MCQ 13 Marks
If in $\text{a}\triangle\text{ABC},\tan\text{B}+\tan\text{C}=6,$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
AnswerCorrect option: C. $\frac16$
In $\triangle\text{ABC},$
$\text{A+B+C}=\pi$
We know that $\tan(\text{A+B+C)}=\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}}{1-\tan\text{A}\tan\text{B}-\tan\text{B}\tan\text{C}-\tan\text{C}\tan\text{A}}$ and $\tan\pi=0.$
$\therefore\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
If $\tan\text{A}+\tan\text{B}+\tan\text{C}=6,\tan\text{A}\tan\text{B}\tan\text{C}=6$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac16$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}=\frac16$
View full question & answer→MCQ 23 Marks
If $\tan\theta_1\tan\theta_2=\text{k},$ then $\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=$
- ✓
$\frac{1+\text{k}}{1-\text{k}}$
- B
$\frac{1-\text{k}}{1+\text{k}}$
- C
$\frac{\text{k}+1}{\text{k}-1}$
- D
$\frac{\text{k}-1}{\text{k}+1}$
AnswerCorrect option: A. $\frac{1+\text{k}}{1-\text{k}}$
$\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}$
$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$
Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:
$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$
$=\frac{1+\text{k}}{1-\text{k}}$
View full question & answer→MCQ 33 Marks
If $\text{A}-\text{B}=\frac\pi4,$ then $(1+\tan\text{A})(1-\tan\text{B})$ is equal to:
Answer$\tan(\text{A - B})=\tan\frac\pi4$
$\Rightarrow\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}=1$
$\Rightarrow\tan\text{A}-\tan\text{B}=1+\tan\text{A}\tan\text{B}\cdots(1)$
Now,
$(1+\tan\text{A})(1-\tan\text{B})=1+\tan\text{A}-\tan\text{B}-\tan\text{A}\tan\text{B}$
$=1+1+\tan\text{A}\tan\text{B}-\tan\text{A}\tan\text{B}$
$=2$
View full question & answer→MCQ 43 Marks
The value of $\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\sin^2\Big(\frac\pi6-\text{x}\Big)$ is:
AnswerCorrect option: A. $\frac{1}{2}\cos2\text{x}$
$\cos^2\Big(\frac\pi6+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$
$=\cos\Big(\frac{\pi}{6}+\text{x}+\frac{\pi}{6})-\text{x}\Big)\cos\Big(\frac{\pi}{6}+\text{x}-\frac{\pi}{6}+\text{x}\Big)$ $\Big[\text{Using}\cos(\text{A+B})\cos(\text{A-B})=\cos^2\text{A}-\sin^2\text{B}\Big]$
$=\cos\frac{2\pi}{6}\cos2\text{x}$
$=\frac12\cos2\text{x}$ $\Big[\text{As}\cos\frac\pi3=\frac12\Big]$
View full question & answer→MCQ 53 Marks
If $\tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=2\text{k},$ then $k =$
- A
$-1$
- B
$\frac12$
- ✓
$\frac{-1}{2}$
- D
AnswerCorrect option: C. $\frac{-1}{2}$
$\tan135^\circ=\tan(90^\circ+45^\circ)$
$=-\tan45^\circ$
$=-1$
Or, $\tan(69^\circ+66^\circ)=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$
$\Rightarrow-1=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$
$\Rightarrow\tan69^\circ+\tan66^\circ-\tan69^\circ+\tan66^\circ=-1$
$\therefore2\text{k}=-1$
$\Rightarrow\text{k}=\frac{-1}{2}$
View full question & answer→MCQ 63 Marks
If $\cot(\alpha+\beta)=0,$ then $\sin(\alpha+2\beta)$ is equal to:
- ✓
$\sin\alpha$
- B
$\cos2\beta$
- C
$\cos\alpha$
- D
$\sin2\alpha$
AnswerCorrect option: A. $\sin\alpha$
Given:
$\cot(\alpha+\beta)=0$
$\Rightarrow\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=0$
$\Rightarrow\cos(\alpha+\beta)=0$
$\Rightarrow\alpha+\beta=\frac\pi2$
$\therefore\sin(\alpha+2\beta)=\sin(\alpha+\alpha+\beta)$
$=\sin\alpha$
View full question & answer→MCQ 73 Marks
If $\text{A+B+C}=\pi,$ then $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$ is equal to:
Answer$\pi=180^\circ$
Using $\tan(180^\circ-\text{A})=-\tan\text{A},$ we get:
$\text{C}=\pi-(\text{A+B})$
Now, $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan[\pi-\text{(A+B)}]}{\tan\text{A}\tan\text{B}\tan[\pi-\text{(A+B)}]}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan\text{(A+B)}}{-\tan\text{A}\tan\text{B}\tan\text{(A+B)}}$
$=\frac{\tan\text{A}+\tan\text{B}-\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}{-\tan\text{A}\tan\text{B}\times\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}-\tan\text{A}-\tan\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=\frac{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=1$
View full question & answer→MCQ 83 Marks
If $\tan(\text{A}-\text{B})=1,\sec(\text{A+B})=\frac{2}{\sqrt{3}},$ then the smallest positive value of $B$ is:
- A
$\frac{25\pi}{24}$
- ✓
$\frac{19\pi}{24}$
- C
$\frac{13\pi}{24}$
- D
$\frac{11\pi}{24}$
AnswerCorrect option: B. $\frac{19\pi}{24}$
Given:
$\tan(\text{A - B})=1$ and $\sec(\text{A+B})=\frac{2}{\sqrt{3}}$
$\Rightarrow\text{A - B}=\frac{\pi}{4}\cdots(1)$ and $\text{A + B}=\frac{\pi}{4}\cdots(2)$
Adding these equations we get:
$2\text{A}=\frac{\pi}{4}+\frac\pi6$
$\Rightarrow\text{A}=\frac{5\pi}{24}$
$\Rightarrow$ Smallest possible value of $\text{B}=\pi-\frac{5\pi}{24}=\frac{19\pi}{24}.$
View full question & answer→MCQ 93 Marks
The value of $\sin^2\frac{5\pi}{12}-\sin^2\frac{\pi}{12}$ is:
- A
$\frac12$
- ✓
$\frac{\sqrt{3}}{2}$
- C
$1$
- D
$0$
AnswerCorrect option: B. $\frac{\sqrt{3}}{2}$
$\frac{5\pi}{12}=75^\circ,\frac{\pi}{12}=15^\circ$
$\sin^275^\circ-\sin^215^\circ$
$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.
