NIMCET Number Series Previous Year Questions (PYQs) – Page 3 of 6

Series: 3, 10, 24, 45, 73, __

Differences:

10−3 = 7
24−10 = 14
45−24 = 21
73−45 = 28

The differences are: 7, 14, 21, 28 → an arithmetic sequence with common difference +7. So the next difference is 35.

Next term: 73 + 35 = 108

Answer: 108

Given
$A_1={3}$,
$A_2={5,7,9}$,
$A_3={11,13,15,17,19}$,
$A_4={21,23,25,27,29,31,33}$, …

Observation:

• $A_k$ contains $2k-1$ numbers
• All elements are consecutive odd numbers
• So the average of $A_k$ is simply its middle term

First term of $A_k$:
$3,5,11,21,\dots$

This follows
$a_k = 3 + 2(k-1)^2$

Middle term of $A_k$:
$\text{Average} = a_k + (2k-2)$

So,
$\text{Average} = 3 + 2(k-1)^2 + 2(k-1)$
$\text{Average} = 3 + 2k(k-1)$

For $k=20$:
$\text{Average} = 3 + 2\times20\times19$
$\text{Average} = 3 + 760$
$\boxed{763}$

Given rule: Add numbers in the base where 137 + 276 = 435 holds.

Find the base \(b\):

\(137_b = 1b^2 + 3b + 7\)
\(276_b = 2b^2 + 7b + 6\)
Sum \(= 3b^2 + 10b + 13\)

\(435_b = 4b^2 + 3b + 5\)

Equate: \(3b^2 + 10b + 13 = 4b^2 + 3b + 5 \;\Rightarrow\; b^2 - 7b - 8 = 0\)

So \(b = 8\) (since base must be > 7). Thus the operations are in base 8 (octal).

Now add in base 8: \(731_8 + 672_8\)

      731₈
    + 672₈
    -------
      (units) 1 + 2 = 3 → write 3
      (eights) 3 + 7 = 10₁₀ = 12₈ → write 2, carry 1
      (sixty-fours) 7 + 6 + 1 = 14₁₀ = 16₈ → write 6, carry 1
      bring down carry → 1
    -------
      1623₈
  

Answer: \(731 + 672 = \mathbf{1623_8}\) (in base 8).