33. Arithmetic Progression Solutions Part 2#

  1. We have

    \[S_{n + 2} - S{n + 1} = a_{n + 2}, ~~~ S_{n + 3} - S_n = a_{n + 1} + a_{n + 2} + a_{n + 3} \]

    Thus, we only have to prove that

    \[a_{n + 1} + a_{n + 2} + a_{n + 3} - 3a_{n + 2} = 0 \]

    Since \(a_{n + 2}\) is the mean between \(a_{n + 1}\) and \(a_{n + 3}\), therefore

    \[a_{n + 1} + a_{n + 3} = 2a_{n + 2} \]

    and, consequently

    \[a_{n + 1} + a_{n + 2} + a_{n + 3} - 3a_{n + 2} = 0. \]

  2. According to our notation we have

    \(S_k = a_{(k - 1)n + 1} + a_{(k - 1)n+2} + ... + a_{kn}\) and \(S_{k + 1} = a_{kn + 1} + a_{kn+2} + ... + a_{(k + 1)n}\)

    \[\therefore S_{k + 1} - S_k = \left[a_{kn + n} - a_{kn} + ... + \left[a_{kn + 2} - a_{(k - 1)n + 2}\right]\right] + \left[a_{kn + 1} - a_{(k - 1)n + 1}\right] \]

    But since \(a_m - a_l = (m - l)d\), we have

    \(S_{k + 1} - Sk = nd + ... + nd + nd = n^2d.\)

  3. Let the given progression be \(a_1, a_2, ..., a_n\). Let \(a_{\overline{k}}\) designate the \(kth\) term from the end of the progression. Then

    \(a_{\overline{k}} = a_n - (k - 1)d, ~~~ a_k = a_1 + (k - 1)d\)

    Considering the product \(a_ka_{\overline{k}}\), we have

    \(a_ka_{\overline{k}} = a_1a_n - (k - 1)^2d^2 + (k - 1)(a_n - a_1)\) \(= a_1a_n - (k - 1)^2d^2 + (k - 1)(d - 1)d^2\).

    And so

    \(a_ka_{\overline{k}} = a_1a_n + d^2\left\{(k - 1)(n - 1) - (k - 1)^2\right\}\).

    It only remains to prove that the expression \(P_k = (k - 1)(n - 1) - (k - 1)^2\) increases with an increase in \(n\) to \(\frac{n}{2}\) or \(\frac{n + 1}{1}\).

    We have \(P_k = (k - 1)(n - k), ~~~ P_{k + 1} = k (n - k - 1)\)

    Hence, \(P_{k + 1} - P_k = n -2k\). Consequently, \(P_{k + 1} > P_k\) if \(n - 2k > 0\) i.e. if \(k < \frac{n}{2}\), and thus, our proposition is proved.

  4. Let the \(nth\) term of the required progression by \(a_n\), its common difference be \(d\). Then

    \[S_x = \frac{a_1 + a_x}{2}x, ~~~ S_{kx} = \frac{a_1 + a_{kx}}{2}kx \]
    \[\Rightarrow \frac{S_{kx}}{S_x} = \frac{2a_1 -d + kxd}{2a_1 - d + kx}.k \]

    For the last relation to be independent of \(x\) is it necessary and sufficient that \(2a_1 - d = 0\),

    i.e. the common difference of the required progression must equal the doubled first term.

  5. We can prove the following proposition \(a_k + a_l = a_{k^{\prime}} + a_{l^{\prime}}\) if \(k + l = k^{\prime} + l^{\prime}\).

    Indeed,

    \(a_k = a_1 + (k - 1)d, a_l = a_1 + (l - 1)d, a_{k^{\prime}} = a_1 + (k^{\prime} - 1)d, a_{l^{\prime}} = a_1 + (l^\prime - 1)d\)

    \(a_k + a_l = 2a_1 + (k + l - 2)d, a_{k^\prime} + a_{l^\prime} = 2a_1 + (k^{\prime}) + l^{\prime} - 2\), thus, \(a_k + a_l = a_{k^{\prime}} + a_{l^{\prime}}\) if \(k + l = k^{\prime} + l^{\prime}\).