Now, $\sin75^\circ=\sin(45^\circ+30^\circ)$
$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\cos75^\circ=\cos(45^\circ+30^\circ)$
$=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Hence,
$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$
$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8}$
$=\frac{4\sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 103 Marks
If $\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ$ is equal to:
- A
$\frac{\sqrt{3}}{4}$
- B
$\frac{\sqrt{3}}{2}$
- ✓
$\sqrt{3}$
- D
$1$
AnswerCorrect option: C. $\sqrt{3}$
$\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ$
$=\tan60^\circ(1-\tan20^\circ\tan40^\circ)+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ-\tan60^\circ\tan20^\circ\tan40^\circ+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ$
$=\sqrt{3}$
View full question & answer→MCQ 113 Marks
The maximum value of $\sin^2\Big(\frac{2\pi}{3}+\text{x}+\sin^2\Big(\frac{2\pi}{3}-\text{x}\Big)$ is:
- A
$\frac12$
- ✓
$\frac32$
- C
$\frac14$
- D
$\frac34$
AnswerCorrect option: B. $\frac32$
$\frac{2\pi}{3}=120^\circ$
Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$
$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$
$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$
$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$
$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$
$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$
$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$
$=\frac32-\sin^2\text{x}$.
For $f(x)$ to be maximum, $\sin^2\text{x}$ must have minimum value, which is $0.$
$\therefore\frac32$ is the maximum value of $f(x).$
View full question & answer→MCQ 123 Marks
If $\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x}),$ then $\sin2\text{x}=$
- A
$\pm\frac34$
- B
$\pm\frac43$
- ✓
$\pm\frac13$
- D
AnswerCorrect option: C. $\pm\frac13$
$\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x})$
As we know that $\sin\text{x}=-\cos\Big(\frac\pi2+\text{x}\Big)$
$\Rightarrow-\cos\Big(\frac\pi2+\pi\cos\text{x}\Big)=\cos(\pi\sin\text{x})$
$\Rightarrow\frac{-\pi}{2}-\pi\cos\text{x}=\pi\sin\text{x}$
$\Rightarrow\pi\sin\text{x}-\pi\cos\text{x}=\frac12$
$\Rightarrow\sin\text{x}-\cos\text{x}=\frac12$
Squaring both sides we get,
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\frac14$
$\Rightarrow1+\sin2\text{x}=\frac{1}{4}$
$\Rightarrow\sin2\text{x}=\frac13$
$\therefore\sin2\text{x}=\pm\frac13$
View full question & answer→MCQ 133 Marks
If $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1},$ then $\alpha+\beta$ is equal to:
- A
$\frac\pi2$
- B
$\frac\pi3$
- C
$\frac\pi6$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1}.$
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{\text{x}}{\text{x}+1}+\frac{1}{2\text{x+1}}}{1-\frac{\text{x}}{\text{x}+1}\times\frac{1}{2\text{x+1}}}$
$=\frac{\frac{\text{x}(2\text{x}+1)+(\text{x}+1)}{(\text{x}+1)(2\text{x}+1)}}{\frac{(\text{x}+1)(2\text{x}+1)-\text{x}}{(\text{x}+1)(2\text{x}+1)}}$
$=\frac{2\text{x}^2+\text{x}+\text{x}+1}{2\text{x}^2+3\text{x}+1+\text{x}}$
$=\frac{2\text{x}^2+2\text{x}+1}{2\text{x}^2+2\text{x}+1}$
$=1$
$\therefore\alpha+\beta=\frac\pi4\ (\tan\frac\pi4=1)$
Hence, the correct answer is option $D.$
View full question & answer→MCQ 143 Marks
If $\cos\text{P}=\frac{1}{7}$ then $\cos\text{Q}=\frac{13}{14},$ where $P$ and $Q$ both are acute angles. Then, the value of $P - Q$ is:
- A
$\frac{\pi}{6}$
- ✓
$\frac\pi3$
- C
$\frac\pi4$
- D
$\frac{5\pi}{12}$
AnswerCorrect option: B. $\frac\pi3$
$\cos\text{P}=\frac17,\cos\text{Q}=\frac{13}{14}$
$\therefore\sin\text{P}=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt{3}}{7}$ and $\sin\text{Q}=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt{3}}{14}$
Hence, $\tan\text{P}=4\sqrt{3},\tan\text{Q}=\frac{3\sqrt{3}}{13}$
$\cos(\text{P - Q})=\cos\text{P}\cos\text{Q}+\sin\text{P}\sin\text{Q}$
$=\frac{1}{7}\times\frac{13}{14}+\frac{4\sqrt{3}}{7}\times\frac{3\sqrt{3}}{14}$
$=\frac{13+36}{98}$
$=\frac{49}{98}$
$\therefore\cos(\text{P - Q})=\frac12$
$\Rightarrow\text{P - Q}=\cos^{-1}\frac12$
$\Rightarrow\text{P - Q}=60^\circ$
