Quantitative Aptitude - SPLessons

Sequences Problems

Chapter 46

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Sequences Problems

shape Introduction

Sequences is defined as the arrangement of progressive terms in definite manner according to some rule. The terms of the sequence are written as \(T_{1}, T_{2}, T_{3}, ……, T_{n}, …..\)

Here: \(T_{n}\) is the \(n\)th term of the sequence (\(n\) is a positive integer).


Example: A sequence whose \(n\)th term is found by substituting \(n\) = 1, 2, 3, ……

The succession of terms \(T_{n}\) = \(n^2\) + 3\(n\) + 1

5, 11, 19, 29, … are obtained.


shape Methods

General term of the sequence is the nth term of the sequence.

1. Finite sequence: It is defined as the sequence which has a last term.
Eg: Sequence of prime numbers less than 20.


2. Infinite sequence: It is defined as the sequence which does not have a last term i.e. infinite.


Types of sequence:

Arithmetic sequence: If the difference between any term and its previous term is a constant then a sequence is known as arithmetic sequence.


  • The successive terms takes the form \(a, a + d, a + 3d, …….\)
    Where, \(a\) is the first term.
    \(d\) is the common difference.

  • \(n\)th term is given for arithmetic progression as
    \(T_{n}\) = \(a + (n – 1)\)

  • Sum of \(n\)th terms is given as
    \(S_{n}\) = \(\frac{n}{2}[2a + (n – 1)d]\)



Arithmetic mean:

It is commonly known as average.

Sum of all given elements divided by the total number of elements is known as mean(average).

Mean = \(\frac{Sum of elements}{number of elements}\)

= \(\frac{a_{1} + a_{2} + a_{3} + …. + a_{n}}{n}\)


Geometric sequence:

A sequence is said to be geometric sequence if the ratio of any term to the preceding term is a constant, called common ratio. the sequence is

\(a\), \(ar\), \(ar^2\), \(ar^3\), …….

Where, \(a\) is the first term.

\(r\) is the common ratio


  • \(n\)th term is given by
    \(T_{n}\) = \(ar^{n -1}\)

  • Sum of first \(n\) terms is given by
    \(S_{n}\) = a[latex]\frac{(r^n – 1)}{r – 1}[/latex] for \(r\) 1
    \(S_{n}\) = \(n * a\) for \(r\) = 1


Geometric mean:


  • \(n\)th root of the product of the scores is known as geometric mean.

  • The geometric mean of the scores: 1, 2, 3, and 10 is the fourth root of 1 x 2 x 3 x 10 which is the fourth root of 60 which equals 2.78.

  • The formula is written as:
    Geometric mean = \((a_{1} * a_{2} * …… a_{n})^{\frac{1}{n}}\)



Infinite Geometric series:

Sum of infinite geometric series, where \(\mid{r}\mid\) 1, sum of an infinite geometric progression tends to infinity.


Harmonic sequence:


  • A sequence formed by the terms \(a_{1}, a_{2}, a_{3}, …….., a_{n}, ……..\) for which the reciprocals of the terms, \(\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, …….., \frac{1}{a_{n}}, …..\) forms an arithmetic sequence is called a harmonic sequence.

  • The \(n\)th term of harmonic sequence is given by
    \(T_{n}\) = \(\frac{1}{a + (n – 1)d}\)

    Where \(a\) and \(d\) are the first term and common difference of the corresponding arithmetic sequence.


Harmonic mean:


  • The harmonic mean of \(a\) and \(b\) is \(\frac{2ab}{a + b}\).

  • The concept of harmonic mean is used to find the average speed of a moving body over a particular distance.



Exponential growth and decay:


  • Exponential growth and decay problems are like percent change problems.

  • Must perform a percent change over and over again.

  • Use exponents on these repeated percentages change questions.


Example 1:
The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the \(n^{th}\) term and the value of the \(50^{th}\) term

Solution:

    Use the value of the common difference d = 3 and the first term \(a_{1}\) = 6 in the formula for the n th term given above.

    \(a_{n}\) = \(a_{1}\) + (n – 1)d

    = 6 + 3 (n – 1) = 3n + 3


Example 2:
Find the terms \(a^{2}\), \(a^{3}\), \(a^{4}\) and \(a^{5}\) of a geometric sequence if \(a^{1}\) = 10 and the common ratio r = – 1.

