Chapter 8 - Sequences and Series

$S_1 = 3(1) + 2(1)^2 = 5 = a_1$.
$S_2 = 3(2) + 2(2)^2 = 6 + 8 = 14 = a_1 + a_2$.
$a_2 = S_2 - S_1 = 14 - 5 = 9$.
$d = a_2 - a_1 = 9 - 5 = 4$.

Let the GP be $a, ar, ar^2, ar^3, ar^4$.
Given 3rd term $ar^2 = 4$.
Product of first 5 terms = $a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} = (ar^2)^5$.
Substituting the value, Product = $4^5$.

Given $9a_9 = 13a_{13}$.
$9(a + 8d) = 13(a + 12d)$.
$9a + 72d = 13a + 156d$.
$4a + 84d = 0 \implies a + 21d = 0$.
$a_{22} = a + 21d = 0$.

x, y, z in GP $\implies y = xr, z = xr^2$.
x, 2y, 3z in AP $\implies 2(2y) = x + 3z \implies 4y = x + 3z$.
$4(xr) = x + 3(xr^2)$.
Dividing by x (since distinct, x $\neq$ 0): $4r = 1 + 3r^2$.
$3r^2 - 4r + 1 = 0 \implies (3r - 1)(r - 1) = 0$.
$r = 1$ or $r = 1/3$.
Since numbers are distinct, $r \neq 1$. Thus $r = 1/3$.

x, y, z in GP $\implies y^2 = xz$.
x+3, y+3, z+3 in HP $\implies y+3 = \frac{2(x+3)(z+3)}{x+3+z+3}$.
Solving this system leads to $y=2$ (implies $x=z=2$) or $y=3$ (implies $x=1, z=9$ or similar).
Given the options, $y=2$ is the intended answer.

$S_n = 3n^2 - n$.
$S_1 = 3(1)^2 - 1 = 2 = a_1$.
$S_2 = 3(2)^2 - 2 = 12 - 2 = 10$.
$a_2 = S_2 - S_1 = 10 - 2 = 8$.
$d = a_2 - a_1 = 8 - 2 = 6$.

$S_{2n} = 3S_n \implies \frac{2n}{2}[2a+(2n-1)d] = 3\frac{n}{2}[2a+(n-1)d]$.
$2(2a+2nd-d) = 3(2a+nd-d) \implies 4a+4nd-2d = 6a+3nd-3d$.
$nd+d = 2a \implies 2a = d(n+1)$.
Ratio $\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]} = 3 \frac{d(n+1)+(3n-1)d}{d(n+1)+(n-1)d} = 3 \frac{4nd}{2nd} = 6$.

Let common ratio be $r$. $x_2=x_1r, x_3=x_1r^2$ and $y_2=y_1r, y_3=y_1r^2$.
Slope $m_1 = \frac{y_2-y_1}{x_2-x_1} = \frac{y_1(r-1)}{x_1(r-1)} = \frac{y_1}{x_1}$.
Slope $m_2 = \frac{y_3-y_2}{x_3-x_2} = \frac{y_1r(r-1)}{x_1r(r-1)} = \frac{y_1}{x_1}$.
Slopes are equal, so points are collinear.

Using $AM \ge GM$: $\frac{4^x + 4^{1-x}}{2} \ge \sqrt{4^x \cdot 4^{1-x}} = \sqrt{4} = 2$.
$4^x + 4^{1-x} \ge 4$.

$S_n = An^2 + Bn$. Given $S_n = qn^2$, so $B=0$.
This implies $a = d/2$. Also $A = d/2 = q \implies d=2q, a=q$.
$S_q = \frac{q}{2}[2a + (q-1)d] = \frac{q}{2}[2q + (q-1)2q] = q[q + q^2 - q] = q^3$.