Hence, the correct answer is option $B.$
View full question & answer→MCQ 153 Marks
$\tan3\text{A}-\tan2\text{A}-\tan\text{A}$ is equal to:
- ✓
$\tan3\text{A}\tan2\text{A}\tan\text{A}$
- B
$-\tan3\text{A}-\tan2\text{A}\tan\text{A}$
- C
$\tan\text{A}\tan2\text{A}\tan2\text{A}\tan3\text{A}-\tan3\text{A}\tan\text{A}$
- D
AnswerCorrect option: A. $\tan3\text{A}\tan2\text{A}\tan\text{A}$
$3\text{A}=2\text{A}+\text{A}$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})=\frac{\tan2\text{A}+\tan\text{A}}{1-\tan2\text{A}\tan\text{A}}$
$\Rightarrow\tan3\text{A}-\tan3\text{A}\tan2\text{A}\tan\text{A}=\tan2\text{A}+\tan\text{A}$
$\Rightarrow\tan3\text{A}-\tan2\text{A}-\tan\text{A}=\tan3\text{A}\tan2\text{A}\tan\text{A}$
View full question & answer→MCQ 163 Marks
If $\tan\text{A}=\frac{\text{a}}{\text{a}+1}$ and $\text{B}=\frac{1}{2\text{a}+1},$ then the value of $A + B$ is:
- A
$0$
- B
$\frac{\pi}{2}$
- C
$\frac\pi3$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
$\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$
$=\frac{\frac{\text{a}}{\text{a}+1}+\frac{1}{2\text{a}+1}}{1-\frac{\text{a}}{(\text{a}+1)(2\text{a}+1)}}$
$=\frac{2\text{a}^2+\text{a}+\text{a}+1}{2\text{a}^2+3\text{a}+1-\text{a}}$
$=\frac{2\text{a}^2+2\text{a}+1}{2\text{a}^2+2\text{a}+1}$
$=1$
$\therefore \text{ A+B}=\tan^{-1}(1)=\frac\pi4.$
View full question & answer→MCQ 173 Marks
If $\cos(\text{A}-\text{B})=\frac35$ and $\tan\text{A}\tan\text{B}=2,$ then
- ✓
$\cos\text{A}\cos\text{B}=\frac15$
- B
$\cos\text{A}\cos\text{B}=-\frac15$
- C
$\sin\text{A}\sin\text{B}=-\frac15$
- D
$\sin\text{A}\sin\text{B}=-\frac15$
AnswerCorrect option: A. $\cos\text{A}\cos\text{B}=\frac15$
$\tan\text{A}\tan\text{B}=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=2\ (\text{Given})\cdots(1)$
Also,
$\cos(\text{A - B})=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac35+\sin\text{A}\sin\text{A}=\frac35$
$\therefore\sin\text{A}\sin\text{B}=\frac35-\cos\text{A}\cos\text{B}\cdots(2)$
Substituting eq $(2)$ in eq $(1),$ we get:
$\Rightarrow\frac{\frac35-\cos\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}=2$
$\Rightarrow3\cos\text{A}\cos\text{B}=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac15$
View full question & answer→MCQ 183 Marks
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$ is equal to:
- ✓
$\tan55^\circ$
- B
$\cot55^\circ$
- C
$-\tan35^\circ$
- D
$-\cot35^\circ$
AnswerCorrect option: A. $\tan55^\circ$
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$
$=\frac{1+\tan10^\circ}{1-\tan10^\circ}$ $[$Dividing the numerator and denominator by $\cos10^\circ]$
$=\frac{\tan45^\circ+\tan10^\circ}{1-\tan45^\circ\times\tan10^\circ}$
$=\tan(45^\circ+10^\circ)$ $\Big[$Using $\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$=\tan55^\circ$
View full question & answer→MCQ 193 Marks
If $3\sin\text{x}+4\cos\text{x}=5,$ then $4\sin\text{x}-3\cos\text{x}=$
Answer$3\sin\text{x}+4\cos\text{x}=5$
$\frac{3}{5}\sin\text{x}+\frac{4}{5}\cos\text{x}=1$
Let $\cos\alpha=\frac35$ and $\sin\alpha=\frac45.$
$\therefore\cos\alpha\sin\text{x}+\sin\alpha\cos\text{x}=1$
$\Rightarrow\sin(\alpha+\text{x})=\sin\frac\pi2$
$\Rightarrow\alpha+\text{x}=\frac\pi2$
$\Rightarrow\text{x}=\frac{\pi}{2}-\alpha\cdots(1)$
We have to find the value of $4\sin\text{x}-3\cos\text{x}.$
$4\sin\Big(\frac\pi2-\alpha\Big)-3\cos\Big(\frac\pi3-\alpha\Big) ...\{$From eq $(1)\}$
$=4\cos\alpha-3\sin\alpha$
$=4\times\frac35-3\times\frac45$ $\Big(\because\cos\alpha=\frac35$ and $\sin\alpha=\frac{4}{5}\Big)$
$=0$
View full question & answer→MCQ 203 Marks
If $\tan\theta=\frac12$ and $\tan\phi=\frac13,$ then the value of $\theta+\phi$ is:
- A
$\frac\pi6$
- B
$\pi$
- C
$0$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\theta=\frac12$ and $\tan\phi=\frac13.$
Now,
$\tan(\theta+\phi)=\frac{\tan+\tan\phi}{1-\tan\theta\tan\phi}$
$=\frac{\frac12+\frac13}{1-\frac12\times\frac13}$
$=\frac{\frac{5}{6}}{\frac56}$
$=1$
$\therefore\theta+\phi=\frac\pi4$ $\Big(\tan\frac\pi4=1\Big)$
Hence, the correct answer is option $D.$
View full question & answer→MCQ 213 Marks