    And so we have \(a_i + a_{i + 2} = 2a_{i + 1}\)

    The given sum is transformed as follows

    \[S = \frac{1}{2}\sum_{i = 1}^na_ia_{i + 2} \]
    \[S = \frac{1}{2}\sum_{i = 1}^n(a_{i + 1}^2 - d^2) \]
    \[S = \frac{1}{2}n\left\{a_1^2 + a_1d(n + 1) + \frac{(n - 1)(2n + 5)}{6}d^2\right\}. \]

  6. We have

    \[\frac{1}{a_1a_n} = \frac{1}{a_1 + a_n}.\frac{a_1 + a_n}{a_1a_n} = \frac{1}{a_1 + a_n}\left(\frac{1}{a_n} + \frac{1}{a_1}\right), \]
    \[\frac{1}{a_2a_{n - 1}} = \frac{1}{a_2 + a_{n - 1}}.\frac{a_2 + a_{n - 1}}{a_na_{n - 1}} = \frac{1}{a_2 + a_{n - 1}}\left(\frac{1}{a_2} + \frac{1}{a_{n - 1}}\right) \]
    \[... \]
    \[\frac{1}{a_na_1} = \frac{1}{a_1 + a_n}.\frac{a_1 + a_n}{a_1a_n} = \frac{1}{a_1 + a_n}\left(\frac{1}{a_1} + \frac{1}{a_n}\right) \]

    But \(a_1 + a_n = a_2 + a_{n - 1} + a_3 + a_{n - 1} = ... .\)

    Therefore, adding our equalities termwise, we find

    \[\frac{1}{a_1a_n} + \frac{1}{a_2a_{n - 1}} + ... + \frac{1}{a_na_1} = \frac{2}{a_1 + a_n}\left(\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}\right) \]

  7. From the first equality we have

    (1)#\[\frac{x_1 + x_n}{2}n = a, ~~~ nx_1 + d\frac{2(n - 1)}{1.2} = a\]

    On the other hand, \(x_k^2 = x_1^2 + 2x_1d(k - 1) + d^2(k - 1)^2\)

    Therefore, from the second relation we get

    \[\sum_{k = 1}^nx_k^2 = ax_1^2 + 2x_1d \sum_{k = 1}^n(k - 1) + d^2 \sum_{k - 1}^n(k - 1)^2 = b^2 \]

    Hence,

    (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 (1) and diving by \(n\), we find

    (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 (2) from (1), we get

    \[\frac{d^2n(n^2 - 1)}{12} = \frac{b^n - a^2}{n} \]

    Consequently

    \[d = \pm\frac{2\sqrt{3(b^n - a^2)}}{n\sqrt{n^2 - 1}} \]

    Substituting \(d\) into equality (1), we find \(x_1\), and, consequently, we can construct the whole arithmetic progression.

  8. It is obvious that

    \[tan^{-1}a_k + tan^{-1}(-a_{k - 1}) = tan^{-1}\frac{a_k - a_{k - 1}}{1 + a_ka_{k - 1}} = tan^{-1}\frac{r}{1+ a_ka_{k - 1}}. \]

    Now we find easily that our sun is equal to

    \[tan^{-1}\frac{a_{n + 1} - a_1}{1 + a_1a_{n + 1}}. \]

  9. \(d = \alpha_2 - \alpha_1 = \alpha_3 - \alpha_2 = ... = \alpha_n - \alpha_{n - 1}\)

    \[T_1 = \frac{\sin d}{\sin \alpha_1 - \sin\alpha_2} = \frac{\sin(\alpha_2 - \alpha_1)}{\sin\alpha_1.\sin\alpha_2} \]
    \[= \cot\alpha1 - \cot\alpha_2 \]
    \[\therefore S = \cot\alpha_1 - \cos\alpha_n \]

  10. This problem is similar to 49 and has been left as an exercise for the reader.

  11. Expanding the sum, we get

    \[S = \log a + \log \frac{a^2}{b} + \log \frac{a^3}{b^2} + ... \]
    \[= \log a + 2\log a - \log b + 3\log a - 2\log b + ... \]

    This is an A. P with common difference \(d = \log a - \log b\) and \(a = \log a\)

    \[\therefore S_n = \frac{n}{2}\left[\log \frac{a^{n + 1}}{b^{n - 1}}\right] \]