Algebra - Arithmetic Progressions

Sum of the first six terms of an arithmetic sequence is $9$. Sum of the first twelve terms is $90$. Find the sum of the thirteenth and the seventeenth terms of this sequence.

Let first term of the arithmetic sequence $ = a$

and common difference $= d$

Sum of $n$ terms of an arithmetic sequence $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$

Sum of first six terms $= S_6 = \dfrac{6}{2} \left[2a + 5d\right] = 6a + 15d$

Given: $\;$ $S_6 = 9$

i.e. $\;\;\;$ $6a + 15d = 9$ $\;\;\; \cdots \; (1)$

Sum of first twelve terms $= S_{12} = \dfrac{12}{2} \left[2a + 11d\right] = 12a + 66d$

Given: $\;$ $S_{12} = 90$

i.e. $\;\;\;$ $12a + 66d = 90$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously gives

$a = \dfrac{-7}{2}$ $\;$ and $\;$ $d = 2$

Now, $\;$ $n^{th}$ $\;$ term of an arithmetic sequence $= t_n = a + \left(n - 1\right) d$

$\therefore \;$ $13^{th}$ $\;$ term of the arithmetic sequence $= t_{13} = a + 12d$

and $\;$ $17^{th}$ $\;$ term of the arithmetic sequence $= t_{17} = a + 16d$

$\therefore \;$ Sum of the thirteenth and the seventeenth terms of the sequence

$= t_{13} + t_{17} = a + 12 d + a + 16 d = 2a + 28d$

Substituting the values of $a$ and $d$ gives

$t_{13} + t_{17} = 2 \times \left(\dfrac{-7}{2}\right) + 28 \times 2 = 49$

Matrices

Solve for matrices $A$ and $B$:
$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ and $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$

Given matrix equations:

$2A + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$ $\;\;\; \cdots \; (1)$; $\;\;$ $A - 2B = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (2)$

We have from equation $(2)$

$A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + 2 B$ $\;\;\; \cdots \; (3)$

$\implies$ $2A = \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes

$\begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} + 4 B + B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix}$

i.e. $\;$ $5 B = \begin{bmatrix} 3 & -4 \\ 2 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 6 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 3 - 8 & -4 - 6 \\ 2 - 2 & 7 - 2 \end{bmatrix} = \begin{bmatrix} -5 & -10 \\ 0 & 5 \end{bmatrix}$

$\implies$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$ $\;\;\; \cdots \; (5)$

Substituting the value of $B$ from equation $(5)$ in equation $(3)$ gives

$A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + 2 \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} -2 & -4 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $A = \begin{bmatrix} 4 - 2 & 3 - 4 \\ 1 + 0 & 1 + 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

$\therefore \;$ $A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$; $\;$ $B = \begin{bmatrix} -1 & -2 \\ 0 & 1 \end{bmatrix}$

Algebra - Airthmetic Progressions

The eighth term of an A.P is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{th}$ term of the A.P.

Let the first term of the A.P be $\;$ $t_1 = a$ $\;$ and the common difference $= d$

$n^{th}$ $\;$ term of A.P $\;$ $= t_n = a + \left(n - 1\right) d$

$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$;

$4^{th}$ term $= t_4 = a + 3d$;

$8^{th}$ term $= t_8 = a + 7d$; $\;\;$ and

$11^{th}$ term $= t_{11} = a + 10d$

As per the question,

$t_8 = \dfrac{1}{2} \times t_2$

i.e. $\;$ $a + 7d = \dfrac{1}{2} \times \left(a + d\right)$

i.e. $\;$ $2a + 14 d = a + d$ $\;\;\;$

i.e. $\;$ $a + 13d = 0$ $\implies$ $a = -13d$ $\;\;\; \cdots \;\;\; (1)$

Also, as per the problem

$t_{11} = \left(\dfrac{1}{3} \times t_4\right) + 1$

i.e. $\;$ $a + 10 d = \dfrac{1}{3} \times \left(a + 3d\right) + 1$

i.e. $\;$ $3a + 30d = a + 3d + 3$

i.e. $\;$ $2a + 27d = 3$ $\;\;\; \cdots \;\;\; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$2 \times \left(-13 d\right) + 27 d = 3$ $\;\;\;$ $\implies$ $d = 3$

Substituting the value of $d$ in equation $(1)$ gives

$a = -13 \times 3 = -39$

Now, $15^{th}$ term of A.P $= t_{15} = a + 14d$

i.e. $\;$ $t_{15} = -39 + \left(14 \times 3\right) = 3$

Trigonometry - Trigonometric Functions

If $\;$ $\text{cosec} \theta - \sin \theta = a^3$, $\;$ $\sec \theta - \cos \theta = b^3$,
then prove that $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Given: $\;$ $\text{cosec } \theta - \sin \theta = a^3$

i.e. $\;$ $\dfrac{1}{\sin \theta} - \sin \theta = a^3$

i.e. $\;$ $a^3 = \dfrac{1 - \sin^2 \theta}{\sin \theta}$

i.e. $\;$ $a^3 = \dfrac{\cos^2 \theta}{\sin \theta}$ $\;\;\; \cdots \; (1)$

and $\;$ $\sec \theta - \cos \theta = b^3$

i.e. $\;$ $\dfrac{1}{\cos \theta} - \cos \theta = b^3$

i.e. $\;$ $b^3 = \dfrac{1 - \cos^2 \theta}{\cos \theta}$

i.e. $\;$ $b^3 = \dfrac{\sin^2 \theta}{\cos \theta}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ gives

