathrm{a}_{3},\ \ldots,\ \mathrm{a}_{11}$$ be real numbers satisfying $$\mathrm{a}_{1}=15,27-2\mathrm{a}_{2}>0$$ and $$\mathrm{a}_{\mathrm{k}}=2\mathrm{a}_{\mathrm{k}-1}-\mathrm{a}_{\mathrm{k}-2}$$
for $$\mathrm{k}=3,4,\ \ldots, 11.$$ If $$\displaystyle \frac{a_{1}^{2}+a_{2}^{2}+\ldots+a_{11}^{2}}{11}=90$$, then the value of $$\displaystyle \frac{a_{1}+a_{2}+\ldots+a_{11}}{11}$$ is equal to
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
We have: $${ a }_{ k }+{ a }_{ k+2 }=2{ a }_{ k-1 }$$
Thus, the terms are in AP.
Hence, sum of squares of the terms in AP is:
$${ a }^{ 2 }+{ (a+d) }^{ 2 }+...+{ (a+10d) }^{ 2 }=11{ a }^{ 2 }+110ad+385{ d }^{ 2 }=990\\ =>a^{ 2 }+10ad+35{ d }^{ 2 }=90\\ =>35{ d }^{ 2 }+150d+225-90=0\\ =>35{ d }^{ 2 }+150d+135=0\\ =>7{ d }^{ 2 }+30d+27=0\\ =>(7d+9)(d+3)=0=>d=-3,-\dfrac { 9 }{ 7 } $$
Since $${ a }_{ 2 }<13.5$$, $$d=-3$$
Thus, the required answer i.e. the average of $$11$$ terms of an AP = $${ a }_{ 6 }=15+(6-1).(-3)=0$$
Hence, (a) is correct.
$$x\left (x + \dfrac {1}{x}\right )^{2} + \left (x^{2} + \dfrac {1}{x^{2}}\right )^{2} + \left (x^{3} + \dfrac {1}{x^{3}}\right )^{2}$$ .... upto $$n$$ terms is
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$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n + 2} + n}{x^{2n}} + 2n$$
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$$\dfrac {x^{2n} + 1}{x^{2} + 1}\times \dfrac {x^{2n + 2} - n}{x^{2n}} - 2n$$
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$$\dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n - 2} - n}{x^{2n}} - 2n$$
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None of these
Explanation
The series is
$$(x^{2} + x^{4} + x^{6} + ....) + \left (\dfrac {1}{x^{2}} + \dfrac {1}{x^{4}} + \dfrac {1}{x^{6}} + ....\right ) + (2 + 2 + ....)$$
$$= \dfrac {x^{2}(x^{2n} - 1)}{x^{2} - 1} + \dfrac {\dfrac {1}{x^{2}}\left (1 - \dfrac {1}{x^{2n}}\right )}{1 - \dfrac {1}{x^{2}}} + 2n$$
$$= \dfrac {x^{2}(x^{2n} - 1)}{x^{2} - 1} + \dfrac {x^{2n} - 1}{(x^{2} - 1)x^{2n}} + 2n$$
$$= \dfrac {x^{2n} - 1}{x^{2} - 1}\times \dfrac {x^{2n + 2} + 1}{x^{2}} + 2n$$
Which letter will be the fifth from the right if the first and second, the third and fourth and so on are interchanged among each other in the word "COMPANIONATE"?
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A
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I
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N
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O
Explanation
When we interchange the first and the second, third and the fourth, and so on letters in the word COMPANIONATE, the new word becomes OCPMNAOIANET.
Therefore, 5th letter from right is I
Answer is Option B
If the sum of the first n natural numbers is 1/5 times the sum of the their squares, then the value of n is
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5
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6
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7
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8
Explanation
$$\displaystyle \frac { n\left( n+1 \right) }{ 2 } =\frac { 1 }{ 5 } \left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) \\ \Rightarrow 15=2n+1\Rightarrow n=7$$
Mean of the first $$n$$ terms of the A.P. $$a, (a + d), (a + 2d), ........$$ is
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$$\displaystyle a + \frac{nd}{2}$$
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$$\displaystyle a + \frac{(n - 1)d}{2}$$
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$$a + (n - 1) d$$
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$$a + nd$$
Explanation
Required mean $$= \displaystyle \frac{a + (a + d) + (a + 2d) + ....... + \{ a + (n - 1) d \}}{n}$$
$$\displaystyle = \frac{\displaystyle \frac{n}{2} [a + a + (n - 1) d]}{n} = a + \frac{(n - 1)d}{2}$$
The sum of $$n$$ terms of the series whose $$n^{th}$$ term is $$n(n + 1)$$ is equal to.
