38. Geometric Progressions Solutions Part 2

51. Let \(r\) be the common ratio then \(b = ar^{n - 1}\). And we have \(p\) as

    \[p = a.ar.ar^2. ... .ar^{n - 1} \]
    \[p = a^nr^{\frac{n(n - 1)}{2}} \]
    \[p^2 = a^{2n}r^{n(n - 1)} \]
    \[p^2 = (a^2r^{n - 1})^n \]
    \[p^2 = (a.ar^{n - 1})^n \]
    \[p^2 = (ab)^n \]

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

    \[S_1 = \frac{a_1(1 - r^{n - 1})}{1 - r}~~~S_2 = \frac{a_2(1 - r^{n -1})}{1 -r} \]
    \[\therefore \frac{S_1}{S_2} = \frac{a_1}{a_2} = \frac{a_1r^{n - 1}}{a_2r^{n - 1}} \]

    Thus, ratio of sums to \(n\) terms is equal to the \(n\text{th}\) terms.

53. Let \(a\) be the first term and \(r\) be the common ratio. Then, we have

    \[S_1 = a + ar + ar^2 + ... + ar^{n - 1}\]
    \[S_2 = a + ar + ar^2 + ... + ar^{n - 1} + ar^n + ar^{n + 1} + ... + ar^{2n - 1}\]
    \[S_3 = a + ar + ar^2 + ... + ar^{2n - 1} + ar^{2n} + ... + ar^{3n - 1}\]
    \[S_2 - S_1 = ar^n + ar^{n + 1} + ... + ar^{2n - 1}\]
    \[S_2 - S_1 = \frac{ar^n(r^{n} - 1)}{r - 1}\]

    Similarly,

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

    and

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

    Thus, \((S_2 - S_1)^2 = S_1(S_3 - S_2)\).

54. We need to compute following:

    \[S_1 + S_2 + ... + S_{2n - 1}\]

    From formula for sum of geometric series we have following:

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

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

56. The sum would be \(S = 3.2 - 4 + 3.2^2 - 4 + ... + 3.2^{100} - 4\).

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

    \(S = 6(2^{100} - 1) - 400\)

57. \(t_n = 1 + 2 + 2^2 + ... + 2^n = 2^n - 1\).

    Sum of \(n\) terms = \(S = 1 + 3 + 7 + 15 + ... + 2^n - 1\)

    \[S = 2^1 - 1 + 2^2 - 1 + 2^3 - 1 + 2^n - 1\]
    \[S = 2^{n + 1} - 2 - n\]

58. Clearly, the common ratio is \(3x\). Thus, from the formula for sum of infinite series we have sum as

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

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

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

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

    Thus, applying the formula, we have

    \[S = \frac{\frac{1}{5}}{1 - \frac{1}{5}} + \frac{\frac{1}{7}}{1 - \frac{1}{7}}\]
    \[S = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}\]

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

    \(S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\)

    Thus, value is \(9^\frac{1}{2} = 3\).

62. Sum within braces is

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

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

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

    The value of exponent is \(\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1\). Thus final value is \(a\).

64. \(0.\dot{5}\dot{4} = 0.54 + 0.0054 + 0.000054 + ... ~~\text{to}~~ \infty\)

    Thus, we have

    \[0.\dot{5}\dot{4} = \frac{54}{100} + \frac{54}{10000} + \frac{54}{1000000} + ...\]
    \[= \frac{54}{100}.\frac{1}{1 - \frac{1}{100}} = \frac{54}{99} = \frac{6}{11}\]

65. We have

    \[.\dot{6} = .6 + .06 + .006 + ... ~~\text{to}~~ \infty\]
    \[= \frac{6}{10} + \frac{6}{100} + \frac{6}{1000} + ...\]
    \[= \frac{6}{10}.\frac{1}{1 - 10} = \frac{6}{9} = \frac{2}{3}.\]

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

    Now, \(S_{\infty} = \frac{a}{1 - r} \because |r| < 1.\)

    Solving these two we have,

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

    Substituting first in second we have

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

    Thus, \(a = 48\) or \(a = 16\). Thus, we can have our desired series as \(16, 8, 4, 2, ...\) or \(48, -24, 12, -6\).

67. Let \(a\) be the first term and \(r\) be the common ratio. Then, we have

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

    Substituting first ins second, we have

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

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

68. Let \(a\) be the first term and \(r\) be the common ratio. Then, we have

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

    Let us consider a term \(t_n = ar^{n - 1}\). Then, sum of succeeding terms is

    \(S = \frac{ar^n}{1 - r}\). Equating these we have

    \[\frac{ar^n}{1 - r} = ar^n - 1\]
    \[\Rightarrow \frac{r}{1 - r} = 1 \Rightarrow r = \frac{1}{2}.\]

    Similarly we can prove for \(>\) or \(<\).