$\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \div \dfrac{\cos^2 \theta}{\sin \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sin \theta}{\cos^2 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^3 \theta}{\cos^3 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \tan^3 \theta$

i.e. $\;$ $\dfrac{b}{a} = \tan \theta$

$\implies$ $\sin \theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (3)$

and $\;$ $\cos \theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (4)$

Substituting the values of $\sin \theta$ and $\cos \theta$ from equations $(3)$ and $(4)$ in equation $(1)$ gives

$a^3 = \dfrac{a^2}{a^2 + b^2} \div \dfrac{b}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^3 = \dfrac{a^2}{a^2 + b^2} \times \dfrac{\sqrt{a^2 + b^2}}{b}$

i.e. $\;$ $a \; b = \dfrac{1}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Hence proved

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$

LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$

$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$

$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$

$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$

$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right) = 2 \sec 2 \alpha$

LHS $=\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{1}{\cos \left(\dfrac{\pi}{4} + \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) + \alpha \right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{2 \sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{\sin \left[2 \left(\dfrac{\pi}{4} - \alpha\right)\right]}$

$= \dfrac{2}{\sin \left(\dfrac{\pi}{2} - 2 \alpha\right)}$

$= \dfrac{2}{\cos 2 \alpha}$

$= 2 \sec 2 \alpha = $ RHS

Algebra - Word Problems: Derivation of Equations

Two cyclists left the same point simultaneously and traveled in the same direction. The speed of the first was $15$ km/h and that of the second was $12$ km/h. Half an hour later, another cyclist left the same point and traveled in the same direction. Some time later, he overtook the second cyclist and another hour and a half later he overtook the first cyclist. Find the speed of the third cyclist.

Let cyclists $C_1$ and $C_2$ leave point $A$ simultaneously and travel in the same direction.

After $\dfrac{1}{2}$ hr, let cyclist $C_3$ start from point $A$.

Speed of $C_1 = u_1 = 15$ kmph

and speed of $C_2 = u_2 = 12$ kmph

Let $C_1$ and $C_2$ start from point $A$ at time $t = 0$

In time $= \dfrac{1}{2}$ hr, distance covered by $C_1 = 15 \times \dfrac{1}{2} = 7.5$ km

In time $= \dfrac{1}{2}$ hr, distance covered by $C_2 = 12 \times \dfrac{1}{2} = 6$ km

Let speed of third cyclist $C_3 = u_3$ kmph

Let $C_3$ overtake $C_2$ after a time $T$ hr

Distance covered by $C_2$ in time $T$ $= \left(12 \times T\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_2$ in time $T$ $= \left(12T + 6\right)$ km $\;\;\; \cdots \; (1)$

Distance covered by $C_3$ in time $T$ $= u_3 \times T$ km $\;\;\; \cdots \; (2)$

$\because \;$ $C_3$ overtakes $C_2$, we have from equations $(1)$ and $(2)$

$12 T + 6 = u_3 T$ $\implies$ $T = \dfrac{6}{u_3 - 12}$ $\;\;\; \cdots \; (3)$

Now, $C_3$ overtakes $C_1$ after a time $\left(T + 1.5\right)$ hr

Distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr is $= 15 \left(T + 1.5\right) = \left(15 T + 22.5\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr

$= \left(15T + 22.5 + 7.5\right) = 15 T + 30$ km $\;\;\; \cdots \; (4)$

Distance covered by $C_3$ in time $\left(T + 1.5\right)$ hr

$= u_3 \times \left(T + 1.5\right) = u_3 T + 1.5 u_3$ km $\;\;\; \cdots \; (5)$

$\because \;$ $C_3$ overtakes $C_1$, we have from equations $(4)$ and $(5)$

$15 T + 30 = u_3 T + 1.5 u_3$

i.e. $\;$ $T = \dfrac{1.5 u_3 - 30}{15 - u_3}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(3)$ and $(6)$,

$\dfrac{6}{u_3 - 12} = \dfrac{1.5 u_3 - 30}{15 - u_3}$

i.e. $\;$ $1.5 u_3^2 - 42 u_3 + 270 = 0$ $\;\;\; \cdots \; (7)$

Solving quadratic equation $(7)$ gives $\;$ $u_3 = 10$ $\;$ or $\;$ $u_3 = 18$

Speed of $C_3$ has to be greater than the speeds of $C_1$ and $C_2$ for $C_3$ to overtake them.

$\implies$ $u_3 = 10$ is not an acceptable solution.

$\therefore \;$ Speed of cyclist $C_3$ $= u_3 = 18$ kmph