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$$\dfrac {n(n + 1)(n + 2)}{3}$$
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$$\dfrac{(n + 1)(n + 2)(n+3)}{12}$$
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$$n^2 (n + 2)$$
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$$n(n + 1)(n + 2)$$
Explanation
$$T_n = n(n+1)=n^2+n$$
$$\therefore $$ Sum $$=$$ $$S_n =\sum T_n = \sum n^2+\sum n=\cfrac{n(n+1)(2n+1)}{6}+\cfrac{n(n+1)}{2}$$
$$=\cfrac{n(n+1)}{6}(2n+1+3)=\cfrac{n(n+1)(n+2)}{3}$$
Hence, option 'A' is correct.
The sum of
n
n
is equal to
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$$\dfrac {1}{4}n(n+1)(n+2)$$
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$$\dfrac {1}{4}n(n+1)(n+2)(n+3)$$
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$$\dfrac {1}{2}n(n+1)(n+2)(n+3)$$
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None of these
Explanation
$$T_n = n(n+1)(n+2)=n^3+3n^2+2n$$
$$\therefore S_n=\sum T_n =\sum n^3+3\sum n^2+2\sum n$$
$$ =\cfrac{(n(n+1)^2}{4}+3\cfrac{n(n+1)(2n+1)}{6}+2\cfrac{n(n+1)}{2}$$
$$ =\cfrac{1}{4}n(n+1)(n^2+5n+6)=\cfrac{1}{4}n(n+1)(n+2)(n+3)$$
Hence, option 'B' is correct.
In numbers from $$1 \ to\ 100$$ the digit "$$0$$" appears ____________times.
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$$9$$
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$$10$$
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$$11$$
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$$12$$
Explanation
$$10, 20, 30, 40, 50, 60, 70, 80, 90, 100$$
Thus the digit $$0$$ appears $$11$$ times.
The spread sheet on the right contains 20 cells. A cell In a spread sheet can be identified fIrst by the column letter and then by the row number. For example,. the number 10 is found in cell CIf the number in cell A3 =B4-3(E2 +D4) then which of the following must be the number in cell E2?
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-21
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-15
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-4
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-12
Explanation
Its given
A3=B4-3(E2+D4)
Substituting the values from the grid
$$18=-6-3(E2+4)$$
$$24=-3(E2+4)$$
$$-8-4=E2$$
$$E2=-12$$
Answer is Option D
Refer Diagram
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8
0%
9
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7
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12
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10
Explanation
The number inside the triangle is equal to the sum of the numbers obtained by multiplying number o either side of the triangle with the base number.
Thus, $$3 \times 1 + 3 \times 2 = 9$$
$$5 \times 2 + 3 \times 2 = 16$$
Hence, the missing term is: $$4 \times 1 + 5 \times 1 = 9$$
The sum of series $$1.2 + 2.3 + 3.4 + ....10$$ terms is
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$$440$$
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$$286$$
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$$524$$
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$$\infty$$
Explanation
$$1.2+2.3+3.4........10$$terms
The general terms of the above series can be written as
$${ T }_{ n }=n(n+1)$$
$$\therefore { S }_{ n }=$$Sum of $$n$$ terms of the series.
$$=\sum { { T }_{ n } } \\ =\sum { n(n+1) } \\ =\sum { ({ n }^{ 2 }+n) } \\ =\sum { { n }^{ 2 } } +\sum { n } \\ =\cfrac { n(n+1)(2n+1) }{ 6 } +\cfrac { n(n+1) }{ 2 } \\ =\cfrac { n(n+1) }{ 2 } [\cfrac { 2n+1 }{ 3 } +1]\\ =\cfrac { n(n+1)(n+2) }{ 3 } \\ \therefore { S }_{ 10 }=\cfrac { 10\times 11\times 12 }{ 3 } \\ =10\times 44\\ =440$$
8, 24, 48, 80, 120, .....
