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Aptitude Number System Theory

Arithmetic progression(AP)

Arithmetic progression(AP) or arithmetic sequence is a sequence of numbers in which each term after the first is obtained by adding a constant, d to the preceding term. The constant d is called common difference.

Formula

  • An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), ...

    where a = the first term , d = the common difference


  • Example

    1, 3, 5, 7, .....?

    Arithmetic progression (AP) with a = 1 and d = 2

    a, (a + d), (a + 2d), (a + 3d),(a + 4d)........

    1, 3, 5, 7, (1+4(2))

    1, 3, 5, 7, 9

    Ans : 9


    Exercise

    44489.7, 13, 19, 25, ... ?
    36
    31
    Explanation:

    Arithmetic progression (AP)
    a = 7 and d= 6
    a, (a + d), (a + 2d), (a + 3d),(a + 4d)........
    7, 13, 19, 25, (7+4(6))
    7, 13, 19, 25,31

    n$^{th}$ term of an arithmetic progression

    Formula

  • tn = a + (n – 1)d

    where t$_{n}$ = n$^{th}$ term, a= the first term , d= common difference


  • Example

    Find 10$^{th}$ term in the series 1, 3, 5, 7, ...

    a = 1

    d = 3 – 1 = 2

    10$^{th}$ term, t$_{10}$ = a + (n-1)d = 1 + (10 – 1)2 = 1 + 18 = 19


    Exercise

    44490.Find 16th term in the series 7, 13, 19, 25, ...
    77
    97
    Explanation:

    a = 7
    d = 13 – 7 = 6
    n=16
    16$^{th}$term, t$_{16}$ = a + (n-1)d = 7 + (16 – 1)6 = 7 + 90 = 97
    16$^{th}$term = 97

    Number of terms of an arithmetic progression

    Formula

  • n = $\dfrac{(l−a)}{d} + 1$

    where n = number of terms, a= the first term , l = last term, d= common difference


  • Example

    Find the number of terms in the series 8, 12, 16, . . .72

    a = 8

    l = 72

    d = 12 – 8 = 4

    n = $\dfrac{(l−a)}{d} + 1$ = n = $\dfrac{(72−8)}{4} + 1$

    =n = $\dfrac{(64)}{4} + 1$ = 16+1 = 17

    Sum of first n terms in an arithmetic progression

    Sn= $\dfrac{n}{2}$ [ 2a+(n−1)d ] = $\dfrac{n}{2}$ (a+l)

    where a = the first term,

    d= common difference,

    l = t$_n$ = n$^{th}$ term = a + (n-1)d


    Example

    Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms

    a = 4

    d = 7 – 4 = 3

    Sum of first 20 terms, S$_{20}$

    = $\dfrac{n}{2}$ [ 2a+(n−1)d ]

    = $\dfrac{20}{2}$ [ (2 * 4) + (20−1)3 ]

    = 10[8+(19×3)]

    = 10(8+57)

    = 650


    Exercise

    44491.Find 6 + 9 + 12 + . . . + 30
    162
    252
    Explanation:

    a = 6
    l = 30
    d = 9 – 6 = 3
    n = $n = \dfrac{(l - a)}{d} + 1$
    = $\dfrac{(30 - 6)}{3} + 1 $
    = $\dfrac{24}{3} + 1 $
    = 8 + 1
    = 9
    Sum, S
    =$\dfrac{n}{2}(a+l)$
    =$\dfrac{9}{2}(6+30)$
    =$\dfrac{9}{2} \times 36 $
    =$9 \times 18$
    =162

    Arithmetic Mean

    If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c. In this case,

    b = $\dfrac{1}{2}(a + c)$

  • The Arithmetic Mean (AM) between two numbers a and b = $\dfrac{1}{2}(a + b)$
  • If a, a1, a2 ... an, b are in AP we can say that a1, a2 ... an are the n Arithmetic Means between a and b.


    Additional Notes on AP

    To solve most of the problems related to AP, the terms can be conveniently taken as

    3 terms: (a – d), a, (a +d)

    4 terms: (a – 3d), (a – d), (a + d), (a +3d)

    5 terms: (a – 2d), (a – d), a, (a + d), (a +2d)


    $T_n$ = $S_n$ - $S_n-1$


    If each term of an AP is increased, decreased , multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.


    In an AP, sum of terms equidistant from beginning and end will be constant.


    Harmonic Progression(HP)

    Non-zero numbers $a_1,\ a_2,\ a_3,........ a_n$ are in Harmonic Progression(HP) if $\dfrac{1}{a_1}, \dfrac{1}{a_2},\dfrac{1}{a_3}, ..... \dfrac{1}{a_n}$ are in AP. Harmonic Progression is also known as harmonic sequence.


    Example

    $\dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{10},$...... is a harmonic progression (HP)


  • Three non-zero numbers a, b, c will be in HP, if $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP

    If a, (a+d), (a+2d), . . . are in AP, nthterm of the AP = a + (n - 1)d
    Hence, if $\dfrac{1}{a}, \dfrac{1}{a+d}, \dfrac{1}{a+2d},$are in HP, nthterm of the HP = $\dfrac{1}{a + (n - 1)d}$

    If a, b, c are in HP, b is the Harmonic Mean(HM) between a and c
    In this case, b = $\dfrac{2ac}{a + c}$

    The Harmonic Mean(HM) between two numbers a and b =$\dfrac{2ab}{a + b}$

    If a, a1, a2 ... an, b are in HP we can say that a1, a2 ... an are the n Harmonic Means between a and b.

    If a, b, c are in HP, $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

  • Geometric progression(GP)

    Geometric Progression(GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant.


