Question 1 1 / -0
Solution
The series alternates the addition of $$5$$ with the subtraction of $$2$$. So, the number is $$20$$.
Question 2 1 / -0
Solution
This is a continuous series of adding $$1$$ to the previous number, in which every third number is not repeated. So, the missing number is $$5$$.
Question 3 1 / -0
Solution
The series is a perfect square in increasing order of $$12, 13, 14, 15...$$ So, the value of $$x$$ is $$196$$.
Question 4 1 / -0
Solution
Now, $$\displaystyle { 2 }^{ n }\tan { \left( { 2 }^{ n }a \right) } +{ 2 }^{ n }\cot { \left( { 2 }^{ n }a \right) } $$ $$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \sin { { 2 }^{ n-1 }a } }{ \cos { { 2 }^{ n-1 }a } } +2\frac { \cos { { 2 }^{ n }a } }{ \sin { { 2 }^{ n }a } } \right] $$ $$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \cos { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } +\sin { { 2 }^{ n }a\sin { { 2 }^{ n-1 }a+\cos { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } } } }{ \sin { { 2 }^{ n }a\cos { { 2 }^{ n-1 }a } } } \right] $$ $$\displaystyle ={ 2 }^{ n-1 }\left[ \frac { \cos { { 2 }^{ n-1 }a } \left( 1+\cos { { 2 }^{ n }a } \right) }{ \sin { { 2 }^{ n }a } \cos { { 2 }^{ n-1 }a } } \right] $$ $$\displaystyle ={ 2 }^{ n-1 }\cot { { 2 }^{ n-1 }a } $$ Proceeding in similar way in last, we get $$\displaystyle \tan { a } +2\cot { 2a } $$ $$\displaystyle =\frac { \sin { a } }{ \cos { a } } +2\frac { \cos { 2a } }{ \sin { 2a } } $$ $$\displaystyle =\frac { \cos { 2a\cos { a } +\sin { 2a } \sin { a } +\cos { 2a\cos { a } } } }{ \sin { 2a\cos { a } } } $$ $$\displaystyle =\frac { \cos { a } \left( 1+\cos { 2a } \right) }{ 2\sin { a } { cos }^{ 2 }a } =\frac { 2{ cos }^{ 2 }a }{ 2\sin { a } } $$ $$\displaystyle =\frac { \cos { a } }{ \sin { a } } =\cot { a } $$
Question 5 1 / -0
Solution
The series adds $$3$$ to each number to get the next number. So, the next number is $$11 + 3 = 14$$
Question 6 1 / -0
Solution
The series adds $$4$$ to each number to get the next number. So, the next number is $$29 + 4 = 33$$
Question 7 1 / -0
Solution
$$\overline { 2 } +\overline { 8 } +\overline { 18 } +\overline { 32 } ......24$$terms
$$=\sqrt { 2 } +\sqrt { 8 } +\sqrt { 18 } +\sqrt { 32 } .....24$$terms
$$\\ =\sqrt { 2 } (1+\sqrt { 4 } +\sqrt { 9 } +\sqrt { 16 } +........24$$terms)
$$ =\sqrt { 2 } (1+2+3+4+.....24)$$
Sum of natural number$$=\cfrac { n(n+1) }{ 2 } $$
$$ =\sqrt { 2 } \cfrac { (24)(25) }{ 2 } \\ =300\sqrt { 2 } =300\overline { 2 } $$
Answer$$(C)$$
Question 8 1 / -0
Question 9 1 / -0
Solution
The series alternates the addition of $$4$$ with the subtraction of $$3$$. So, the missing number is $$61$$.
Question 10 1 / -0
Solution
This is continuous series multiplied by $$5$$. $$9 \times 5 = 45$$ $$45 \times 5 = 225$$ $$225 \times 5 = 1,125$$ $$1,125 \times 5= 5,625.$$ So, the missing integer is $$225$$.
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