Arithmetic Progression Solutions Part 2 *************************************** 51. We have .. math:: S_{n + 2} - S{n + 1} = a_{n + 2}, ~~~ S_{n + 3} - S_n = a_{n + 1} + a_{n + 2} + a_{n + 3} Thus, we only have to prove that .. math:: a_{n + 1} + a_{n + 2} + a_{n + 3} - 3a_{n + 2} = 0 Since :math:`a_{n + 2}` is the mean between :math:`a_{n + 1}` and :math:`a_{n + 3}`, therefore .. math:: a_{n + 1} + a_{n + 3} = 2a_{n + 2} and, consequently .. math:: a_{n + 1} + a_{n + 2} + a_{n + 3} - 3a_{n + 2} = 0. 52. According to our notation we have :math:`S_k = a_{(k - 1)n + 1} + a_{(k - 1)n+2} + ... + a_{kn}` and :math:`S_{k + 1} = a_{kn + 1} + a_{kn+2} + ... + a_{(k + 1)n}` .. math:: \therefore S_{k + 1} - S_k = \left[a_{kn + n} - a_{kn} + ... + \left[a_{kn + 2} - a_{(k - 1)n + 2}\right]\right] + \left[a_{kn + 1} - a_{(k - 1)n + 1}\right] But since :math:`a_m - a_l = (m - l)d`, we have :math:`S_{k + 1} - Sk = nd + ... + nd + nd = n^2d.` 53. Let the given progression be :math:`a_1, a_2, ..., a_n`. Let :math:`a_{\overline{k}}` designate the :math:`kth` term from the end of the progression. Then :math:`a_{\overline{k}} = a_n - (k - 1)d, ~~~ a_k = a_1 + (k - 1)d` Considering the product :math:`a_ka_{\overline{k}}`, we have :math:`a_ka_{\overline{k}} = a_1a_n - (k - 1)^2d^2 + (k - 1)(a_n - a_1)` :math:`= a_1a_n - (k - 1)^2d^2 + (k - 1)(d - 1)d^2`. And so :math:`a_ka_{\overline{k}} = a_1a_n + d^2\left\{(k - 1)(n - 1) - (k - 1)^2\right\}`. It only remains to prove that the expression :math:`P_k = (k - 1)(n - 1) - (k - 1)^2` increases with an increase in :math:`n` to :math:`\frac{n}{2}` or :math:`\frac{n + 1}{1}`. We have :math:`P_k = (k - 1)(n - k), ~~~ P_{k + 1} = k (n - k - 1)` Hence, :math:`P_{k + 1} - P_k = n -2k`. Consequently, :math:`P_{k + 1} > P_k` if :math:`n - 2k > 0` i.e. if :math:`k < \frac{n}{2}`, and thus, our proposition is proved. 54. Let the :math:`nth` term of the required progression by :math:`a_n`, its common difference be :math:`d`. Then .. math:: S_x = \frac{a_1 + a_x}{2}x, ~~~ S_{kx} = \frac{a_1 + a_{kx}}{2}kx .. math:: \Rightarrow \frac{S_{kx}}{S_x} = \frac{2a_1 -d + kxd}{2a_1 - d + kx}.k For the last relation to be independent of :math:`x` is it necessary and sufficient that :math:`2a_1 - d = 0`, i.e. the common difference of the required progression must equal the doubled first term. 55. We can prove the following proposition :math:`a_k + a_l = a_{k^{\prime}} + a_{l^{\prime}}` if :math:`k + l = k^{\prime} + l^{\prime}`. Indeed, :math:`a_k = a_1 + (k - 1)d, a_l = a_1 + (l - 1)d, a_{k^{\prime}} = a_1 + (k^{\prime} - 1)d, a_{l^{\prime}} = a_1 + (l^\prime - 1)d` :math:`a_k + a_l = 2a_1 + (k + l - 2)d, a_{k^\prime} + a_{l^\prime} = 2a_1 + (k^{\prime}) + l^{\prime} - 2`, thus, :math:`a_k + a_l = a_{k^{\prime}} + a_{l^{\prime}}` if :math:`k + l = k^{\prime} + l^{\prime}`. And so we have :math:`a_i + a_{i + 2} = 2a_{i + 1}` The given sum is transformed as follows .. math:: S = \frac{1}{2}\sum_{i = 1}^na_ia_{i + 2} .. math:: S = \frac{1}{2}\sum_{i = 1}^n(a_{i + 1}^2 - d^2) .. math:: S = \frac{1}{2}n\left\{a_1^2 + a_1d(n + 1) + \frac{(n - 1)(2n + 5)}{6}d^2\right\}. 