The value of $\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ+\text{A})\cos(54^\circ-\text{A})$ is:
- A
$\sin2\text{A}$
- ✓
$\cos2\text{A}$
- C
$\cos3\text{A}$
- D
$\sin3\text{A}$
AnswerCorrect option: B. $\cos2\text{A}$
$\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ+\text{A})\cos(54^\circ-\text{A})$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin[90^\circ-(54^\circ+\text{A})]\sin[90^\circ-(54^\circ-\text{A})]$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin(36^\circ-\text{A})\sin(36^\circ-\text{A})$
$=\cos(36^\circ-\text{A}-36^\circ+\text{A})$
$[$ Using $\cos(\text{A - B)}$ formula $]$
$=\cos2\text{A}$
View full question & answer→MCQ 223 Marks
If $\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a},$ then $\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)=$
- A
$a^2 + 1$
- B
$a^2 + 2$
- ✓
$a^2 - 2$
- D
AnswerCorrect option: C. $a^2 - 2$
Given:
$\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}$
$\Rightarrow\Big[\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)\Big]^2=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)+2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{\tan45^\circ-\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\times\frac{\tan45^\circ+\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{1^\circ-\tan\text{x}}{1+\tan\text{x}}\times\frac{1^\circ+\tan\text{x}}{1-\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big(\frac{1-\tan^2\text{x}}{1-\tan^2\text{x}}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2$
View full question & answer→MCQ 233 Marks
If $\text{A+B+C}=\pi,$ then $\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})$ is equal to:
Answer$\pi=180^\circ$
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$
We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$
$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$
Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}\cos\text{B}^2+\cos\text{B}\sin\text{A}\sin\text{B}-\sin\text{B}\sin\text{A}\cos\text{B}-\sin^2\cos\text{A}}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}(\cos^2\text{B}+\sin^2\text{B})}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$
View full question & answer→Question 243 Marks
Find the sum of odd integers from 1 to 2001.
AnswerLet the number of terms is n.
Now the sum of the series is:
1 + 3 + 5 + ... + 2001
Here,
$\text{l}=2001$ and $\text{d}=2$
Therefore,
$\text{l}=\text{a}+(\text{n}-1)\text{d}$
$2001=1+(\text{n}-1)\text{d}$
$2(\text{n}-1)=2000$
$\text{n}-1=1000$
$\text{n}=1001$
Therefore the sum of the series is:
$\text{s}=\frac{1001}{2}[2+(1001-1)2]$
$=1001^2$
$=10021001$
View full question & answer→Question 253 Marks
In an A.P. the first term is $2$ and the sum of the first five terms is one fourth of the next five terms. Show that $20\ th$ term is $−112$.
AnswerHere the first term a $= 2.$
Let the common difference is d.
Now,
$\frac{5}{2}[2\text{a}+(5-1)\text{d}]=\frac{1}{4}\Big[\frac{5}{2}[2(\text{a}+\text{d}5)+(5-1)\text{d}]\Big]$
$\frac{5}{2}[2.2+4\text{d}]=\frac{5}{8}[2.2+14\text{d}]$
$10+10\text{d}=\frac{5}{2}+\frac{35}{4}\text{d}$
$\frac{5}{4}\text{d}=-7.5$
$\text{d}=-6$
The $20^{th}$ term will be:
$\text{a}+(\text{n}-1)\text{d}=2+(20-1)(-6)$
$=-112$
Hence it is shown.
View full question & answer→Question 263 Marks
In a cricket team tournament 16 teams participated. A sum of ₹ 8000 is to be awarded among themselves as prize money. If the last place team is awarded ₹ 275 in prize money and the award increases by the same amount for successive finishing places, then how much amount will the first place team receive?
AnswerSuppose the award increases by ₹ x.$\text{d}=\text{x}$
In cricket team tournament 16 teams participated. $\text{n}=16$ The last place team is awarded ₹ 275 in prize money$\text{a}_1=275$
Sum of ₹ 8000 is to be awarded as prize money$\text{S}=8000$
$\frac{16}{2}[\text{a}_1+\text{a}_1+(16-1)\times]=8000$
$2\text{a}_1+15\text{x=1000}$
$15\text{x}=450$
$\text{x}=30$
The amount received by first place team$=\text{a}_{16}$
$=\text{a}_1+(16-1)\text{d}$
$=275+15\times30$
$=725$
The amount received by first place team is ₹ 725.