69. Given, \(y = 1 + x + x^2 + ...\)

    From the formula for infinite G. P.

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

70. \(c = ar^2\) and \(b = ar\). Thus, we have \(c > 4b - 3a = ar^2 > 4ar - 3a\)

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

    Thus, \(r > 3\) or \(r < 1\).

71. \(1 + 2x + 4x^2 + ... + 32x^5 = \frac{1 - k^6}{1 - k}\)

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

    \(\therefore k = 2x \Rightarrow \frac{k}{x} = 2.\)

72. Simplifying the given relation, we have

    \[(b^4 - 2b^2ac + a^2c^2) + (c^4 - 2c^2bd + b^2d^2) + (a^2d^2 - 2abcd - b^2c^2) \le 0\]
    \[\Rightarrow (b^2 - ac)^2 + (c^2 - bd)^2 + (ad - bc)^2 \le 0\]

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

73. On simplification, we have

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

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

    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.

    \(\therefore a_1a_3 = a_2^2\) or \(a_1, a_2, a_3\) are in G. P. Similarly, others are also in G. P.

    Hence proved.

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

    Sum of the roots of the equations

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

    Substituting \(S_1\) in \(S_2\), we have

    \(3r^2 = 12\) or \(r = 2 \therefore k = 1\).

    Product of the roots

    \(P_1 = \alpha\beta = k^2r = a, P_2 = \gamma\delta = k^2r^5 = b\)

    Putting for \(k\) and \(r\), we have

    \(a = 2, b = 32\).

75. Given \(d = 2, r = \frac{1}{2}.\)

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

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

    \(a + 2nd = a + 4n\).

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

    Let \(T_{n+1}\) and \(t_{n + 1}\) are mid terms of A. P. and G. P.

    \(T_{n + 1} = a + nd = a + 2n\)

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

    Given, \(T_{n + 1} = t_{n + 1}\)

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

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

    Hence, mid term is

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

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

    \[T_1 = S_1 = 3 - \frac{9}{16} = \frac{39}{16}\]
    \[S_2 = 3 - \frac{27}{256} = T_1 + T_2\]
    \[\therefore T_2 = \frac{117}{256}\]
    \[r = \frac{T_2}{T_1} = \frac{3}{16}\]

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

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

    or \(r^2 - 4r + 1 = 0\)

    or \(r = 2 \pm \sqrt{3}\)

    Since, it is an increasing G. P. therefore \(r = 2 + \sqrt{3}\).

78. Given, \((4x + 1)^2 = (2x + 1)(8x + 1)\)

    \(2x = 0 \Rightarrow x = 0\)

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

79. Let \(r\) be the common ratio, then, we have

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

    \(Discriminant > 0\) i.e. \((1 - x)^2 - 4 > 0\)

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

80. Let the numbers be \(a\) and \(b\) then \(A = \frac{a + b}{2}\) or \(a + b = 2A\)

    Also, \(G = \sqrt{ab} \Rightarrow G^2 = ab\)

    Thus, \(a\) and \(b\) are roots of

    \(t^2 - 2At + G^2 = 0\)

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

81. Let \(A\) be the A. M. and \(G\) be the G. M., then, we have

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

    \(\therefore a + b = 2mk, ab = n^2k^2\)

    Hence, \(a, b\) are roots of

    \(x^2 - 2mkx + k^2n^2 = 0\)

    \(\therefore x = 2mk \pm 2k\sqrt{m^2 - n^2}\)

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

82. \(x = \frac{1}{1 - a}, y = \frac{1}{1 - b}, z = \frac{1}{1 - c}\)

    \(\therefore \frac{1}{x} = 1 - a, \frac{1}{y} = 1 - b, \frac{1}{z} = 1 - c\)

    Since \(a, b, c\) are in A. P., therefore \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A. P.

83. For \(p, a =1, r = -\tan^2x\)

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

    For \(q, a =1, r = -\cot^2y\)

    \[\therefore q = \frac{a}{1 - r} = \frac{1}{1 + \cot^2y} = \sin^2y\]
    \[S = \frac{1}{1 - \tan^2x\cot^2y}\]
    \[= \frac{1}{1 - \frac{1 - \cos^2x}{\cos^2x}\frac{1 - \sin^2y}{\sin^2y}}\]
    \[= \frac{pq}{p + q - 1} = \frac{1}{\frac{1}{p} + \frac{1}{q} - \frac{1}{pq}}\]

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

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

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

85. \(\cos^2x = |\cos^2x|\)

    Sum of infinite series is \(S = \frac{1}{1 - |\cos x|}\) where \(|\cos x| < 1\).