56. We have .. math:: \frac{1}{a_1a_n} = \frac{1}{a_1 + a_n}.\frac{a_1 + a_n}{a_1a_n} = \frac{1}{a_1 + a_n}\left(\frac{1}{a_n} + \frac{1}{a_1}\right), .. math:: \frac{1}{a_2a_{n - 1}} = \frac{1}{a_2 + a_{n - 1}}.\frac{a_2 + a_{n - 1}}{a_na_{n - 1}} = \frac{1}{a_2 + a_{n - 1}}\left(\frac{1}{a_2} + \frac{1}{a_{n - 1}}\right) .. math:: ... .. math:: \frac{1}{a_na_1} = \frac{1}{a_1 + a_n}.\frac{a_1 + a_n}{a_1a_n} = \frac{1}{a_1 + a_n}\left(\frac{1}{a_1} + \frac{1}{a_n}\right) But :math:`a_1 + a_n = a_2 + a_{n - 1} + a_3 + a_{n - 1} = ... .` Therefore, adding our equalities termwise, we find .. math:: \frac{1}{a_1a_n} + \frac{1}{a_2a_{n - 1}} + ... + \frac{1}{a_na_1} = \frac{2}{a_1 + a_n}\left(\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}\right) 57. From the first equality we have .. math:: :label: 1 \frac{x_1 + x_n}{2}n = a, ~~~ nx_1 + d\frac{2(n - 1)}{1.2} = a On the other hand, :math:`x_k^2 = x_1^2 + 2x_1d(k - 1) + d^2(k - 1)^2` Therefore, from the second relation we get .. math:: \sum_{k = 1}^nx_k^2 = ax_1^2 + 2x_1d \sum_{k = 1}^n(k - 1) + d^2 \sum_{k - 1}^n(k - 1)^2 = b^2 Hence, .. math:: :label: 2 nx_1^2 + 2x_1d\frac{n(n - 1)}{1.2} + d^2\frac{n(n - 1)}{2n - 1}{6} = b^2 Squaring both member of :eq:`1` and diving by :math:`n`, we find .. math:: :label: 3 nx_1^2 + 2x_1d\frac{n(n - 1)}{1.2} + d^2\frac{n(n - 1)^2}{4} = \frac{a^2}{n}. Subtracting :eq:`2` from :eq:`1`, we get .. math:: \frac{d^2n(n^2 - 1)}{12} = \frac{b^n - a^2}{n} Consequently .. math:: d = \pm\frac{2\sqrt{3(b^n - a^2)}}{n\sqrt{n^2 - 1}} Substituting :math:`d` into equality :eq:`1`, we find :math:`x_1`, and, consequently, we can construct the whole arithmetic progression. 58. It is obvious that .. math:: tan^{-1}a_k + tan^{-1}(-a_{k - 1}) = tan^{-1}\frac{a_k - a_{k - 1}}{1 + a_ka_{k - 1}} = tan^{-1}\frac{r}{1+ a_ka_{k - 1}}. Now we find easily that our sun is equal to .. math:: tan^{-1}\frac{a_{n + 1} - a_1}{1 + a_1a_{n + 1}}. 59. :math:`d = \alpha_2 - \alpha_1 = \alpha_3 - \alpha_2 = ... = \alpha_n - \alpha_{n - 1}` .. math:: T_1 = \frac{\sin d}{\sin \alpha_1 - \sin\alpha_2} = \frac{\sin(\alpha_2 - \alpha_1)}{\sin\alpha_1.\sin\alpha_2} .. math:: = \cot\alpha1 - \cot\alpha_2 .. math:: \therefore S = \cot\alpha_1 - \cos\alpha_n 60. This problem is similar to 49 and has been left as an exercise for the reader. 61. Expanding the sum, we get .. math:: S = \log a + \log \frac{a^2}{b} + \log \frac{a^3}{b^2} + ... .. math:: = \log a + 2\log a - \log b + 3\log a - 2\log b + ... This is an A. P with common difference :math:`d = \log a - \log b` and :math:`a = \log a` .. math:: \therefore S_n = \frac{n}{2}\left[\log \frac{a^{n + 1}}{b^{n - 1}}\right]