Solution:

    Use the definition of a geometric sequence

    \(a^{2}\) = \(a^{1}\) x \(r\) = 10(-1) = -10

    \(a^{3}\) = \(a^{2}\) x \(r\) = -10(-1) = 10

    \(a^{4}\) = \(a^{3}\) x \(r\) = 10(-1) = -10

    \(a^{5}\) = \(a^{4}\) x \(r\) = -10(-1) = 10


Example 3:
Calculate the harmonic mean of the numbers 13.5, 14.5, 14.8, 15.2 and 16.1

Solution:

    x \(\frac{1}{x}\)
    13.2 0.0758
    14.2 0.0704
    14.8 0.0676
    15.2 0.0658
    16.1 0.0621
    Total \(\sum (\frac{1}{x})\) = 0.3417


    H.M of \(X\) = \( \overline{X}\) = \( \frac{n}{\sum (\frac{1}{x})}\)

    H.M. of \(X\) = \( \overline{X}\) = \(\frac{5}{0.3417}\) = 14.63

shape Formulae

Arithmetic Progression:-

1. \(S_{n}\) = \(\frac{n}{2}[2a + (n – 1)d]\)

2. \(T_{n}\) = \(a + (n – 1)\)

3. \(d\) = \(T_{n} – T_{n – 1}\)

4. \(S_{n}\) = \(\frac{n}{2}[T_{1} + T_{2}]\)

5. Mean = \(\frac{Sum of elements}{number of elements}\)

= \(\frac{a_{1} + a_{2} + a_{3} + …. + a_{n}}{n}\)


Geometric Progression:-

6. \(T_{n}\) = \(ar^{n -1}\)

7. Sum of first \(n\) terms is given by

\(S_{n}\) = a\(\frac{(r^n – 1)}{r – 1}\) for \(r\)>1

\(S_{n}\) = \(n * a\) for \(r\) = 1

8. Geometric mean = \((a_{1} * a_{2} * …… a_{n})^{\frac{1}{n}}\)

shape Samples

1. Find the 100 th term of the sequence 3, 6, 9, 12, …….

Solution:

    Given sequence 3, 6, 9, 12, …….

    This is a arithmetic progression.

    So, First term = \(a\) = 3

    Common difference (\(d\)) = 6 – 3 = 3

    \(n\)th term sequence = \(T_{100}\) = \(a + (n – 1)\) i.e.

    \(T_{100}\) = 3 + (100 – 1) = 300


2. A ball dropped from a wall of height of 44m bounce back one third of the distnace it falls. How much distance it travel before coming to rest?

Solution:

    Given that

    A ball dropped from a wall of height = 44m

    Here, it is in the form of geometric progression.

    Where, First term =\(a\) = 44

    Common ratio = \(\frac{1}{3}\)

    Number of times it bounces back to come at rest will be infinite, then the total distance travel by ball is nothing but sum of series of infinite.

    \(s_{∞}\) = \(\frac{44}{1 – \frac{1}{3}}\) = 22 x 3 = 66

    Therefore, distance = 132m.


3. Three numbers are in geometric progression such that their product is 512. Find the middle number?

Solution:

    Let the three numbers are \(\frac{a}{r}\), \(a\), \(ar\)

    Given that

    Product of the three numbers = 512 i.e.

    \(\frac{a}{r} * a * ar\)

    \(a^3\) = 512

    \(a\) = 8.

    Therefore, middle term = \(a\) = 8.


4. Quantitative Comparison:
0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2, ……..
By using the above series answer the question


Quantity A Quantity B
53rd term of
the sequence
13


A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

    Given series is

    0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2, ……..

    This can also be written as

    \(\frac{1}{4}. \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{5}{4}, \frac{6}{4}, ……\)

    Since each term given is equivalent to the term number.

    \(n\)th term would be equal to \(\frac{n}{4}\)

    Therefore, the 53rd term = \(\frac{n}{4}\) = \(\frac{53}{4}\) = 13.25

    Hence, 13.25 > 13.

    So, correct option is A.


5. Quantitative Comparison:
1, 1, 2, 3, 5, 8, 13, 21, 34, ……
By using the above Fibonacci sequence answer the question


Quantity A Quantity B
11th term in
the sequence
90


A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

    Given series is

    1, 1, 2, 3, 5, 8, 13, 21, 34, ……

    Each term after the first two is found by adding the previous two terms in this Fibonacci sequence.

    Nine terms are provided in the sequence.

    Tenth term is calculated by adding the eighth and ninth terms = 21 + 34 = 55.

    Now, tenth term = 55.

    By adding the ninth and tenth terms, eleventh term is determined i.e.

    34 + 55 = 89.

    Quantity B is greater (Since 89 < 90).

    Therefore, correct option B.