Geometric Progressions Solutions Part 2
***************************************
51. Let :math:`r` be the common ratio then :math:`b = ar^{n - 1}`. And we have
    :math:`p` as

    .. math::
       p = a.ar.ar^2. ... .ar^{n - 1}

    .. math::
       p = a^nr^{\frac{n(n - 1)}{2}}

    .. math::
       p^2 = a^{2n}r^{n(n - 1)}

    .. math::
       p^2 = (a^2r^{n - 1})^n

    .. math::
       p^2 = (a.ar^{n - 1})^n

    .. math::
       p^2 = (ab)^n

52. Let :math:`a_1` and :math:`a_2` be the first terms of the two G. P.s are
    :math:`r` be the common ratio. Let :math:`S_1` and :math:`S_2` be the sum
    to :math:`n` terms, we have

    .. math::
       S_1 = \frac{a_1(1 - r^{n - 1})}{1 - r}~~~S_2 = \frac{a_2(1 - r^{n
       -1})}{1 -r}

    .. math::
       \therefore \frac{S_1}{S_2} = \frac{a_1}{a_2} = \frac{a_1r^{n -
       1}}{a_2r^{n - 1}}

    Thus, ratio of sums to :math:`n` terms is equal to the :math:`n\text{th}`
    terms.

53. Let :math:`a` be the first term and :math:`r` be the common ratio. Then, we
    have

    .. math::

       S_1 = a + ar + ar^2 + ... + ar^{n - 1}

    .. math::

       S_2 = a + ar + ar^2 + ... + ar^{n - 1} + ar^n + ar^{n + 1} + ... + ar^{2n - 1}

    .. math::

       S_3 = a + ar + ar^2 + ... + ar^{2n - 1} + ar^{2n} + ... + ar^{3n - 1}

    .. math::

       S_2 - S_1 = ar^n + ar^{n + 1} + ... + ar^{2n - 1}

    .. math::

       S_2 - S_1 = \frac{ar^n(r^{n} - 1)}{r - 1}

    Similarly,

    .. math::

       S_3 - S_2 = \frac{ar^{2n}(r^{n} - 1)}{r - 1}

    and

    .. math::

       S_1 = \frac{a(r^n - 1)}{r - 1}


    Thus, :math:`(S_2 - S_1)^2 = S_1(S_3 - S_2)`.

54. We need to compute following:

    .. math::

       S_1 + S_2 + ... + S_{2n - 1}

    From formula for sum of geometric series we have following:

    .. math::

       = \frac{a(r - 1)}{r - 1} + \frac{a(r^2 - 1)}{r - 1} + ... +
       \frac{a(r^{2n - 1} - 1)}{r - 1}

    .. math::

       = \frac{a}{r-1}\left[r + r^2 + ... + r^{2n - 1} - (2n - 1)\right]

    .. math::

       = \frac{a}{r-1}\left[\frac{r(1 - r^{2n - 1})}{r - 1} - (2n -
       1)\right]

55. Given, :math:`S_n = a.2^n - b` :math:`\therefore S_{n - 1} =
    a.2^{n - 1} - b`. Thus, :math:`t_n = S_n - S_{n - 1} = a2^{n -
    1}`. Since the ratio of terms will be 2 as evident from
    :math:`t_n` the series is in G. P.

56. The sum would be :math:`S = 3.2 - 4 + 3.2^2 - 4 + ... +
    3.2^{100} - 4`.

    From the formula for sum of a geometric series, we have,

    :math:`S = 6(2^{100} - 1) - 400`

57. :math:`t_n = 1 + 2 + 2^2 + ... + 2^n = 2^n - 1`.