    Formula

    A geometric progression(GP) is given by a, ar, ar2, ar3, ...

    where a = the first term , r = the common ratio


    Example

    1, 3, 9, 27, ... is a geometric progression(GP) with a = 1 and r = 3


    Exercise

    44492.2, 4, 8, 16, ... is a geometric progression(GP) with a = ? and r = ?
    a = 2 and r = 2
    a = 2 and r = 1

    nth term of a geometric progression(GP)

    Formula

  • $t_n = ar^{n-1}$

    where tn = nth term, a= the first term , r = common ratio, n = number of terms


  • Example

    Find the 10th term in the series 2, 4, 8, 16, ...

    a = 2, r = $\dfrac{4}{2}$ , n = 10

    n = 10

    10th term, t10

    =$ar^{n-1} = 2 \times 2^{10-1}$

    = $2 \times 2^{9} = 2 \times 512 = 1024 $


    Exercise

    44493.Find 5th term in the series 5, 15, 45, ...
    350
    405
    Explanation:

    a = 5, r - $\dfrac{15}{5}$ = 3, n=5
    5th term, t5
    =$ar^{n-1} = 5 \times 3^{5-1}$
    = $5 \times 3^4 = 5 \times 81$ = 405

    Sum of first n terms in a Geometric progression(GP)

    Formula

  • $S_n =\begin{cases} \dfrac{a(r^n - 1)}{r - 1}\ & (\text{if } r \gt 1) \\ \\ \dfrac{a(1 - r^n)}{1 - r}\ & (\text{if } r \lt 1) \\\end{cases}$

    where a= the first term,

    r = common ratio,

    n = number of terms


  • Example

    Find 4 + 12 + 36 + ... up to 6 terms

    a = 4, r = $\dfrac{12}{4}$ = 3 , n = 6

    Here r > 1. Hence,

    $S_6$= $ \dfrac{a(r^n - 1)}{r - 1} $= $\dfrac{4(3^6 - 1)}{3 - 1} $

    =$ \dfrac{4(729 - 1)}{2}$=$ \dfrac{4 \times 728}{2} $= $2 \times 28$

    = 1456


    Exercise

    44494.Find 1 + $\dfrac{1}{2} + \dfrac{1}{4}$ +..........up to 5 terms
    $1\dfrac{15}{16}$
    $2\dfrac{16}{15}$
    Explanation:

    a = 1, r =$\dfrac{\left(\dfrac{1}{2}\right)}{1} $= $\dfrac{1}{2}$, n = 5

    Here r < 1. Hence,

    $S_6$ = $ \dfrac{a(1 - r^n)}{1 - r} $= $ \dfrac{1\left[1 - \left(\dfrac{1}{2}\right)^5 \right]}{\left(1 - \dfrac{1}{2}\right)} $

    = $\dfrac{\left(1 - \dfrac{1}{32} \right)}{\left(\dfrac{1}{2}\right)}$ = $\dfrac{\left(\dfrac{31}{32}\right) }{\left(\dfrac{1}{2}\right)} $=$ \dfrac{31}{16}$ = $1\dfrac{15}{16}$


    Sum of an infinite geometric progression(GP)

    Formula

  • $S_\infty$=$ \dfrac{a}{1 - r}$ (if -1 < r < 1)

    where a= the first term , r = common ratio


  • Example

    Find 1 + $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}+.....\infty$

    a = 1,r = $\dfrac{\left(\dfrac{1}{2}\right)}{1} $= $\dfrac{1}{2}$

    Here -1 < r < 1. Hence,

    $S_\infty$ = $\dfrac{a}{1 - r}$ = $\dfrac{1}{\left(1 - \dfrac{1}{2}\right)}$ = $ \dfrac{1}{\left(\dfrac{1}{2}\right)} $= 2


    Geometric Mean

    If three non-zero numbers a, b, c are in GP, b is the Geometric Mean(GM) between a and c. In this case, b = $\sqrt{ac}$

  • The Geometric Mean(GM) between two numbers a and b = $\sqrt{ab}$

    (Note that if a and b are of opposite sign, their GM is not defined.)


  • Additional Notes on GP

    To solve most of the problems related to GP, the terms of the GP can be conveniently taken as

    3 terms:$\dfrac{a}{r}$, a, ar

    5 terms: $\dfrac{a}{r^2}$, $\dfrac{a}{r}$, a, ar, ar2


    If a, b, c are in GP, $\dfrac{a-b}{b-c}$ = $\dfrac{a}{b}$


    In a GP, product of terms equidistant from beginning and end will be constant.


    Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two Numbers

    If GM, AM and HM are the Geometric Mean, Arithmetic Mean and Harmonic Mean of two positive numbers respectively, then

    GM$^2$ = AM × HM

    Some Interesting Properties to Note

    Three numbers a, b and c are in AP if b $= \dfrac{a + c}{2}$

    Three non-zero numbers a, b and c are in HP if b = $\dfrac{2ac}{a + c}$

    Three non-zero numbers a, b and c are in HP if $\dfrac{a - b}{b - c} = \dfrac{a}{c}$


    Let A, G and H be the AM, GM and HM between two distinct positive numbers. Then

    (1) A > G > H

    (2) A, G and H are in GP


    If a series is both an AP and GP, all terms of the series will be equal. In other words, it will be a constant sequence.


    Power Series : Important Formulas

    1+1+1+⋯ n terms = ∑1=n

    1+2+3+⋯+n =∑n = $\dfrac{n(n+1)}{2}$

    $1^2 + 2^2 + 3^2 + ........ +n^2 = ∑ n^2$ = $\dfrac{n(n+1)(2n + 1)}{6}$

    $1^3 + 2^3 + 3^3 + ....... + n^3 = ∑ n^3$ = $\dfrac{n^2(n+1)^2}{4}$ = $\left[\dfrac{n(n+1)}{2}\right]^2 $

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