    \(E = e^{S\log_e 4} = 4^S\)

    \(E\) satisfied theq equation \(t^2 - 20t + 64 = 0 \therefore t = 16, 4\)

    \(4^S = 4^1\) or \(4^2\) or \(S = 1\) or \(2\)

    or \(\frac{1}{1 - |\cos x|} = 1 ~~\text{or}~~ 2\)

    or \(1 - |\cos x| = 1 ~~\text{or}~~\frac{1}{2}\)

    \(\Rightarrow |\cos x| = 0 ~~\text{or}~~ \frac{1}{2}\)

    \(\therefore \cos x = 0 ~~\text{or}~~ \pm\frac{1}{2}\)

    \(x = \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3}\).

86. The given equation may be written as

    \(8^{1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty} = 8^2\)

    \(1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty = 2\)

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

    \(-\pi < x < \pi, x \ne 0\), we have \(|\cos x| < 1\).

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

    or \(|\cos x| = 1/2\) i.e. \(\cos x = \pm 1/2\).

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

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

    Putting \(n = 1, 2, 3, ..., \infty\) and adding

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

88. \(S_{\infty} = \frac{\sin^2x}{1 - \sin^2x} = \tan^2x\)

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

    and the roots of the equation \(x^2 - 9x + 8 = 0\) are \(1\) and \(8\).

    \(2^{\tan^2x} = 1 = 2^0, 2^{\tan^2x} = 8 = 2^3\)

    \(\therefore \tan^2x = 0, \tan^2x = 3\)

    \(\therefore \tan x = 0, \tan x = \pm \sqrt{3}\)

    \(\therefore x = \frac{\pi}{3}\) is the only value of \(x\) satisfying \(0 < x < \frac{\pi}{2}\)

    \(\therefore \frac{\cos x}{\cos x + \sin x} = \frac{1}{1 + \tan x} = \frac{1}{1 + \sqrt{3}}\)

89. \(S_{\lambda = 1 + \frac{1}{\lambda} + \frac{1}{\lambda^2} + ... \infty} = \frac{\lambda}{\lambda - 1}\)

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

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

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

    Above is true as \(a, b, c\) are in A. P.

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

    \[\frac{2a}{a - be^x} - 1 = \frac{2b}{b - ce^x} - 1 = \frac{2c}{c - de^x} - 1\]
    \[\Rightarrow \frac{a - be^x}{a} = \frac{b - ce^x}{b} = \frac{c - de^x}{c}\]
    \[\Rightarrow 1 - \frac{b}{a}e^x = 1 - \frac{c}{b}e^x = 1 - \frac{d}{c}e^x\]
    \[\frac{b}{a} = \frac{c}{b} = \frac{d}{c}\]

    Thus, \(a, b, c\) are in G. P.

92. Since, \(x, y, z\) are in G. P. \(y^2 = xz\)

    and \(2\tan^{-1}x = \tan^{-1}y + \tan^{-1}z\)

    \[\frac{2y}{1 - y^2} = \frac{x + z}{1 - xz} \Rightarrow 2y = x + z\]
    \[4y^2 = (x + z)^2 \Rightarrow 4zx = (x + z)^2 \Rightarrow (x - z)^2 = 0 \Rightarrow x = z\]
    \[\therefore x = y = z\]

93. Given, \(b - a = c - b\) and \((c - b)^2 = a(b - a)\)

    or \((b - a)^2 = a(b - a) \Rightarrow b = 2a\)

    but \(c = 2b - a = 3a\).

    \(\therefore a : b : c = 1 : 2 : 3\)

94. \(\log \frac{a}{2b}, \log \frac{2b}{3c}, \log \frac{3c}{a}\) are in A. P.

    \(\therefore 2\log \frac{2b}{3c} = \log \frac{a}{2b} + \log \frac{3c}{a}\)

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

    \(\Rightarrow \frac{4b^2}{9c^2} = \frac{3c}{2b}\) or \(8b^3 = 27c^3\) \(\therefore 2b = 3c\)

    Also, \(a, b, c\) are in G. P. \(\therefore b^2 = ac\)

    \(\frac{9c^2}{4} = ac\) \(\therefore a = \frac{9}{4}c\)

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

    \(\cos A = \frac{b^2 + c^2 - a^2}{2bc} = -\frac{29}{48} < 0\)

    Therefore, \(\angle A\) is obstuse so the triangle is obtuse angled triangle.

95. Given,

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

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

96. \(a^t = \log_t a . \log_b t = \log_b a\)

    \(t = \log_a(\log_b a)\)