Determinants

Using properties of determinants prove that $ \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} = \left(a+b+c\right) \left(a^2 + b^2 + c^2\right) $

$ \begin{aligned} \text{Let } \Delta & = \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow aC_1 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2 & b-c & c+b \\ a^2 + ac & b & c-a \\ a^2 - ab & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow C_1 + bC_2 + cC_3 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2-bc+c^2+bc & b-c & c+b \\ a^2+ac+b^2+c^2-ac & b & c-a \\ a^2-ab+b^2+ab+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2+c^2 & b-c & c+b \\ a^2+b^2+c^2 & b & c-a \\ a^2+b^2+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_2 \rightarrow R_2-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_3 \rightarrow R_3-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix} \\ & \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left[-bc+\left(a+b\right)\left(a+c\right)\right]}{a} \left[\text{Note: Expanding along }C_1\right] \\ & = \dfrac{\left(a^2+b^2+c^2\right)\left(a^2+ab+ac+bc-bc\right)}{a} \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left(a^2+ab+ac\right)}{a} \\ & = \left(a^2+b^2+c^2\right) \left(a+b+c\right) \end{aligned} $

Sequences and Series

The sum of three consecutive numbers of a G.P is 56. If we subtract 1, 7 and 21 from these numbers in order, the resulting numbers form an A.P. Find the numbers.

Let the three consecutive numbers in G.P be $\dfrac{a}{r}$, a, $ar$
where first term $= a$; common ratio $= r$
$\therefore$ $\dfrac{a}{r}+a+ar = 56$ $\implies$ $\dfrac{a}{r} + ar = 56 - a$ $\cdots$ (1)
On subtracting 1, 7 and 21 from these numbers in order gives $\dfrac{a}{r}-1$, $a-7$ and $ar-21$
Then, $\dfrac{a}{r}-1$, $a-7$, $ar-21$ are in A.P
Common difference $= a-7-\left(\dfrac{a}{r} - 1\right) = ar - 21 - \left(a-7\right)$
i.e. $a-\dfrac{a}{r}-6=ar-a-14$
i.e. $\dfrac{a}{r}+ar-2a= 8$
i.e. $\dfrac{a}{r} + ar = 8 + 2a$ $\cdots$ (2)
$\therefore$ From equations (1) and (2) we have
$56-a = 8 + 2a$ i.e. $3a = 48$ $\implies$ $a=16$
Therefore, from equation (2) we have,
$\dfrac{16}{r} + 16r = 8 + 32 = 40$
i.e. $16r^2 -40r + 16 = 0$
i.e. $2r^2 -5r+2 = 0$
i.e. $\left(2r-1\right) \left(r-2\right) = 0$
i.e. $r = 2$ or $r = \dfrac{1}{2}$
Therefore, when $a=16$, $r=2$, the numbers are $\dfrac{16}{2}$, 16, $16\times2$ i.e. 8, 16, 32
When $a=16$, $r = \dfrac{1}{2}$, the numbers are $\dfrac{16}{1/2}$, 16, $16 \times \dfrac{1}{2}$ i.e. 32, 16, 8

Sequences and Series

Two arithmetic progressions have the same common difference. The first term of one A.P is $-5$ and that of the other is $-9$. Find the difference between their $7^{th}$ terms.

Let common difference $= d$
Case 1: First term $= a = -5$
$\therefore$ $7^{th}$ term $= t_7 = a + 6d = -5 + 6d$
Case 2: First term $= A = -9$
$\therefore$ $7^{th}$ term $= T_7 = A + 6d = -9 + 6d$
$\therefore$ Difference between the two $7^{th}$ terms $= t_7 - T_7 = -5 + 6d - \left(-9+6d\right) = -5+6d+9-6d = 4$

Sequences and Series

If $\left(p+q\right)^{th}$ term and $\left(p-q\right)^{th}$ terms of a G.P are a and b respectively, prove that the $p^{th}$ term is $\sqrt{ab}$.

