Chapter 8 - Sequences and Series

Let the two numbers be $a$ and $b$.
$A = \frac{a+b}{2} \implies 2A = a+b$.
Since $G_1, G_2$ are two geometric means between $a$ and $b$, the sequence $a, G_1, G_2, b$ is in G.P.
Let $r$ be the common ratio. Then $G_1 = ar, G_2 = ar^2, b = ar^3$.
LHS = $\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = \frac{(ar)^2}{ar^2} + \frac{(ar^2)^2}{ar} = \frac{a^2r^2}{ar^2} + \frac{a^2r^4}{ar} = a + ar^3 = a + b$.
Since $a+b = 2A$, LHS = $2A$ = RHS.

The general term is $T_k = \sec\theta_k \sec\theta_{k+1} = \frac{1}{\cos\theta_k \cos\theta_{k+1}}$.
Multiply and divide by $\sin d = \sin(\theta_{k+1} - \theta_k)$.
$T_k = \frac{\sin(\theta_{k+1} - \theta_k)}{\sin d \cos\theta_k \cos\theta_{k+1}} = \frac{\sin\theta_{k+1}\cos\theta_k - \cos\theta_{k+1}\sin\theta_k}{\sin d \cos\theta_k \cos\theta_{k+1}}$
$= \frac{1}{\sin d} (\frac{\sin\theta_{k+1}}{\cos\theta_{k+1}} - \frac{\sin\theta_k}{\cos\theta_k}) = \frac{1}{\sin d} (\tan\theta_{k+1} - \tan\theta_k)$.
Sum = $\frac{1}{\sin d} \sum_{k=1}^{n-1} (\tan\theta_{k+1} - \tan\theta_k)$
This is a telescoping sum: $\frac{1}{\sin d} [(\tan\theta_2 - \tan\theta_1) + (\tan\theta_3 - \tan\theta_2) + \dots + (\tan\theta_n - \tan\theta_{n-1})]$
$= \frac{\tan\theta_n - \tan\theta_1}{\sin d}$.

Given $S_p = q$ and $S_q = p$.
$\frac{p}{2}[2a + (p-1)d] = q \implies 2ap + p(p-1)d = 2q$ ... (i)
$\frac{q}{2}[2a + (q-1)d] = p \implies 2aq + q(q-1)d = 2p$ ... (ii)
Subtracting (ii) from (i): $2a(p-q) + d[p^2-p - q^2+q] = 2(q-p)$.
$2a(p-q) + d[(p-q)(p+q) - (p-q)] = -2(p-q)$.
Dividing by $(p-q)$: $2a + d(p+q-1) = -2$.
Now, $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] = \frac{p+q}{2}(-2) = -(p+q)$.
To find $S_{p-q}$, we need $a$ and $d$.
From $2a + d(p+q-1) = -2$, we have $2a = -2 - d(p+q-1)$.
Substitute into (i): $(-2 - d(p+q-1))p + p(p-1)d = 2q$.
$-2p - pd(p+q-1) + p^2d - pd = 2q$.
$-2p - p^2d - pqd + pd + p^2d - pd = 2q$.
$-2p - pqd = 2q \implies -pqd = 2(p+q) \implies d = -\frac{2(p+q)}{pq}$.
$S_{p-q} = \frac{p-q}{2}[2a + (p-q-1)d] = \frac{p-q}{2}[-2 - d(p+q-1) + d(p-q-1)]$
$= \frac{p-q}{2}[-2 + d(p-q-1 - p - q + 1)] = \frac{p-q}{2}[-2 - 2qd] = -(p-q)(1+qd)$.
$= -(p-q)(1 + q(-\frac{2(p+q)}{pq})) = -(p-q)(1 - \frac{2(p+q)}{p}) = -(p-q)(\frac{p-2p-2q}{p}) = \frac{(p-q)(p+2q)}{p}$.

Let the A.P. be $A, A+D, \dots$ and G.P. be $X, XR, \dots$.
For A.P.: $a = A + (p-1)D$, $b = A + (q-1)D$, $c = A + (r-1)D$.
So $b-c = (q-r)D$, $c-a = (r-p)D$, $a-b = (p-q)D$.
For G.P.: $a = XR^{p-1}$, $b = XR^{q-1}$, $c = XR^{r-1}$.
LHS = $a^{b-c} b^{c-a} c^{a-b} = (XR^{p-1})^{(q-r)D} (XR^{q-1})^{(r-p)D} (XR^{r-1})^{(p-q)D}$.
Exponent of $X$: $D(q-r + r-p + p-q) = 0$. So $X^0 = 1$.
Exponent of $R$: $D[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$.
Expanding the term in bracket: $(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) = 0$.
So $R^0 = 1$.
Thus, LHS = $1 \cdot 1 = 1$.

If $a, b, c$ are in G.P., then $b = ar, c = ar^2$.
$\frac{a-b}{b-c} = \frac{a-ar}{ar-ar^2} = \frac{a(1-r)}{ar(1-r)} = \frac{1}{r} = \frac{a}{b}$.

In an A.P., $a_k + a_{n-k+1} = a + l$.

$a_3 = ar^2 = 4$.
Product of first 5 terms = $a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} = (ar^2)^5 = 4^5 = 1024$.

A constant sequence (e.g., 2, 2, 2, ...) is both an A.P. (d=0) and a G.P. (r=1).

A progression is a sequence with a specific pattern/formula. Random sequences (like digits of pi) are not progressions.

$a_n = \frac{a_{n-k} + a_{n+k}}{2}$.

Example: $1, 2, 4, 8$ and $1, 3, 9, 27$. Sum: $2, 5, 13, 35$. $5/2 \neq 13/5$.

If $S_n = An^2 + Bn$, then $a_n = S_n - S_{n-1}$ is a linear expression in $n$, which characterizes an A.P.

(a) Common ratio $r = 1/4$. It is a G.P.
(b) Prime numbers. It is a sequence.
(c) Common difference $d = -5$. It is an A.P.

(a) Sum of squares: $\frac{n(n+1)(2n+1)}{6}$.
(b) Sum of cubes: $(\frac{n(n+1)}{2})^2$.
(c) Sum of first $n$ even numbers: $2 \sum n = n(n+1)$.
(d) Sum of first $n$ natural numbers: $\frac{n(n+1)}{2}$.