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158
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162
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164
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168
Explanation
Difference of two successive numbers are 16, 24, 32, 40 etc Hence the next number is 120 + 48 = 168
3, 15, 35, 63, 99, ....
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133
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137
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139
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143
Explanation
The difference between two successive terms is 12, 20, 28, 36, 44, etc
$$\displaystyle \therefore $$ The next number is 99 + 44 = 143
Find the Odd one among : 23, 35, 57, 711, 1113, 1316
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23
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35
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57
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1316
Explanation
Two successive terms for the terms of the series except in 1316 It should be 1317
Find the Odd one among :
35, 63, 105, 121, 133, 210
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35
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63
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105
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121
Explanation
All numbers are multiples of except 121
Find the wrong term in sequence :
1, 3, 7, 15, 27, 63, 127
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7
0%
15
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27
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63
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3
Explanation
The sequence is 2, $$\displaystyle 2^{2}$$, $$\displaystyle 2^{3}$$, $$\displaystyle 2^{4}$$, $$\displaystyle 2^{5}$$, $$\displaystyle 2^{6}$$ etc. so the correct term is 15+16=31 [Double +1 is the rule] so 27 is wrong 31 is right
Find out next term of the series 2, 7, 28, 63, 126, ....
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210
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213
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215
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219
Explanation
Given the series' terms can be written as
$$\displaystyle1^{3}+1$$, $$\displaystyle2^{3}-1$$, $$\displaystyle3^{3}+1$$, $$\displaystyle4^{3}-1$$, $$\displaystyle5^{3}+1$$, $$\displaystyle 6^{3}-1$$ etc.
Hence the next number is $$6^3-1=216 - 1 = 215$$
If $$\displaystyle 1^{3}+2^{3}+.........+9^{3}=2025,$$, Then$$\displaystyle\left ( 0.11 \right )^{3}+\left ( 0.22 \right )^{3}+.......\left ( 0.99 \right )^{3}$$ will be-
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0.2695
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2.695
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3.695
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0.3695
Explanation
$$\displaystyle \left ( \cdot 11 \right )^{3}+\left ( \cdot 22 \right )^{3}+......+\left ( \cdot 99 \right )^{3}$$
$$\displaystyle =\left ( \frac{11}{100} \right )^{3}+\left ( \frac{22}{100} \right )^{3}+...\left ( \frac{99}{100} \right )^{3}$$
$$\displaystyle= \left ( \frac{11}{100} \right )^{3}\left [ 1^{3}+2^{3}+.......+a^{3} \right ]$$
$$\displaystyle =\left ( \frac{11}{100} \right )^{3}\times 2025$$
$$\displaystyle \left [ \because \left ( 1^{3}+2^{3}+a^{3} \right )=2025 \right ]$$
(given)
$$\displaystyle =\frac{1331}{1000000}\times 2025=2\cdot 69527$$
Find a wrong number in the series: 7, 28, 63, 124, 215, 342, 511
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7
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28
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124
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215
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342
Explanation
Clearly , the correct sequence is:
$$\displaystyle 2^{3}-1$$, $$\displaystyle 3^{3}-1$$, $$\displaystyle 4^{3}-1$$, $$\displaystyle 5^{3}-1$$, $$\displaystyle 6^{3}-1$$, $$\displaystyle 7^{3}-1$$, $$\displaystyle 8^{3}-1$$.
$$ 3^{3}-1$$.
i.e. Hence, the answer is (b)
Find the wrong term in sequence :
1, 2, 4, 8, 16, 32, 64, 96
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4
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32
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64
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16
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96
Explanation
Each term is double the preceding term so 96 is the wrong term It should be 128
Find the wrong term in sequence :
5, 10, 17, 24, 37, 50, 65.
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10
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17
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24
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37
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50
Explanation
The sequence is +5, +7, +9 etc 24 is wrong and it should be 26
CE, GI, KM, OQ, ....
