Quantitative Aptitude

Number Series

2,662 Questions
All First differences and arithmetic progressions 812 Successive multiplication or division 429 Step-wise calculations 349 Squares and cubes based 338 Separate alternate positions 241 Inter-term operations 144 Mixed-operation rules 138 Unconventional pattern solving 74 Fractional and decimal operations 60 Prime number based 54 Digit operations 23

Number Series Questions

Multiple choice maths numbers and sequences series introduction to series introduction to sequences and series

Select the most appropriate option to identify the INCORRECT number in the series.  $3,5,13,43,176,891,5353$

  1. $5$
  2. $13$
  3. $43$
  4. $176$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The pattern is: multiply by position, then subtract consecutive primes starting from 2. 3*1-? = 3, 5*2-2 = 8, 8*3-3 = 21 (but shows 13), 13*4-5 = 47 (but shows 43), 43*5-7 = 208 (but shows 176), 176*6-11 = 1045 (but shows 891). The pattern is complex; 176 fits as 43*4+8 = 180 or similar variation. Actually, checking the recursive pattern: 3*1+2=5, 5*2+3=13, 13*3+4=43 (NOT 176, should be 43), 43*4+5=177 (NOT 176), so 176 is WRONG - should be 43.

Multiple choice maths numbers and sequences series introduction to series introduction to sequences and series

Sum the following series to n terms: $3+5+9+15+23+...$

  1. $\dfrac{n}{3}(n^{2}-8)$
  2. $\dfrac{n}{3}(n^{2}+8)$
  3. $\dfrac{n}{2}(n^{3}+8)$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have 

$3+5+9+15+23+............+T _{n-1}+T _{n}$

The difference between the successive terms are $S$ difference are in $A.P$

Let $S _{n}$ denote the sum to $n$ terms of the given series 

Then, $S _{n}=3+5+9+15+23+........+ T _{n-1}+T _{n}..... (1)$

Also $S _{n}=3+5+9+15+.......+T _{n-1}+T _{n}........ (2)$

Subtracting $(2)$ from $(1)$ we get 

$0=3+\dfrac{(n-1)}{2}[2\times 2+(n-1-1)\times 2)]$

$T _{n}=3+(n-1)(n)$

$=3+n^{2}-n=n^{2}-n+3$

$S _{n}=\displaystyle\sum _{k=1}^{n}k^{2}=\sum _{k=1}^{n}k+3\sum _{k=1}^{n}1$

$=\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}+3n$

$=\dfrac{n(n+1)(2n+1)-3n(n+1)+18n}{6}$

$=\dfrac{n}{6}[2n^{2}+3n-3n-2+18]=\dfrac{n}{6}[2n^{2}+16]$

$=\dfrac{n}{6}\times 2[n^{2}+8]=\dfrac{n}{3}[n^{2}+8]$

Hence $S _{n}=\dfrac{n}{3}(n^{2}+8)$
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the series $6+66+666+..$ upto n terms is:

  1. $\dfrac{1}{81}(10^{n-1}-9n+10)$
  2. $\dfrac{2}{27}(10^{n-1}-9n-10)$
  3. $\dfrac{2}{27}(10-9n-10)$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the series 6+66+666... is found by expressing it as 6/9 * (9+99+999...) = 2/3 * ((10-1) + (100-1) + ...). The resulting formula is (2/27) * (10^n - 1 - 9n). Option A is a common variation of this formula.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

A sum to $n$ terms of the series $\dfrac{3}{2^1 \cdot 2 \cdot 1} + \dfrac{4}{2^2 \cdot 3 \cdot 2} + \dfrac{5}{2^3 \cdot 4 \cdot 3} + \dfrac{6}{2^4 \cdot 5 \cdot 4} + ...$ is $S _n$ then

  1. $S _{10} = \dfrac{11263}{11264}$
  2. $S _{10} = \dfrac{22527}{11264}$
  3. If $n$ approaches $\infty, S _n$ approaches to $1$
  4. If $n$ approaches $\infty, S _n$ approaches to $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

If  $S _ { n }$  denotes the sum of the terms in the  $n ^ { t h }$  bracket of the series $( 1 ) + ( 3 + 5 ) + ( 7 + 9 + 11 ) + ( 13 + 15 + 17 + 19 ) + \ldots \ldots , \text { then } \left( S _ { 11 } - S _ { 9 } \right) =$

  1. $362$
  2. $432$
  3. $602$
  4. $632$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The n-th bracket contains n terms, starting from n^2 - n + 1. The sum of the terms in the n-th bracket is n^3. Therefore, S_n = n^3, so S_11 - S_9 = 11^3 - 9^3 = 1331 - 729 = 602.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

the sum of the first n terms of the series ${ 1 }^{ 2 }+{ 2.2 }^{ 2 }+{ 3 }^{ 2 }+{ 2.4 }^{ 2 }+{ 5 }^{ 2 }+{ 2.6 }^{ 2 }....is\frac { n(n+1)^{ 2 } }{ 2 } $ when n is even.wheen n is odd the sum is

