# 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 $$a = \log a$$

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