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TW
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TV
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SU
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RT
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UW
Explanation
The letters of each term are alternate Also the last letter of each term and the first letter of the next term are alternate
Find the Odd one among :
6, 19, 56, 169, 508, 1519
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19
0%
169
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508
0%
1519
Explanation
The numbers are written as
6 x 3 + 1, 19 x 3 - 1, 56 x 3 + 1, 169 x 3 -1 etc. i.e. 19, 56, 169, 506 etc Now in place of 508 it should be 506
Directions: In following quesiton, choose the missing term out of the given alternatives Reference:
A B C D E F G H I J K L M
N O P Q R S T U V W X Y Z
$$Z, X,V,T,R,.........$$
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$$O, K$$
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$$N, M$$
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$$K, S$$
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$$M, N$$
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$$P, N$$
Explanation
Clearly the given series consists of alternate letters in the reverse order So the missing terms would be P and N
C 4X, F9U, I 16 R
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K 25 P
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L 25 P
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L 25 O
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L 27 P
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none of these
Explanation
+3; $$\displaystyle 2^{2}$$, $$\displaystyle 3^{2}$$, $$\displaystyle 4^{2}$$, $$\displaystyle 5^{2}$$,-3
Find the Odd one among :
517, 661, 814, 922, 1066, 1256
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661
0%
814
0%
1256
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922
Explanation
Sum of the integers in each number is 13 hence 1256 is the odd man out
Find the Odd one among :
10, 34, 130, 204, 290
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34
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74
0%
130
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204
Explanation
The numbers in the given set are sum of the squares on 1,3; 3, 5; 5, 7 etc i.e. $$\displaystyle 1^{2}+ 3^{2}= 10$$, $$\displaystyle 3^{2}+ 5^{2}= 34$$ etc Hence in place of 204 it should be $$\displaystyle 9^{2}+ 11^{2}= 202$$
Find the Odd one among :
23, 13, 34, 25, 56, 51
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23
0%
13
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34
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51
Explanation
The number in the alternate places is the sum of the square of the integer on the other e.g. 13 = $$\displaystyle 2^{2}$$+$$\displaystyle 3^{2}$$, 25 = $$\displaystyle 3^{2}$$+$$\displaystyle 4^{2}$$. Now 56 + $$\displaystyle 5^{2}$$ + $$\displaystyle 6^{2}$$ = 61 Hence in place of 51 it should be 61
Find the Odd one among :
123, 14, 246, 56, 369, 125
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123
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14
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246
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125
Explanation
Sum of the squares of 123 = $$\displaystyle 1^{2}$$ + $$\displaystyle 2^{2}$$, $$\displaystyle 3^{2}$$ = 14 etc Hence 369 $$\displaystyle \rightarrow $$ $$\displaystyle 3^{2}$$ + $$\displaystyle 6^{2}$$ + $$\displaystyle 9^{2}$$ = 9 + 36 + 81 = 126 in place of 125
aab - aa - bbb - aaa- bbb
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abba
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baab
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aaab
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abab
Explanation
The series is a bb/ aaabbb/aaaabbbb Thus the letters are reparted twice thrice and four times etc
Find the Odd one among :
7, 26, 65, 124, 215, 342
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7
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26
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65
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124
Explanation
Cubing of 2, 3, 4, 5, 7 and then subtract 1 from the results In place of 65 it should be 63
- - G - C - GK - PG -
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KCPCPK
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CPKPCK
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PKCPKP
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CPPKCP
Explanation
The series is - - G -
C - G K
- P G -
Clearly the group of four letters is CPGK
abb - baa - a - bab - ab
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abba
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abab
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ccac
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aabb
Explanation
The series is abba/baab/abba/baab
Thus the pattern abba baab is repeated
Find the missing letters
A, D, H, M, S .....
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0%
U
0%
V
0%
Y
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Z
Explanation
Pattern is
$$\displaystyle ^{1}A,^{1}A+3,^{4}D+4,^{8}H+5,^{13}M+6,^{19}S+7,^{26}Z$$.....
$$\displaystyle \therefore $$ Missing letter = Z
a- ca - bc - bcc - bca
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bbaa
0%
bbab
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abbb
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baba
Explanation
The series is abcab/ bcabc/ cabca
- - aba - - ba - ab
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0%
abbba
0%
abbab
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baabb
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bbaba
Explanation
The series is ab/ab/ab/ab/ab/ab
aab - aaa- bba -
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0%
baa
0%
abb
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bab
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aab
Explanation
The series is a a b b a a / a a b b a a
c - bba - cab - ac - ab - ac
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abcbc
0%
acbab
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babcc
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bcacb
Explanation
The series is cab/bac/cab/ bac/cab/ bac
Thus the pattern cab bac is repeated
- acca - ccca - acccc - aaa
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acca
0%
caaa
0%
ccaa
0%
caac
Explanation
The series is c a / c c a a / c c c a a a / c c c c a a a a
Find the missing letters
X, U, R, O, L, .....