View full question & answer→Question 273 Marks
In an A.P., show that $\text{a}_{\text{m}+\text{n}}+\text{a}_{\text{m}-\text{n}}=2\text{a}_\text{m}.$
AnswerIt is given that the sequence $<\text{a}_\text{n}>$ is an A.P.
$\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}\ .....(1)$
Similarly from (1)
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}+\text{n}-1)\text{d}\ .....(2)$
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}-\text{n}-1)\text{d}\ .....(3)$
Adding (2) and (3)
$\text{a}_{\text{m+n}}+\text{a}_{\text{m}-\text{n}}=(\text{a}+(\text{m+n}-1)\text{d})(\text{a}+(\text{m}-\text{n}-1)\text{d})$
$=2\text{a}+(\text{m+n}-1+\text{m}-\text{n}-1)\text{d}$
$=\text{2a}+\text{2d}(\text{m}-1)$
$=2(\text{a}+(\text{m}-1)\text{d})$
$=\text{2a}_\text{m}$
Hence proved.
View full question & answer→Question 283 Marks
If in A.P. is such that $\frac{\text{a}_4}{\text{a}_7}=\frac{2}{3},$ find $\frac{\text{a}_6}{\text{a}_8}.$
Answer$\frac{\text{a}_4}{\text{a}_7}=\frac{2}{3}$ [Given]
$\Rightarrow\frac{\text{a}+3\text{d}}{\text{a}+6\text{d}}=\frac{2}{3}$
$\Rightarrow3\text{a}+9\text{d}=2\text{a}+12\text{d}$
$\Rightarrow\text{a}=3\text{d}\ .....(1)$
$\frac{\text{a}_6}{\text{a}_8}=\frac{\text{a}+5\text{d}}{\text{a}+7\text{d}}$ $[\because3\text{d from}(1)]$
$\Rightarrow\frac{8\text{d}}{10\text{d}}$
$\Rightarrow\frac{4}{5}$
$\frac{\text{a}_6}{\text{a}_8}=\frac{4}{5}$
View full question & answer→Question 293 Marks
We know that the sum of the interior angles of a triangle is $180^\circ $. Show that the sums of the interior angles of polygons with $3, 4, 5, 6, ...$ sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
AnswerWe know that sum of interior angles of a polygon with n sides is given by,
$\text{a}_\text{n}=180^\circ(\text{n}-2)$
Sum of interior angles of a polygon with 3 sides is given by,
$\text{a}_3=180^\circ(3-2)=180^\circ\ .....(1)$
Sum of interior angles of a polygon with 7 sides is given by,
$\text{a}_4=180^\circ(4-2)=360^\circ\ .....(2)$
Sum of interior angles of a polygon with 5 sides is given by,
$\text{a}_5=180^\circ(5-2)=540^\circ\ .....(3)$
From $eq^n (2),\ eq^n (2)$ and $eq^n\ (3)$ we get,
$\text{a}_4=360^\circ=180^\circ+180^\circ=\text{a}_4+180^\circ=\text{a}_4+\text{d}$
$\text{a}_5=540^\circ=180^\circ+360^\circ=\text{a}_5+2\text{d}$
Hence the sums of the interior angles of polygons with $3, 4, 5, 6, ... $ sides form an arithmetic progression.
Sum of interior angles of $21$ sided polygon
$=180^\circ(21-2)$
$=3420^\circ$
View full question & answer→Question 303 Marks
A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.
AnswerLet the man save ₹ 200 in n numbers of year.
Then,
$\text{ATQ}$
$32+36+40+\ ...\ +=200$
It rorms a series of n terms, with $\text{a}=32$ and $\text{d}=4$
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}(\text{n}-1)\text{d}]$
$\Rightarrow200=\frac{\text{n}}{2}[2(32)+(\text{n}-1)4]$
$\Rightarrow400=60\text{n}+4\text{n}^2$
$\Rightarrow\text{n}^2+15\text{n}-100=0$
$\Rightarrow\text{n}=5$ or $-20$ [It can't be negative]
$\therefore\text{n}=5$
The man will save ₹ 200 in year
View full question & answer→Question 313 Marks
The sum of first 7 terms of an A.P. is 10 and that next 7 terms is 17. find the progression.
AnswerGiven,
$\text{a}_7=10$
$\text{s}_{14}-\text{s}_7=17\ .....{(1)}$
$\therefore\text{s}_{14}=17+\text{s}_7=17+10=27\ .....{(2)}$
From (1) and (2)
$\text{s}_7=\frac{7}{2}[2\text{a}+(7-1)\text{d}]$ $\Big[$Using $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\Big]$
$\Rightarrow10=7\text{a}+21\text{d}\ .....(3)$
and
$\text{s}_{14}=\frac{14}{2}[2\text{a}+13\text{d}]$
$\Rightarrow27=28\text{a}+182\text{d}\ .....{4}$
Solving (3) and (4)
$\text{a}=1$ and $\text{d}=\frac{1}{7}$
$\therefore$ The required A.P is
$1,\ 1+\frac{1}{7},\ 2+\frac{2}{7},\ 1+\frac{3}{7},...,\ +\infty$
or $1,\ \frac{8}{7},\ \frac{9}{7},\ \frac{10}{7},\ \frac{11}{7},\ ...,\ \infty$
View full question & answer→Question 323 Marks
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
Answer$\text{a}_4+\text{a}_8=24$ [Given]
$\Rightarrow(\text{a+3d})+(\text{a}+7\text{d})=24$
$\Rightarrow\text{a}+5\text{d}=12\ .....(1)$
$\text{a}_6+\text{a}_10=34$
$\Rightarrow(\text{a}+5\text{d})+(\text{a}+9\text{d})=34$
$\Rightarrow\text{a}+7\text{d}=17\ .....(2)$
From (1) amd (2)
$\text{a}=\frac{-1}{2}$ and $\text{d}=\frac{5}{2}$
$\therefore$ 1st term is $\frac{-1}{2}$ and common diffrence is $\frac{5}{2}.$
View full question & answer→Question 333 Marks
Insert five numbers between $8$ and $26$ such that the resulting sequence is an A.P.