    Sum of :math:`n` terms = :math:`S = 1 + 3 + 7 + 15 + ... + 2^n -
    1`

    .. math::

       S = 2^1 - 1 + 2^2 - 1 + 2^3 - 1 + 2^n - 1

    .. math::

       S = 2^{n + 1} - 2 - n

58. Clearly, the common ratio is :math:`3x`. Thus, from the formula
    for sum of infinite series we have sum as

    .. math::

       S_{\infty} = \frac{1}{1 - 3x}

59. Common ratio is :math:`\frac{-1}{3}`, thus sum for infinite terms
    of the given series is:

    .. math::

       S_{\infty} = \frac{1}{1 - (\frac{1}{3})} = \frac{9}{4}

60. There are two serieses having common ratios :math:`\frac{1}{5}`
    and :math:`\frac{1}{7}` having first terms as :math:`\frac{1}{5}`
    and :math:`\frac{1}{7}` respectively, as well.

    Thus, applying the formula, we have

    .. math::

       S = \frac{\frac{1}{5}}{1 - \frac{1}{5}} +
       \frac{\frac{1}{7}}{1 - \frac{1}{7}}

    .. math::

       S = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}

61. Value of exponent is: :math:`S = \frac{1}{3} + \frac{1}{9} +
    \frac{1}{23} + ... ~~\text{to}~~ \infty`


    :math:`S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}`

    Thus, value is :math:`9^\frac{1}{2} = 3`.

62. Sum within braces is

    .. math::

       S = \frac{1}{1 - \frac{2x}{1 + x^2}} = \frac{1 + x^2}{(1 -
       x)^2}

    Thus, final sum is :math:`S = \frac{1}{(1 - x^2)}`.

63. :math:`L. H. S. = a^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +
    ... ~~\text{to}~~ \infty}`

    The value of exponent is :math:`\frac{\frac{1}{2}}{1 -
    \frac{1}{2}} = 1`. Thus final value is :math:`a`.

64. :math:`0.\dot{5}\dot{4} = 0.54 + 0.0054 + 0.000054 + ... ~~\text{to}~~
    \infty`

    Thus, we have

    .. math::

       0.\dot{5}\dot{4} = \frac{54}{100} + \frac{54}{10000} +
       \frac{54}{1000000} + ...

    .. math::

       = \frac{54}{100}.\frac{1}{1 - \frac{1}{100}} = \frac{54}{99} =
       \frac{6}{11}

65. We have

    .. math::

       .\dot{6} = .6 + .06 + .006 + ... ~~\text{to}~~ \infty

    .. math::

       = \frac{6}{10} + \frac{6}{100} + \frac{6}{1000} + ...

    .. math::

       = \frac{6}{10}.\frac{1}{1 - 10} = \frac{6}{9} = \frac{2}{3}.

66. Given, :math:`S_{\infty} = 32` and :math:`a + ar = 24`, where,
    :math:`a` is the first term and :math:`r` is the common ratio.

    Now, :math:`S_{\infty} = \frac{a}{1 - r} \because |r| < 1.`

    Solving these two we have,

    .. math::

       a = 32(1 - r) ~~\text{and}~~ a(1 + r) = 24

    Substituting first in second we have

    .. math::

       32(1 - r^2) = 24 ~~\text{i.e.}~~ r^2 = \frac{1}{4} \Rightarrow r =
       \pm\frac{1}{2}

    Thus, :math:`a = 48` or :math:`a = 16`. Thus, we can have our
    desired series as :math:`16, 8, 4, 2, ...` or :math:`48, -24, 12,
    -6`.

67. Let :math:`a` be the first term and :math:`r` be the common
    ratio. Then, we have

    .. math::

       \frac{a}{1 - r} = 4 ~~\text{and}~~ \frac{a^2}{1 - r^2} =
       \frac{16}{3}

    Substituting first ins second, we have

    .. math::

       \frac{16(1 - r)^2}{1 - r^2} = \frac{16}{3} \Rightarrow
       \frac{1 - r}{1 + r} = \frac{1}{3}

    Solving, we have :math:`r = \frac{1}{2}` and then :math:`a =
    2`. Thus, we have our series as :math:`2, 1, \frac{1}{2},
    \frac{1}{4}, ...`.