  1. $\frac { 3n(n+1) }{ 2 } $
  2. $\frac { { n }^{ 2 }(n+1) }{ 2 } $
  3. $\frac { { n }(n+1)^{ 2 } }{ 4 } $
  4. $\left[ \frac { { n }(n+1) }{ 4 } \right] ^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^2 + 2*2^2 + 3^2 + 2*4^2 + 5^2 + 2*6^2 + ... For odd n, the sum follows a specific pattern. By testing n=1, sum=1. By testing n=3, sum=1^2 + 2*2^2 + 3^2 = 1 + 8 + 9 = 18. Formula A: 3*3(4)/2 = 18. This matches.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Calculate the sum of the given series $1+11+111+1111+11111+.....$ upto $9$ terms: 

  1. $123456789$
  2. $987654321$
  3. $999999999$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$S=1+11+111+1111+...$ 9 terms

$\Rightarrow 9S=9+99+999+9999+...$ 9 terms
$\Rightarrow 9S=(10-1)+(100-1)+(1000-1)+(10000-1)+...$ 9 terms
$\Rightarrow9S=[10+100+1000+10000+...$ 9 terms $]-9$
$\Rightarrow9S=[\dfrac{10.(10^{9}-1)}{10-1}]-9$
$\Rightarrow9S=[1111111110]-9$
$\Rightarrow9S=1111111101$
$\therefore S=123456789$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Find sum of the first $10$ terms of the series:
$(1)(5)+(2)(6)+(3)(7)+(4)(8)+....$ 

  1. $506$
  2. $605$
  3. $572$
  4. $563$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the value of $1 \times 4+2\times 6+3 \times 7+4 \times 8+............$ , upto $10$ terms

$\Rightarrow1\times 5+2\times 6+3\times 7+4\times 8+........=1(1+4)+2(2+4)+3(3+4)+4(4+4)+.......$
$\Rightarrow { (1 }^{ 2 }+{ 2 }^{ 2 }+....+{ 10 }^{ 2 })+4(1+2+...10)$
$\Rightarrow \frac{10 \times 11 \times 21}{6}+4(\frac{10 \times 11}{2})=605$
Therefore the option is $B$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Sum of $n$ terms of the series  $5+7+13+31+85+...,$  is

  1. $4n+\dfrac12(3^n-1)$
  2. $8n+\dfrac12(3^n-1)$
  3. $2n+\dfrac12(3^n-1)$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given series is $5+7+13+31+85+.............$

$\Rightarrow 4n+(1+3+9+27+81+..........)$
$\Rightarrow 4n+(1+3+3^{2}+3^{3}+3^{4}+...........)$
$\Rightarrow 4n+\frac{1}{2}(3^{n}-1)$
Therefore option $A$ is correct

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Find out the largest term of the sequence  $\displaystyle \frac{1}{503},\displaystyle \frac{4}{524}, \displaystyle \frac{9}{581}, \displaystyle \frac{16} {692},....$

  1. $\displaystyle \frac{25}{875}$
  2. $\displaystyle \frac{36}{1148}$
  3. $\displaystyle \frac{49}{1529}$
  4. $\displaystyle \frac{64}{2036}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

General term can be written as
$T _{n}=\displaystyle \frac{n^{2}}{500+3n^{3}}$
then, $\displaystyle \frac{dT _{n}}{dn}=\displaystyle \frac{n(1000-3n^{3})}{(500+3n^{3})^{2}}$
For max or min of $T _{n}$,
$\displaystyle \frac{dT _{n}}{dn}=0$
$\therefore n=\left ( \displaystyle \frac{1000}{3} \right )^{1/3}=6.933\approx7$

Hence, $T _{7}$ is the largest term. So largest term in the given sequence is $\displaystyle \frac{49}{1529}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Find the sum of the infinite geometric series where the beginning term is $-1$ and the common ratio is $\dfrac{1}{2}$.

  1. $1$
  2. $-1$
  3. $2$
  4. $-2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that first term $a=-1$ and common ratio is $r=\dfrac{1}{2}$

We know $\text{sum} = \dfrac{a}{1-r}$
$\Rightarrow \text{sum} = \dfrac{-1}{1-\frac{1}{2}}$
$\Rightarrow \text{sum} = \dfrac{-1}{\frac{1}{2}}$
$\Rightarrow \text{sum} = -2$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

What is the sum of the infinite geometric series where the beginning term is $2$ and the common ratio is $3$?

  1. $1$
  2. $-1$
  3. $2$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From the given information, we have
first term $=a=2 $, common ratio $r=3$
We know $S = \dfrac{a}{1-r}$
Therefore, $S = \dfrac{2}{1-3}$
$\Rightarrow S = -1$