AnswerLet $A_1, A_2, A_3, A_4, A_5 $ be five numbers between $8$ and $26.$
Let d be the common difference.
Then, we have:
$26 = A_7$
$\Rightarrow 26 = 8 + (7−1)d$
$\Rightarrow 26 = 8 + 6d$
$\Rightarrow d = 3$
$A_1 = 8 + d = 8 + 3 = 11$
$A_2 = 8 + 2d = 8 + 6 = 14$
$A_3 = 8 + 3d = 8 + 9 = 17$
$A_4 = 8 + 4d = 8 + 12 = 20$
$A_5 = 8 + 5d = 8 + 15 = 23$
Therefore, the five numbers are $11, 14, 17, 20, 23.$
View full question & answer→Question 343 Marks
The first and the last terms of an A.P. area and I respectively. Show that the sum of $n^{th}$ term from the beginning and $n^{th}$ term from the end is $a + l.$
AnswerThe $n^{th}$ term from starting
$=\text{a}_\text{n}=\text{aa}+(\text{n}-1)\text{d}\ .....(1)$
The $n^{th}$ term from end
$=\text{l}-(\text{n}-1)\text{d}\ .....{(2)}$
Adding $(1)$ and $(2)$ we get
Sum of $n^{th}$ term from begining and $n^{th}$ term from the end
$=\text{a}+(\text{n}-1)\text{d}+\text{l}-(\text{n}-1)\text{d}$
$=\text{a}+\text{l}$
Hence proced.
View full question & answer→Question 353 Marks
How many numbers of two digit are divisible by $3?$
AnswerThe first two digit number divisible by $3$ is $12.$
and last two digit number divisible by $3$ is $99.$
So, the required series is $12, 15, 18, ...99.$
Let there be $n$ terms then $n^{th}$ term $= 99$
$\Rightarrow99=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow99=12+(\text{n}-1)3$
$\Rightarrow\text{n}=30$
30 two digit number are divisible by $3.$
View full question & answer→Question 363 Marks
Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
AnswerLet the 3 nimber in A.P are
$\text{a}-\text{d}+\text{a}+\text{a}+\text{d}=27$
$3\text{a}=27$
$\therefore\text{a}=9\ .....{(1)}$
and
$(\text{a}-\text{d})(\text{a})(\text{a}+\text{d})=648$
$(9-\text{d})9(9-\text{d})=648\ [\because\text{a}=9]$
$9^2-\text{d}^2=72$
$\therefore\text{d}=3\ .....(2)$
$\therefore$ the given sequeance is 6, 9, 12.
View full question & answer→Question 373 Marks
If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?
Answer$\text{a}_{5}=\text{a}+4\text{d}=30\ .....(1)$ [Given]
$\text{a}_{12}=\text{a}+11\text{d}=65\ .....(2)$ [Given]
From (1) and (2)
$\text{d}=5$ and $\text{a}=10$
Then,
Sum of irst 20 terms is
$\text{s}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\text{s}_{\text{20}}=\frac{\text{20}}{2}[2\times10+(\text{20}-1)\text{5}]$
$=1150$
Sum of first 20 tems is 1150.
View full question & answer→Question 383 Marks
Find the sum of n terms of the A.P. whose $k^{th}$ terms is $5k + 1.$
AnswerHere,
$\text{a}_\text{k}=5\text{k}+1$
$\text{a}_1=5+1=6$
$\text{a}_2=5(2)+1=11$
$\text{a}_3=5(3)+1=16$
$\text{d}=11-6=16-11=5$
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2(6)+(\text{n}-1)(5)]$
$=\frac{\text{n}}{2}[12+5\text{n}-5]$
$\text{s}_\text{n}=\frac{\text{n}}{2}(5\text{n}+7)$
View full question & answer→Question 393 Marks
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.
AnswerGiven,
$\text{n}=60$
$\text{a}=7$
$\text{l}=125$
$\therefore\text{a}+(\text{n}-1)\text{d}=125$
$7+(59)\text{d}=125$
$\text{d}=2$
$\therefore\text{a}_{32}=\text{a}+(32-1)\text{d}$
$=7+(31)2$
$=67$
View full question & answer→Question 403 Marks
Find the sum of all integers between $100$ and $550,$ wgich are divisible by $9.$
AnswerThe series fromed bt all the integers between $100$ and $550$ which are divisible $9$ is
$108, 117, 123, ... , 549$
Let there be n terms in the A.P then, the $n^{th}$ term is $549$
$549=\text{a}(\text{n}-1)\text{d}$
$549=108+(\text{n}-1)9$
$\Rightarrow\text{n}=50$
Then,
$\text{s}_\text{n}=\frac{\text{n}}{2}=[\text{a}+\text{l}]$
$\text{s}_{50}=\frac{50}{2}[108+549]$
$=16425$
View full question & answer→Question 413 Marks
The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
AnswerGiven:
$\text{a}_{10}=41=\text{a}+9\text{d}\ .....(1)$
$\text{a}_{18}=73=\text{a}+17\text{d}\ .....(2)$
Solving (1) and (2)
$\text{a}+9\text{d}=41$
$\text{a}+17\text{d}=73$
we get $\text{a}=5$ and $\text{d}=4$
$\therefore\text{a}_{26}=\text{a}+(26-1)\text{d}$
$=5+25(4)$
$=105$
26th term od the given A.P. is 105.
