Mathematics · Quantitative Aptitude

Sequences and Series

226 Questions

Sequences and series involve ordered lists of numbers and the sum of their terms. The questions primarily test knowledge of arithmetic progressions, geometric progressions, and infinite series. It is an essential part of quantitative aptitude that requires strong pattern recognition skills.

Arithmetic progressionGeometric progressionInfinite seriesSum of termsNumber sequences

Sequences and Series Questions

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the following geometric series:
$ \sqrt{7}, \sqrt{21}, 3\sqrt{7},...$ to n terms

  1. $ \sqrt{7}\left ( \dfrac{3^{-n/2}-1}{\sqrt{3}-1} \right )$
  2. $ \sqrt{6}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  3. $ \sqrt{7}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  4. $ \sqrt{5}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $ S _n$ denote the sum of n terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, ...,$ 


Cleraly, the given series is a $G.P.$ with first term$=a=\sqrt7$ and common ratio$=r=\sqrt 3$

Then,
$ S _n = \sqrt{7} \left { \dfrac{\left ( \sqrt{3} \right )^{n}-1}{\sqrt{3}-1} \right } = \sqrt{7} \left ( \dfrac{3^{n/2}-1}{3^{1/2}-1} \right )$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the first 100 terms -5, -4, -3, -2, -1, 0, 1, 2 ............. using Gauss method

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that $a = -5, n = 100, a _n =?, d = 1$
we know that Gauss formula is, $S _n = \dfrac{n}{2}$   [First term + Last term]
To find nth term,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of first 31 terms of an A.P. whose third term is 12 and fourth term is 16.

  1. 1,983

  2. 1,984

  3. 1,985

  4. 1,986
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that, $a _3 = 12; a _4 = 16$
Common difference, $d = a _4 - a _3 = 16 - 12 = 4$
$a _3 - a _2 = d$
$12 - 4 = a _2 $ $\Rightarrow  8$
$d = a _2 - a _1$ 
$a = 4$
We know the formula,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

The sum of the first 12 terms is 100. The first term is 20. Find the last term. (use Gauss method)

  1. $\dfrac{-20}{6}$
  2. $\dfrac{-10}{6}$
  3. $\dfrac{-15}{5}$
  4. $\dfrac{-30}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that $S _n = 100, n = 12, a = 20$. fast term = ?
we know that, $S _n = \dfrac{n}{2}$  [First term + Last term]
$100 = \dfrac{12}{2}$   [20 + Last term]
$100 = 120 + 6 (\text{Last term})$
Last term $= \dfrac{-20}{6}$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

The sum of infinity of the series $\displaystyle 1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+$..... is

  1. $\displaystyle\frac{16}{35}$
  2. $\displaystyle\frac{11}{8}$
  3. $\displaystyle\frac{35}{16}$
  4. $\displaystyle\frac{8}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $\displaystyle S=1+\frac { 4 }{ 5 } +\frac { 7 }{ { 5 }^{ 2 } } +\frac { 10 }{ { 5 }^{ 3 } } +...$   ...(1)
Multiply (1) by $\displaystyle \frac { 1 }{ 5 } $ , we get

$\displaystyle \frac { 1 }{ 5 } S=\frac { 1 }{ 5 } +\frac { 4 }{ { 5 }^{ 2 } } +\frac { 7 }{ { 5 }^{ 3 } } +\frac { 10 }{ { 5 }^{ 4 } } +...$   ...(2)

$(1) -(2)$, gives 
$\displaystyle \left( 1-\frac { 1 }{ 5 }  \right) S=1+\frac { 3 }{ 5 } +\frac { 3 }{ { 5 }^{ 2 } } +\frac { 3 }{ { 5 }^{ 3 } } +...$

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 5 } \left( 1+\frac { 1 }{ 5 } +\frac { 1 }{ { 5 }^{ 2 } } +... \right) $

$\displaystyle \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 1 }{ 1-\dfrac { 1 }{ 5 }  }  \right) \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 5 }{ 4 }  \right) $

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 4 } \Rightarrow S=\frac { 35 }{ 16 } $

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the first 25 terms of the A.P.: 2 + 5 + 8 + 11 + ............ (use Gauss method)

  1. 910

  2. 930

  3. 950

  4. 940
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $a = 2; d = 3$
$n = 25 ; a _{25} = $?
Using Gauss method, we find the value of $a _{25}$
$a _n = a + (n - 1) d$
$a _{25} = 2 + (25 - 1) 3$
$= 2 + (24) 3$
$a _{25} = 74$
Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$= \dfrac{25}{2} [2 + 74]$
$= 12.5 (76)$
$S _n = 950$

Multiple choice mean and median mean maths assumed mean method assumed mean method of finding mean measure of central tendency

Three geometric means between 5 and 3125 are___.

  1. 15,75,375

  2. 25,125,625

  3. 11,44,176

  4. 10,40,160
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The geometric mean is a geometric series which is in the form of arn where a is the first term and r is the common ratio.

T1= 5 and T5 = 3125

Now, T1/T5 = 5/ 3125

=>ar1//ar5 =1/ 625

=> 1/ r4 = 1/625

=> r4 = 625

=> r = 5

So if r=5, then a= 1

T2 = ar2

      = 1(5) 2 = 25

T3 = ar3

      = 1(5) 3 = 125

T4 = ar4

      = 1(5)4 = 625

Therefore, three geometric means between 5 and 3125 are 25, 125 and 625. 

Multiple choice physics measurements and units measurement of area and volume measurement of volume measurement of area, volume and density

There are $n\ sets$ of real numbers. In every set there are four real numbers in AP. Such that the square of the last term is equal to the sum of the square of the first three terms. Then: -

  1. $N = 4$
  2. $N = 1000$
  3. $N = 10$
  4. $N \rightarrow \infty$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the terms of the AP be a - 3d, a - d, a + d, a + 3d (or standard a, a+d, a+2d, a+3d). Setting up the given condition that the square of the last term equals the sum of squares of the first three terms leads to a quadratic in terms of common difference or number of valid sets, which can be extended infinitely as parameters vary, yielding N -> infinity.

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

The population of a bacteria culture doubles in number every 12 minutes. The ratio of the number of bacteria at the end of 1 hour to the number of bacteria at the beginning of that hour is

  1. $8 : 1$
  2. $16 : 1$
  3. $32 : 1$
  4. $60 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a number is multiplies repeatedly by 2 for n times then the last result will be $ { 2 }^{ n }\times number $.
Let  the number of bacteria is $a$. The number of bacteria doubles every $12$   minutes, which means that it doubles $5$ times in an hour.
Therefore,
$(2)(2)(2)(2)(2)a =a\times2^5$
                            $ = 32 a$  at the end of the hour.
Thus, the ratio of the number of bacteria at the end of 1 hour to the number of bacteria at the beginning of that hour is $\cfrac{32a}{a} = 32:1$.

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

When each term of a sequence is connected using $a +$ or $a -$ sign, then it is referred to as the _____ of numbers.

  1. Series

  2. Progression

  3. Arithmetic Progression

  4. Geometric Prpgression
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When each term of a sequence is connected using a $+$ or a $-$ sign, then it is referred to as the series of numbers. For example, $2 + 5 + 8 + 11 + 15 + 18 + ....$ is a series of numbers. Thus the correct answer is '$a$'.