68. Let :math:`a` be the first term and :math:`r` be the common
    ratio. Then, we have

    .. math::

       S = 1 + a + ar + ar^2 + ... ~~\text{to}~~ \infty

    Let us consider a term :math:`t_n = ar^{n - 1}`. Then, sum of
    succeeding terms is

    :math:`S = \frac{ar^n}{1 - r}`. Equating these we have

    .. math::

       \frac{ar^n}{1 - r} = ar^n - 1

    .. math::

       \Rightarrow \frac{r}{1 - r} = 1 \Rightarrow r = \frac{1}{2}.

    Similarly we can prove for :math:`>` or :math:`<`.

69. Given, :math:`y = 1 + x + x^2 + ...`

    From the formula for infinite G. P.

    .. math::

       y = \frac{1}{1 - x} \Rightarrow 1 - x = \frac{1}{y} \Rightarrow
       x = \frac{y - 1}{y}

70. :math:`c = ar^2` and :math:`b = ar`. Thus, we have :math:`c > 4b -
    3a = ar^2 > 4ar - 3a`

    :math:`\Rightarrow r^2 > 4r - 3 \Rightarrow (r - 1)(r - 3) > 0`

    Thus, :math:`r > 3` or :math:`r < 1`.

71. :math:`1 + 2x + 4x^2 + ... + 32x^5 = \frac{1 - k^6}{1 - k}`

    :math:`\frac{1 - (2x)^6}{1 - 2x} = \frac{1 - k^6}{1 - k}`

    :math:`\therefore k = 2x \Rightarrow \frac{k}{x} = 2.`

72. Simplifying the given relation, we have

    .. math::

       (b^4 - 2b^2ac + a^2c^2) + (c^4 - 2c^2bd + b^2d^2) + (a^2d^2 -
       2abcd - b^2c^2) \le 0

    .. math::

       \Rightarrow (b^2 - ac)^2 + (c^2 - bd)^2 + (ad - bc)^2 \le 0

    This is only possible if and only if, :math:`b^2 = ac, c^2 = bd,
    ac = bd` i.e. :math:`\frac{b}{a} = \frac{c}{b} =
    \frac{d}{c}`. Thus, :math:`a, b, c, d` are in G. P.

73. On simplification, we have

    .. math::

       \sum (a_1^2a_3^2 + a_2^4 - 2a_2^2a_1a_3) \le 0

    or :math:`\sum (a_1a_3 - a_2^2)^2 \le 0`.

    In L. H. S. every bracket being square of real number is +ve and
    hence their sum cannot be less than zero. It will be zero if each
    term is zero.

    :math:`\therefore a_1a_3 = a_2^2` or :math:`a_1, a_2, a_3` are
    in G. P. Similarly, others are also in G. P.

    Hence proved.

74. :math:`\alpha, beta, \gamma, \delta` being in increasing G. P.,
    they may be taken as :math:`k, kr, kr^2, kr^3`, where :math:`r >
    1`.

    Sum of the roots of the equations

    .. math::

       S_1 = k(1 + r) = 3, S_2 = kr^2(1 + r) = 12

    Substituting :math:`S_1` in :math:`S_2`, we have

    :math:`3r^2 = 12` or :math:`r = 2 \therefore k = 1`.

    Product of the roots

    :math:`P_1 = \alpha\beta = k^2r = a, P_2 = \gamma\delta = k^2r^5 =
    b`

    Putting for :math:`k` and :math:`r`, we have

    :math:`a = 2, b = 32`.

75. Given :math:`d = 2, r = \frac{1}{2}.`

    If there be odd number of terms then mid-term =
    :math:`\frac{1}{2}(odd + 1)`.

    :math:`T_{2n + 1}` is the mid-term of sequence of :math:`(4n + 1)`
    odd terms.

    :math:`a + 2nd = a + 4n`.

    This middle term is the last term of A. P. and first term of
    following G. P. each of :math:`(2n + 1)` terms with this term
    being common to both.

    Let :math:`T_{n+1}` and :math:`t_{n + 1}` are mid terms
    of A. P. and G. P.