Let the first term of G.P $= A$ and common ratio $= R$
$n^{th}$ term of G.P $= t_n = AR^{n-1}$
$\therefore$ $\left(p+q\right)^{th}$ term $= t_{p+q} = AR^{p+q-1} = a$ (given) $\cdots$ (1)
and $\left(p-q\right)^{th}$ term $= t_{p-q} = AR^{p-q-1} = b$ (given) $\cdots$ (2)
$\therefore$ $\dfrac{t_{p+q}}{t_{p-q}} = \dfrac{AR^{p+q-1}}{AR^{p-q-1}} = \dfrac{a}{b}$
i.e. $R^{p+q-1-p+q+1} = \dfrac{a}{b}$
i.e. $R^{2q} = \dfrac{a}{b}$ $\implies$ $R = \sqrt[2q]{\dfrac{a}{b}} = \left(\dfrac{a}{b}\right)^{\frac{1}{2q}}$ $\cdots$ (3)
$\therefore$ From equations (1) and (3) we have
$A\left[\left(\dfrac{a}{b}\right)^{\frac{1}{2q}}\right]^{p+q-1} = a$
i.e. $A \left(\dfrac{a}{b}\right)^{\frac{p+q-1}{2q}} = a$
$\implies$ $A = a \times \left(\dfrac{b}{a}\right)^{\frac{p+q-1}{2q}}$
i.e. $A = a^{1-\frac{\left(p+q-1\right)}{2q}} b^{\frac{p+q-1}{2q}} = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}}$
$ \begin{aligned} \therefore p^{th} \text{term} = t_p & = AR^{p-1} \\ & = a^{\frac{1-p+q}{2q}} b^{\frac{p+q-1}{2q}} \times \left(\dfrac{a}{b}\right)^{\frac{p-1}{2q}} \\ & = a^{\frac{1-p+q+p-1}{2q}} b^{\frac{p+q-1}{2q} - \frac{p-1}{2q}} \\ & = a^{\frac{1}{2}} b^{\frac{1}{2}} = \sqrt{ab} \end{aligned} $

Sequences and Series

Determine the $12^{th}$ term of a G.P whose $8^{th}$ term is 192 and common ratio is 2. Also find $t_8 : t_{12}$

$8^{th}$ term of G.P $= t_8 = ar^7 = 192$
Common ratio $= r = 2$
$\therefore$ $a \times 2^7 = 192$
i.e. $a = \dfrac{192}{128} = \dfrac{3}{2}$
$\therefore$ $12^{th}$ term $= t_{12} = ar^{11} = \dfrac{3}{2} \times 2^{11} = 3072$
$\dfrac{t_8}{t_{12}} = \dfrac{192}{3072} = \dfrac{1}{16}$

Sequences and Series

Find the $n^{th}$ term and deduce the sum to n terms of the series $4+11+22+37+56+\cdots$

$ \begin{aligned} \text{Let } S_n & = & 4 + & 11 + 22 + 37 + 56 + \cdots + t_{n-1} + t_n \\ & & & \text{Shifting every term one place to the right} \\ S_n & = & & 4 + 11 + 22 + 37 + \cdots \cdots \cdots + t_{n-1} + t_n \end{aligned} $
Subtracting, we get $ \begin{aligned} 0 & = \left[4 + 7 + 11 + 15 + 19 + \cdots \text{to n terms}\right] - t_n \\ \text{i.e. } t_n & = 4 + \left[7 + 11 + 15 + 19 + \cdots \text{to } \left(n-1\right) \text{ terms}\right] \\ & = 4 + \left(\dfrac{n-1}{2}\right)\left[2 \times 7 + \left(n-2\right) \times 4\right] \\ & \left[\text{Note: Sum to n terms of an A.P } = \dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\right] \\ & \left[\text{Here, first term }= a = 7 \; ; \text{common difference }= d = 4 \right] \\ & = 4 + \dfrac{\left(n-1\right) \left(4n+6\right)}{2} \\ & = 4 + \dfrac{4n^2 + 2n - 6}{2} \\ & = 4 + 2n^2 + n - 3 \\ \text{i.e. } t_n & = 2n^2 + n + 1 \end{aligned} $ $ \begin{aligned} \therefore \text{ Sum to n terms} = S_n & = 2 \sum_{k = 1}^{n} k^2 + \sum_{k=1}^{n} k + n \\ & = 2 \times \dfrac{n \left(n+1\right) \left(2n+1\right)}{6} + \dfrac{n \left(n+1\right)}{2} + n \\ & = \dfrac{n}{3} \left(2n^2 + 3n + 1\right) + \dfrac{n \left(n+1\right)}{2} + n \\ & = \dfrac{n}{6} \left(4n^2 + 6n + 2 + 3n + 3 + 6\right) \\ \therefore S_n & = \dfrac{n}{6} \left(4n^2 + 9n + 11\right) \end{aligned} $

Sequences and Series

The product of the third and the eighth terms of a G.P is 243. If the fourth term is 3, find its seventh term.

