Tag: some special sequences

Questions Related to some special sequences

Fill in the blanks:
$10^2 +1^2 + 10^2 = 10^2$
$12^2 + 2^2 + 6^2 = 12^2$
$14^2 + 7^2$ + ____ = ____

  1. $3^2, 14^2$

  2. $2^2, 14^2$

  3. $2^2, 7^2$

  4. $2^2, 12^2$


Correct Option: B
Explanation:

From the pattern, the third number is the division of the first two number.
The fourth number can be obtained by same as the first number.
Then, the missing numbers will be
$14^2 + 7^2 + 2^2 = 14^2$
So, $2^2, 14^2$ are the missing numbers.

Which statement is true about consecutive natural numbers?

  1. The numbers between the difference of square of consecutive numbers is $2n + 1$

  2. The non-perfect square numbers between the square of consecutive numbers is $2n$

  3. The sum of the squares of two consecutive numbers is never a perfect square

  4. $n^{2} - 1$ is the standard form of the difference between two consecutive numbers


Correct Option: B
Explanation:

$1^{2}$$=$$1$

$2^{2}$$=$$4$
$3^{2}$$=$$9$
$4^{2}$$=$$16$
$5^{2}$$=$$25$
$6^{2}$$=$$36$
$7^{2}$$=$$49$
Between $1$ and $4$, there are 2 numbers that is $2$ and $3$.
Between $4$ and $9$, there are 4 numbers that is $5$,$6$,$7$ and $8$.
So, there are $2n$ numbers between square of consecutive numbers where $n$is the smaller number.
Hence, Option B is correct.


Which is the smallest natural number which when added to the difference of square of $17$ and $13$ gives a perfect square?

  1. $1$

  2. $5$

  3. $11$

  4. $24$


Correct Option: A
Explanation:

$17^{2}$$-$$13^{2}$$=$$289-169$$=$$120$

$121$ is a perfect square
So, $121$$-$$120$$=$$1$
Hence, Option A is correct.

State true or false
Square numbers can only have even number of zeros at the end.

  1. True

  2. False


Correct Option: A
Explanation:

$If\quad a\quad number\quad has\quad 0\quad in\quad the\quad unit’s\quad place,then\quad its\quad square\quad end\quad with\quad two\quad 0's.$

The square root of sum of the digits in the square of $121$ is 

  1. $4$

  2. $3$

  3. $6$

  4. $9$


Correct Option: A
Explanation:
The square root of sum of digits in the square of 121 

$ Sol^{n} $ square of $121 = 14641 $

sum of digits $ \Rightarrow 1+4+6+4+1 = 16 $

and square root of 16; 
$ \sqrt{16} = 4 $

Square numbers can only have ____________ at the end.

  1. Odd number of zeros

  2. Even number of zeros

  3. Both (A) and (B)

  4. None of these


Correct Option: B
Explanation:

Square numbers can only have EVEN NUMBER OF ZEROS because zero comes in only the square of $10$ and square of $10$ is $100$ that contains even number of zeros.

Let $S$ be the set of all ordered pairs $(x,y) $ of positive integers satisfying the condition $x^{2}-y^{2}=12345678$. Then:

  1. $S$ is an infinite set

  2. $S$ is the empty set

  3. $S$ has exactly one element

  4. $S$ is a finite set and has at least two elements


Correct Option: B
Explanation:
$x^{2}-y^{2}=12345678 (x,y \, \epsilon \,  1^{+})$

RHS is even, so, x & y should be odd integer but difference square of two odd integer is multiple of 8 but RHS is not multiple of $8$

$\therefore  8 $ is an empty set.

If a number of $n$-digits is perfect square and $n$ is an odd number, then which of the following is the number of digits of its square root?

  1. $\cfrac{n-1}{2}$

  2. $\cfrac{n}{2}$

  3. $\cfrac{n+1}{2}$

  4. $2n$


Correct Option: C
Explanation:

no of digits in a perfect square is $n$

 If $n$ is odd then no of digits in its square roots is $\dfrac{n+1}{2}$

Which of the following can be expressed as the sum of the square of integers? 

  1. 2000

  2. 2003

  3. 2007

  4. 2011


Correct Option: A
Explanation:
Consider, 
$2000=1936+64$ $=(44)^2+8^2$.