    :math:`T_{n + 1} = a + nd = a + 2n`

    :math:`t_{n + 1} = AR^n = T_{2n + 1}\left(\frac{1}{2}\right)^n =
    (a + 4n)\left(\frac{1}{2}\right)^n`

    Given, :math:`T_{n + 1} = t_{n + 1}`

    :math:`a + 2n = (a + 4n)\frac{1}{2^n}`

    .. math::

       \therefore a = \frac{4n - n.2^{n + 1}}{2^n - 1}

    Hence, mid term is

    .. math::

       a + 4n = \frac{n.2^{n + 1}}{2^n - 1}

76. :math:`S_n = 3 - \frac{3^{n + 1}}{4^{2n}}`. Putting :math:`n = 1,
    2` we have

    .. math::

       T_1 = S_1 = 3 - \frac{9}{16} = \frac{39}{16}

    .. math::

       S_2 = 3 - \frac{27}{256} = T_1 + T_2

    .. math::

       \therefore T_2 = \frac{117}{256}

    .. math::

       r = \frac{T_2}{T_1} = \frac{3}{16}

77. Let three number in G. P. are :math:`ar, a, \frac{a}{r}` then
    given that :math:`ar, 2a, \frac{a}{r}` in A. P.

    .. math::

       \therefore 2(2a) = a\left(r + \frac{1}{r}\right)

    or :math:`r^2 - 4r + 1 = 0`

    or :math:`r = 2 \pm \sqrt{3}`

    Since, it is an increasing G. P. therefore :math:`r = 2 +
    \sqrt{3}`.

78. Given, :math:`(4x + 1)^2 = (2x + 1)(8x + 1)`

    :math:`2x = 0 \Rightarrow x = 0`

    But :math:`x = 0` makes :math:`f(x), f(2x)` and :math:`f(4x)`
    equal which is G. P. of :math:`r = 1`.

79. Let :math:`r` be the common ratio, then, we have

    :math:`a + ar + ar^2 = x. ar` or :math:`r^2 + r(1 - x) + 1 = 0`,
    :math:`r` is real.

    :math:`Discriminant > 0` i.e. :math:`(1 - x)^2 - 4 > 0`

    or :math:`(x + 1)(x - 3) > 0` :math:`\Rightarrow x < -1
    ~~\text{or}~~ x > 3`.

80. Let the numbers be :math:`a` and :math:`b` then :math:`A = \frac{a
    + b}{2}` or :math:`a + b = 2A`

    Also, :math:`G = \sqrt{ab} \Rightarrow G^2 = ab`

    Thus, :math:`a` and :math:`b` are roots of

    :math:`t^2 - 2At + G^2 = 0`

    .. math::

       t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = A \pm \sqrt{A^2 -
       G^2}

81. Let :math:`A` be the A. M. and :math:`G` be the G. M., then, we
    have

    .. math::

       \frac{A}{G} = \frac{m}{n}

    .. math::

       \frac{A}{m} = \frac{G}{n} = k

    :math:`\therefore a + b = 2mk, ab = n^2k^2`

    Hence, :math:`a, b` are roots of

    :math:`x^2 - 2mkx + k^2n^2 = 0`

    :math:`\therefore x = 2mk \pm 2k\sqrt{m^2 - n^2}`

    :math:`\therefore a:b = m + \sqrt{m^2 - n^2}: m - \sqrt{m^2 -
    n^2}`.

82. :math:`x = \frac{1}{1 - a}, y = \frac{1}{1 - b}, z = \frac{1}{1 -
    c}`

    :math:`\therefore \frac{1}{x} = 1 - a, \frac{1}{y} = 1 - b,
    \frac{1}{z} = 1 - c`

    Since :math:`a, b, c` are in A. P., therefore :math:`\frac{1}{x},
    \frac{1}{y}, \frac{1}{z}` are in A. P.