Third term of G.P $= t_3 = ar^2$
Eighth term of G.P $= t_8 = ar^7$
$\therefore$ $ar^2 \times ar^7 = 243$ $\implies$ $a^2 r^9 = 243$ $\cdots$ (1)
Fourth term of G.P $= t_4 = ar^3 = 3$ $\cdots$ (2)
Now, equation (1) can be rewritten as
$ar^3 \times ar^3 \times r^3 = 243$ $\cdots$ (3)
In view of equation (2), equation (3) becomes
$3 \times 3 \times r^3 = 243$ $\implies$ $r^3 = \dfrac{243}{9} = 27$ $\implies$ $r = \sqrt[3]{27} = 3$
Substituting the value of r in equation (2) gives
$a \times 3^3 = 3$ $\implies$ $a = \dfrac{1}{9}$
$\therefore$ Seventh term $= t_7 = ar^6 = \dfrac{1}{9} \times 3^6 = 81$

Trigonometric Functions

Show that $\dfrac{1}{2}\left(\cos 4x + \cos 2x\right) = \cos 3x \cos x$

$\cos 4x = \cos \left(3x + x\right)$ and $\cos 2x = \left(3x - x\right)$ $\cos \left(3x + x\right) = \cos 3x \cos x - \sin 3x \sin x$ $\cos \left(3x - x\right) = \cos 3x \cos x + \sin 3x \sin x$ $ \begin{aligned} \therefore \dfrac{1}{2}\left(\cos 4x + \cos 2x\right) & = \dfrac{1}{2} \left[\cos \left(3x + x\right) + \cos \left(3x - x\right)\right] \\ & = \dfrac{1}{2} \left(\cos 3x \cos x - \sin 3x \sin x + \cos 3x \cos x + \sin 3x \sin x\right) \\ & = \dfrac{1}{2} \times 2 \cos 3x \cos x \\ & = \cos 3x \cos x = \text{RHS} \end{aligned} $

Trigonometric Functions

Prove that $\left(1+\cot \theta - \text{cosec} \theta\right) \left(1+\tan \theta + \sec \theta\right) = 2$

$ \begin{aligned} \text{LHS } & = \left(1+\cot \theta - \text{cosec} \theta\right) \left(1+\tan \theta + \sec \theta\right) \\ & = \left(1 + \dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta}\right) \left(1 + \dfrac{\sin \theta}{\cos \theta} + \dfrac{1}{\cos \theta}\right) \\ & = \dfrac{\left(\sin \theta + \cos \theta - 1\right) \left(\cos \theta + \sin \theta + 1\right)}{\sin \theta \cos \theta} \\ & = \dfrac{\left(\sin \theta + \cos \theta\right)^2 - 1^2}{\sin \theta \cos \theta} \\ & = \dfrac{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \\ & = \dfrac{1 + 2 \sin \theta \cos \theta -1}{\sin \theta \cos \theta} \\ & = \dfrac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} \\ & = 2 = \text{RHS} \end{aligned} $

Trigonometric Functions

Prove that $\sqrt{\sec^2 \theta + \text{cosec}^2 \theta} = \tan \theta + \cot \theta$

$ \begin{aligned} \text{LHS} = \sqrt{\sec^2 \theta + \text{cosec}^2 \theta} & = \sqrt{\left(1+\tan^2 \theta\right) + \left(1 + \cot^2 \theta\right)} \\ & \left[\text{Note: }1+\tan^2 \theta = \sec^2 \theta; 1+\cot^2 \theta = \text{cosec}^2 \theta\right] \\ & = \sqrt{\tan^2 \theta + 2 + \cot^2 \theta} \\ & = \sqrt{\tan^2 \theta + 2 \tan \theta \cot \theta + \cot^2 \theta} \\ & \left[\text{Note: } \tan \theta \times \cot \theta = 1 \right] \\ & = \sqrt{\left(\tan \theta + \cot \theta\right)^2} \\ & = \tan \theta + \cot \theta = \text{RHS} \end{aligned} $

Trigonometric Functions

Show that $\left(\text{cosec}\theta - \sin \theta\right) \left(\sec\theta - \cos\theta\right) \left(\tan\theta + \cot\theta\right) = 1$

$ \begin{aligned} \text{LHS} & = \left(\text{cosec}\theta - \sin \theta\right) \left(\sec\theta - \cos\theta\right) \left(\tan\theta + \cot\theta\right) \\ & = \left(\dfrac{1}{\sin\theta} - \sin\theta\right) \left(\dfrac{1}{\cos\theta} - \cos\theta\right) \left(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}\right) \\ & = \left(\dfrac{1 - \sin^2 \theta}{\sin\theta}\right) \left(\dfrac{1 - \cos^2 \theta}{\cos\theta}\right) \left(\dfrac{\sin^2 \theta + \cos^2 \theta}{\sin\theta \cos\theta}\right) \\ & = \dfrac{\cos^2 \theta}{\sin\theta} \times \dfrac{\sin^2 \theta}{\cos\theta} \times \dfrac{1}{\sin\theta \cos\theta} \\ & = 1 \\ & = \text{RHS} \end{aligned} $