View full question & answer→Question 423 Marks
Find the sum of the following arithmetic progression:
$\frac{\text{x}-\text{y}}{\text{x}+\text{y}},\ \frac{3\text{x}-2\text{y}}{\text{x}+\text{y}},\ \frac{5\text{x}-3\text{y}}{\text{x}+\text{y}},\ ...$ to n terms.
Answer$\frac{\text{x}-\text{y}}{\text{x}+\text{y}},\ \frac{3\text{x}-2\text{y}}{\text{x}+\text{y}},\ \frac{5\text{x}-3\text{y}}{\text{x}+\text{y}},\ ...$ to n terms.
$n^{th}$ term is above sequence is $\frac{(2\text{n}-1)\text{x}-\text{ny}}{\text{x}+\text{y}}$
Sum of n terms is given by
$\frac{1}{\text{x}+\text{y}}[\text{x}+3\text{x}+5\text{x}+.....+(2\text{n}-1)\\\text{x}-(\text{y}+2\text{y}+3\text{y}...+\text{ny})]$
$=\frac{1}{\text{x}+\text{y}}\Big[\frac{\text{n}}{2}(2\text{x}+(\text{n}-1)2\text{x})-\frac{\text{n}(\text{n}+1)\text{y}}{2}\Big]$
$=\frac{1}{2(\text{x}+\text{y})}[2\text{n}^2\text{x}-2\text{n}^2\text{y}-\text{ny}]$
View full question & answer→Question 433 Marks
The first term of an A.P. is $2$ and the last term is $50.$ The sum of all these terms is $442.$ Find the common difference.
AnswerGiven,
$\text{a}=2$
$\text{l}=50$
$\therefore\text{l}=\text{a}+(\text{n}-1)\text{d}$
$50=2+(\text{n}-1)\text{d}$
$(\text{n}-1)\text{d}=48\ .....(1)$
$s_n $ of all n terms is given $442$
$\therefore\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$442=\frac{\text{n}}{2}[2+50]$
or $\text{n}=17\ .....{(2)}$
From (1) and (2)
$\text{d}=\frac{48}{\text{n}-1}=\frac{48}{16}=3$
The common difference is $3.$
View full question & answer→Question 443 Marks
Solve:$25 + 22 + 19 + 16 + ... + x = 115$
Answer$25 + 22 + 19 + 16 + ... + x = 115$
Here, sum of the given series of say $n$ terms is $115$
So, the $n^{th}$ term = $x$
Here,
$\text{a}=25$ and $\text{d}=22-25=-3$
$\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{x}=25-3(\text{n}-1)$
$\Rightarrow\text{x}=28-3\text{n}\ ...(1)$
The sum of n terms
$\text{s}_{\text{n}}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\Rightarrow115=\frac{\text{n}}{2}[25+28-3\text{n}]$
$\Rightarrow230=53\text{n}-3\text{n}^2$
$\Rightarrow3\text{n}^2-53\text{n}-3\text{n}^2$
$\Rightarrow3\text{n}^2-30\text{n}-23\text{n}-230=0$
$\Rightarrow\text{n}=10$ or $\frac{23}{3}$
But n can't be function
$\therefore\text{n}=10\ .....(2)$
From (1) and (2)
$\text{x}=28-3\text{n}$
$=28-3(10)$
$=-2$
$\text{x}=-2$
View full question & answer→Question 453 Marks
If $\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P., proved that a, b, c are in A.P.
Answer$\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P.
$\Rightarrow\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)+1,\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big)+1,\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)+1$ are in A.P.
$\Rightarrow\Big(\frac{\text{ac}+\text{ab}+\text{bc}}{\text{bc}}\Big),\ \Big(\frac{\text{ab}+\text{bc}+\text{ac}}{\text{ac}}\Big),\ \Big(\frac{\text{cd}+\text{ac}+\text{ab}}{\text{ab}}\Big)$ are in A.P.
$\Rightarrow\frac{1}{\text{bc}},\ \frac{1}{\text{ac}},\ \frac{1}{\text{ab}}$ are in A.P.
$\Rightarrow\frac{\text{abc}}{\text{bc}},\ \frac{\text{abc}}{\text{ac}},\ \frac{\text{abc}}{\text{ab}}$ are in A.P.
$\Rightarrow\text{a},\text{b},\text{c}$ are in A.P.