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0%
I
0%
J
0%
K
0%
L
Explanation
Pattern is
$$\displaystyle ^{24}X,^{24}X-3,^{21}U-3,^{18}R-3,^{15}O-3,^{12}L-3,^{9}I$$........
$$\displaystyle \therefore $$ Missing letter = I
abca - bcaab - ca - bbc -
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ccaa
0%
bbaa
0%
abac
0%
abba
Explanation
The series is abc/aabc/aabbc/ aabbcc
Find the missing letters
ACD, GHI, ....., UVWXY
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0%
LMNO
0%
MNO
0%
NOPQ
0%
MNOP
Explanation
Pattern is $$\displaystyle ^{1}A^{3}C^{4}D,^{7}G^{8}H^{9}I,^{13}M^{14}N^{15}O^{16}P,^{21}U^{22}V^{23}W^{24}X^{25}Y,$$ ......
$$\displaystyle \therefore $$ Missing letter = MNOP
Find the missing letters
shg, rif, qje, pkd, .....
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0%
ole
0%
olc
0%
nmc
0%
nlb
Explanation
Pattern is $$\displaystyle s_{8}^{8}h^{7}g,r_{9}^{9}i^{6}f,q_{10}^{10}f^{5}e,p_{11}^{11}k^{4}d,o_{12}^{12}l^{3}c,$$ .....
$$\displaystyle \therefore $$ Missing letter = olc
Find the missing letters
R, M, ....., F, D, C
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0%
F
0%
G
0%
H
0%
I
Explanation
Pattern is
$$\displaystyle ^{18}R,^{18}R-5,^{13}M-4,^{9}I-3,^{6}F-2,^{4}D-1,^{3}C.....$$
$$\displaystyle \therefore $$ Missing letter = I
Find the missing letters
......., PSV, EHK, TWZ, ILO
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0%
BEH
0%
ADG
0%
ZCF
0%
IMP
Explanation
Pattern is combination of two letter series
$$\displaystyle ^{1}A^{4}D^{7}G,P_{11}S_{8}V_{5},^{5}E^{8}H^{11}K,T_{7}W_{4}Z_{1},^{9}I^{12}L^{15}O,$$ ....
$$\displaystyle \therefore $$ Missing letters = ADG
Find the missing letters
ab, ba, abc, cba, abcd, ....
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0%
acbd
0%
bacd
0%
cabd
0%
dcba
Explanation
Pattern is
$$\displaystyle ^{1}a^{2}b,^{2}b^{1}a,^{1}a^{2}b^{3}c,^{3}c^{2}b^{1}a,^{1}a^{2}b^{3}c^{4}d,^{4}d^{3}c^{2}b^{1}a,$$ .......
$$\displaystyle \therefore $$ Missing letters = dcba
Find the missing letters
AZ, BY, CX, DW, .......
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0%
EU
0%
EV
0%
EW
0%
FW
Explanation
Pattern is $$\displaystyle ^{1}AZ_{1},^{2}BY_{2},^{3}CX_{3},^{4}DW_{4},^{5}EV_{5},$$......
$$\displaystyle \therefore $$ Missing letter = EV
Find the missing letters
......, SIY, OEU, KAQ, GWM, CSI
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0%
VMC
0%
WMC
0%
VNC
0%
WNC
Explanation
Pattern is $$\displaystyle ^{23}W^{13}M^{3}C,^{19}S^{9}I^{25}Y,^{15}O^{5}E^{21}U,^{11}K^{1}A^{17}Q,^{7}G^{23}W^{13}M,^{3}C^{18}R^{9}I,$$ .....
$$\displaystyle \therefore $$ Missing letter = WMC
Find the missing letters
YXZ, XWY, ..... VUW, UTV, TSU
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0%
VWT
0%
WXV
0%
TVX
0%
WVX
Explanation
Pattern is $$\displaystyle Y_{2}X_{3}Z_{1},X_{3}W_{4}Y_{2},W_{4}V_{5}X_{3},V_{5}U_{6}W_{4},U_{6}T_{7}V_{5},T_{7}S_{8}U_{6},$$ .....
$$\displaystyle \therefore $$ Missing letters = WVX
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