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 $= t_{12} = ar^{11} = \dfrac{3}{2} \times 2^{11} = 3072$
$\dfrac{t_8}{t_{12}} = \dfrac{192}{3072} = \dfrac{1}{16}$
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}
$
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 $= t_7 = ar^6 = \dfrac{1}{9} \times 3^6 = 81$
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} $
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} $
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} $
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} $