View full question & answer→Question 463 Marks
If $\text{s}_\text{n}=\text{n}^2\ \text{p}$ and $\text{s}_\text{m}=\text{m}^2\ \text{p},\ \text{m}\not=\text{n},$ in an A.P., prove that $\text{s}_\text{p}=\text{p}^3.$
AnswerLet a be the first term of the AP and d is the common difference. then
$\text{s}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{n}^2\text{p}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{np}=\frac{1}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$2\text{np}=2\text{a}+(\text{n}-1)\text{d}\ .....{(1)}$
Again
$\text{s}_\text{m}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{m}^2\text{p}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{mp}=\frac{1}{2}[2\text{a}+(\text{m}-1)\text{d}]\ .....{(2)}$
Now subtract (1) from (2)
$2\text{p}(\text{m}-\text{n})=(\text{m}-\text{n})\text{d}$
$\text{d}=2\text{p}$
Therefore
$2\text{mp}=2\text{a}+(\text{m}-1)\times2\text{p}$
$2\text{a}=2\text{p}$
$\text{a}=\text{p}$
The sum up p terms will be:
$\text{s}_\text{p}=\frac{\text{p}}{2}(2\text{a}+(\text{p}-1)\text{d})$
$=\frac{\text{p}}{2}(2\text{p}+(\text{p}-1).2\text{p})$
$=\frac{\text{p}}{2}(2\text{p}+2\text{p}^2-2\text{p})$
$=\text{p}^3$
View full question & answer→Question 473 Marks
Solve:$1 + 4 + 7 + 10 + ... + x = 590.$
Answer$1 + 4 + 7 + 10 + ... + x = 590.$
Here,
$\text{a}=1$
$\text{d}=4-1=3$
Let there be n terms so the $n^{th} $ term $= x$
$\Rightarrow\text{x}=1+(\text{n}-1)3$ $[\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\text{x}=3\text{n}-2\ .....(1)$
and
$\text{s}_{\text{n}}=590 [$given$]$
$\Rightarrow\frac{\text{n}}{2}[\text{a}+\text{l}]=590$
$\Rightarrow\frac{\text{n}}{2}[1+3\text{n}-2]$ $[\because\text{l}=\text{x}=\text{3n}-2]$
$\Rightarrow3\text{n}^2-\text{n}-1080=0$
$\Rightarrow3\text{n}^2-60\text{n}+59(\text{n}-20)=0$
$\Rightarrow3\text{n}(\text{n}-20)+59(\text{n}-20)=0$
$\Rightarrow\text{n}=2 .....(2)$
from$(1)$ and $(2)$
$\text{x}=3\text{n}-2$
$=3(20)-2$
$=58$
$\text{x}=58$
View full question & answer→Question 483 Marks
If the sum of n terms of an A.P. is $\text{np}+\frac{1}{2}\text{n}(\text{n}-1)$ Q, where P and Q are constants, find the common difference.
Answer$\text{s}_\text{n}=\text{n}\text{p}+\frac{1}{2}(\text{n}-1)\text{Q}$ [Given]
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{p}+(\text{n}-1)\text{Q}]\ .....(1)$
We know,
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\ .....(2)$
Where a = first term and d = common diffrence (1) and (2) d = Q
$\therefore$ the common diffrence Q.
View full question & answer→Question 493 Marks
The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by $10\frac{1}{2},$ find the number of terms and the series.
AnswerLet no. of term be 2n
Odd terms sum $=24=\text{T}_1+\text{T}_3+...+\text{T}_{2\text{n}-1}$
Even terms sum $=30=\text{T}_2+\text{T}_4+...+\text{T}_{2\text{n}}$
Subtract above two equtions
$\text{nd}=6$
$\text{T}_{2\text{n}}=\text{T}_1+\frac{21}{2}$
$\text{T}_{2\text{n}}-\text{a}=\frac{21}{2}$
$(2\text{n}-1)\text{d}=\frac{21}{2}$
$12-\frac{21}{2}=\text{d}=\frac{3}{2}$
$\Rightarrow\text{n}=6\times\frac{2}{3}=4$
Total terms $=2\text{n}=8$
Subtite above values in equation of
Sum of even terms or add terms, we get
$\text{a}=\frac{3}{2}$
So series is $\frac{3}{2},3\frac{9}{2}......$
View full question & answer→Question 503 Marks
If $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P., Prove that:
$\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{\text{c}}$ are in A.P.
Answer$\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{\text{c}}$ will be in A.p if $\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{b}+\text{c}}{\text{a}}=\frac{\text{a}+\text{b}}{\text{c}}-\frac{\text{c}+\text{a}}{\text{b}}$
if $\frac{\text{ca}+\text{a}^2-\text{b}^2-\text{cd}}{\text{ab}}=\frac{\text{ab}+\text{b}^2-\text{c}^2-\text{ac}}{\text{bc}}$
$\text{LHS}\Rightarrow\frac{\text{ca}+\text{a}^2-\text{b}^2-\text{cd}}{\text{ab}}$
$\Rightarrow\frac{\text{c}^2\text{a}+\text{a}^2\text{c}-\text{b}^2\text{c}-\text{c}^2\text{b}}{\text{abc}}$
$\Rightarrow\frac{\text{c}(\text{a}-\text{b})[\text{a}+\text{b}+\text{c}]}{\text{bc}}\ .....(1)$
$\text{RHS}\Rightarrow\frac{\text{ab}+\text{b}^2-\text{c}^2-\text{ac}}{\text{bc}}$
$\Rightarrow\frac{\text{a}^2\text{b}+\text{ab}^2-\text{ac}^2-\text{a}^2\text{c}}{\text{abc}}$
$\Rightarrow\frac{\text{a}(\text{b}-\text{c})[\text{a}+\text{b}+\text{c}]}{\text{abc}}\ .....(2)$
and since $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\text{c}(\text{b}-\text{a})=\text{a}(\text{b}-\text{c})\ .....(3)$
$\therefore$ LHS = RHS and the given terms are in A.P
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