83. For :math:`p, a =1, r = -\tan^2x`

    .. math::

       \therefore p = \frac{a}{1 - r} = \frac{1}{1 + \tan^2x} = \cos^2x

    For :math:`q, a =1, r = -\cot^2y`

    .. math::

       \therefore q = \frac{a}{1 - r} = \frac{1}{1 + \cot^2y} =
       \sin^2y

    .. math::

       S = \frac{1}{1 - \tan^2x\cot^2y}

    .. math::

       = \frac{1}{1 - \frac{1 - \cos^2x}{\cos^2x}\frac{1 -
       \sin^2y}{\sin^2y}}

    .. math::

       = \frac{pq}{p + q - 1} = \frac{1}{\frac{1}{p} + \frac{1}{q} -
       \frac{1}{pq}}

84. Let side of outermost equilateral triangle is :math:`a`, then its
    area is :math:`\frac{\sqrt{3}}{4}a^2`. The sides of subsequent
    internal triangles will be :math:`\frac{a}{2}, \frac{a}{4},
    \frac{a}{8}, ...`

    Therefore, total area is :math:`\frac{\sqrt{3}}{4}a^2\left(\frac{1}{4}
    + \frac{1}{16} + \frac{1}{64} + ... \right)`

    .. math::

       = \frac{\sqrt{3}}{4}a^2. \frac{1}{1 - \frac{1}{4}} = 1

85. :math:`\cos^2x = |\cos^2x|`

    Sum of infinite series is :math:`S = \frac{1}{1 - |\cos x|}` where
    :math:`|\cos x| < 1`.

    :math:`E = e^{S\log_e 4} = 4^S`

    :math:`E` satisfied theq equation :math:`t^2 - 20t + 64 = 0
    \therefore t = 16, 4`

    :math:`4^S = 4^1` or :math:`4^2` or :math:`S = 1` or :math:`2`

    or :math:`\frac{1}{1 - |\cos x|} = 1 ~~\text{or}~~ 2`

    or :math:`1 - |\cos x| = 1 ~~\text{or}~~\frac{1}{2}`

    :math:`\Rightarrow |\cos x| = 0 ~~\text{or}~~ \frac{1}{2}`

    :math:`\therefore \cos x = 0 ~~\text{or}~~ \pm\frac{1}{2}`

    :math:`x = \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3}`.

86. The given equation may be written as

    :math:`8^{1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty} =
    8^2`

    :math:`1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty = 2`

    To sum the G. P., we must observe that for

    :math:`-\pi < x < \pi, x \ne 0`, we have :math:`|\cos x| < 1`.

    Hence :math:`\frac{1}{1 - |\cos x|} = 2` or :math:`1 - |\cos x| =
    \frac{1}{2}` by :math:`S_{\infty}` for G. P.

    or :math:`|\cos x| = 1/2` i.e. :math:`\cos x = \pm 1/2`.

87. :math:`T_n = (1 + a + a^2 + ... + a^{n - 1})b^{n - 1}`

    .. math::

       = \frac{1 - a^n}{1 - a}.b^{n - 1}

    .. math::

       = \frac{1}{1 - a}[b^{n - 1} - a(ab)^{n - 1}]

    Putting :math:`n = 1, 2, 3, ..., \infty` and adding

    .. math::

       S_{\infty} = \frac{1}{1 - a}[(1 + b + b^2 + ... \infty) +
       a(1 + ab + a^2b^2 + ... \infty)]

    .. math::

       S_{\infty} = \frac{1}{1 - a}\left[\frac{1}{1 - b} -
       a.\frac{1}{1 - ab}\right]

    .. math::

       S_{\infty} = \frac{1}{(1 - b)(1 - ba)}

88. :math:`S_{\infty} = \frac{\sin^2x}{1 - \sin^2x} = \tan^2x`

    .. math::

       L. H. S. = e^{\tan^2x \log 2} = 2^{\tan^2x}

    and the roots of the equation :math:`x^2 - 9x + 8 = 0` are
    :math:`1` and :math:`8`.

    :math:`2^{\tan^2x} = 1 = 2^0, 2^{\tan^2x} = 8 = 2^3`

    :math:`\therefore \tan^2x = 0, \tan^2x = 3`

    :math:`\therefore \tan x = 0, \tan x = \pm \sqrt{3}`

    :math:`\therefore x = \frac{\pi}{3}` is the only value of :math:`x`
    satisfying :math:`0 < x < \frac{\pi}{2}`

    :math:`\therefore \frac{\cos x}{\cos x + \sin x} = \frac{1}{1 +
    \tan x} = \frac{1}{1 + \sqrt{3}}`

89. :math:`S_{\lambda = 1 + \frac{1}{\lambda} + \frac{1}{\lambda^2} +
    ... \infty} = \frac{\lambda}{\lambda - 1}`


    .. math::

       \therefore \sum_{\lambda = 1}^n (\lambda - 1)S_{\lambda} =
       \sum_{\lambda}^n \lambda = \frac{n(n + 1)}{2}

90. :math:`\frac{T_2}{T_1} = \frac{T_3}{T_2} \Rightarrow 2^{(b - a).x}
    = 2^{(c - b)x}`

    :math:`\Rightarrow (b - a)x = (c - a)x \Rightarrow b - a = c - a
    \forall x, x\ne 0`

    Above is true as :math:`a, b, c` are in A. P.

91. Writing, :math:`a + be^x = 2a - (a - be^x)`, we have

    .. math::

       \frac{2a}{a - be^x} - 1 = \frac{2b}{b - ce^x} - 1 = \frac{2c}{c
       - de^x} - 1

    .. math::

       \Rightarrow \frac{a - be^x}{a} = \frac{b - ce^x}{b} = \frac{c -
       de^x}{c}

    .. math::

       \Rightarrow 1 - \frac{b}{a}e^x = 1 - \frac{c}{b}e^x = 1 -
       \frac{d}{c}e^x

    .. math::

       \frac{b}{a} = \frac{c}{b} = \frac{d}{c}

    Thus, :math:`a, b, c` are in G. P.

92. Since, :math:`x, y, z` are in G. P. :math:`y^2 = xz`

    and :math:`2\tan^{-1}x = \tan^{-1}y + \tan^{-1}z`

    .. math::

       \frac{2y}{1 - y^2} = \frac{x + z}{1 - xz} \Rightarrow 2y = x +
       z

    .. math::

       4y^2 = (x + z)^2 \Rightarrow 4zx = (x + z)^2 \Rightarrow (x -
       z)^2 = 0 \Rightarrow x = z

    .. math::

       \therefore x = y = z

93. Given, :math:`b - a = c - b` and :math:`(c - b)^2 = a(b - a)`

    or :math:`(b - a)^2 = a(b - a) \Rightarrow b = 2a`

    but :math:`c = 2b - a = 3a`.

    :math:`\therefore a : b : c = 1 : 2 : 3`

94. :math:`\log \frac{a}{2b}, \log \frac{2b}{3c}, \log \frac{3c}{a}`
    are in A. P.

    :math:`\therefore 2\log \frac{2b}{3c} = \log \frac{a}{2b} + \log
    \frac{3c}{a}`

    :math:`\log\left(\frac{2b}{3c}\right)^2 =
    \log\left(\frac{a}{2b}.\frac{3c}{a}\right)`

    :math:`\Rightarrow \frac{4b^2}{9c^2} = \frac{3c}{2b}` or
    :math:`8b^3 = 27c^3` :math:`\therefore 2b = 3c`

    Also, :math:`a, b, c` are in G. P. :math:`\therefore b^2 = ac`

    :math:`\frac{9c^2}{4} = ac` :math:`\therefore a = \frac{9}{4}c`

    Thus, sides are :math:`\frac{9}{4}c, \frac{6}{4}c` and
    :math:`c`. Clearly, :math:`a` is greatest side so that
    :math:`\angle A` is greatest.

    :math:`\cos A = \frac{b^2 + c^2 - a^2}{2bc} = -\frac{29}{48} < 0`

    Therefore, :math:`\angle A` is obstuse so the triangle is obtuse
    angled triangle.

95. Given,

    .. math::

       Area = \begin{vmatrix}
       a & c & e & a           \\
       d & e & f & b
       \end{vmatrix}

    Substituting the values and evaluating the determinant will yield
    the desired result.

96. :math:`a^t = \log_t a . \log_b t = \log_b a`

    :math:`t = \log_a(\log_b a)`