If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that
Question: If $a, b, c$ and $d$ in any binomial expansion be the 6 th, 7 th, 8 th and 9th terms respectively, then prove that $\frac{b^{2}-a c}{c^{2}-b d}=\frac{4 a}{3 c}$. Solution: Suppose the binomial expression is $(1+x)^{n}$. Then, the 6 th, 7 th, 8 th and 9 th terms are ${ }^{n} C_{5} x^{5},{ }^{n} C_{6} x^{6},{ }^{n} C_{7} x^{7}$ and ${ }^{n} C_{8} x^{8}$, respectively. Now, we have: $\frac{{ }^{n} C_{6 x^{6}}}{{ }^{n} C_{5} x^{5}}=\frac{b}{a}, \frac{{ }^{n} C_{8 x^{8}}}{{ }^{n} C_{7} x^{7...
Read More →A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall
Question: A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize. Solution: In the given problem, Total amount of money (Sn) = Rs 700 There are a total of 7 prizes and each prize is Rs 20 less than the previous prize. So let us take the first prize as Rsa. So, the second prize will be Rs, third prize will be Rs. Therefore, the prize money will form a...
Read More →If 3rd, 4th 5th and 6th terms in the expansion of
Question: If 3 rd, 4th 5th and 6th terms in the expansion of $(x+a)^{n}$ be respectively $a, b, c$ and $d$, prove that $\frac{b^{2}-a c}{c^{2}-b d}=\frac{5 a}{3 c}$. Solution: We have: $(x+a)^{n}$ The $3 \mathrm{rd}, 4$ th, 5 th and 6 th terms are ${ }^{n} C_{2} x^{n-2} a^{2},{ }^{n} C_{3} x^{n-3} a^{3},{ }^{n} C_{4} x^{n-4} a^{4}$ and ${ }^{n} C_{5} x^{n-5} a^{5}$, respectively. Now, ${ }^{n} C_{2} x^{n-2} a^{2}=a$ ${ }^{n} C_{3} x^{n-3} a^{3}=b$ ${ }^{n} C_{4} x^{n-4} a^{4}=c$ ${ }^{n} C_{5} x...
Read More →If
Question: If $\alpha=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$ and $\beta=\tan ^{-1}\left(-\tan \frac{2 \pi}{3}\right)$, then (a) $4 \alpha=3 \beta$ (b) $3 a=4 \beta$ (c) $\alpha-\beta=\frac{7 \pi}{12}$ (d) none of these Solution: (a) $4 \alpha=3 \beta$ We know that $\tan ^{-1}(\tan x)=x$. $\therefore \alpha=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$ $=\tan ^{-1}\left\{\tan \left(\pi+\frac{\pi}{4}\right)\right\}$ $=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$ $=\frac{\pi}{4}$ and $\beta=\tan ...
Read More →In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E.
Question: In the given figure, the sideBCof ∆ABChas been produced on both sideson the left toDand on the right toE. If ABD= 106 and ACE= 118, find the measure of each angle of the triangle. Solution: Side BC of triangle ABC is produced to D. $\therefore \angle A B C=\angle A+\angle C$ $\Rightarrow 106^{\circ}=\angle A+\angle C \quad \ldots(i)$ Also, side BC of triangle ABC is produced to E. $\angle A C E=\angle A+\angle B$ $\Rightarrow 118^{\circ}=\angle A+\angle B \quad \ldots(i i)$ Adding $(i)...
Read More →The sum of first n terms of an A.P.
Question: The sum of first $n$ terms of an A.P. is $3 n^{2}+4 n$. Find the 25 th term of this A.P. Solution: $S_{n}=3 n^{2}+4 n$ We know $a_{n}=S_{n}-S_{n-1}$ $\therefore a_{n}=3 n^{2}+4 n-3(n-1)^{2}-4(n-1)$ $\Rightarrow a_{n}=6 n+1$ $a_{25}=6(25)+1=151$...
Read More →If in the expansion of
Question: If in the expansion of (1 +x)n, the coefficients of three consecutive terms are 56, 70 and 56, then findnand the position of the terms of these coefficients. Solution: Suppose $r^{t h},(r+1)^{t h}$ and $(r+2)^{t h}$ terms are the three consecutive terms. Their respective coefficients are ${ }^{n} C_{r-1},{ }^{n} C_{r}$ and ${ }^{n} C_{r+1}$. We have: ${ }^{n} C_{r-1}={ }^{n} C_{r+1}=56$ $\Rightarrow r-1+r+1=n \quad\left[\right.$ If ${ }^{n} C_{r}={ }^{n} C_{s} \Rightarrow r=s$ or $\lef...
Read More →Find the number of terms of the A.P. −12, −9, −6, ..., 21.
Question: Find the number of terms of the A.P. 12, 9, 6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained. Solution: First term, $a_{1}=-12$ Common difference, $d=a_{2}-a_{1}=-9-(-12)=3$ $a_{n}=21$ $\Rightarrow a+(n-1) d=21$ $\Rightarrow-12+(n-1) \times 3=21$ $\Rightarrow 3 n=36$ $\Rightarrow n=12$ Therefore, number of terms in the given A.P. is 12.Now, when 1 is added to each of the 12 terms, the sum will increase by 12.So, the sum of al...
Read More →The number of solutions of the equation
Question: The number of solutions of the equation $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$ is (a) 2(b) 3(c) 1(d) none of these Solution: (a) 2 We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\therefore \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}$ $\Rightarrow \frac{2 x+3 x}{1-2 x \times 3 x}=\tan \frac{\pi}{4}$ $\Rightarrow \frac{5 x}{1-6 x^{2}}=1$ $\Rightarrow 5 x=1-6 x^{2...
Read More →If $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$, then $\frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=$
[question] Question. If $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$, then $\frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=$ (a) $\sin ^{2} \alpha$ (b) $\cos ^{2} \alpha$ (c) $\tan ^{2} \alpha$ (d) $\cot ^{2} a$ [/question] [solution] Solution: (a) $\sin ^{2} \alpha$ We know that $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$. $\therefore \cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$ $\Rightarrow \cos ^{-1}\left(\frac...
Read More →$2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to
[question] Question. $2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to (a) $\cot ^{-1} x$ (b) $\cot ^{-1} \frac{1}{x}$ (c) $\tan ^{-1} x$ (d) none of these [/question] [solution] Solution: (c) $\tan ^{-1} x$ Let $\tan ^{-1} x=y$ So, $x=\tan y$ $\therefore 2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}=2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \l...
Read More →The nth term of an A.P is given by
Question: The $n$th term of an A.P is given by $(-4 n+15)$, Find the sum of first 20 terms of this A.P. Solution: $a_{n}=-4 n+15$ $\Rightarrow a_{1}=-4+15=11$ Also, $a_{2}=-8+15=7$ Common difference, $d=a_{2}-a_{1}=7-11=-4$ Now, $S_{20}=\frac{20}{2}[2 \times 11+(20-1)(-4)]$ $=10(22-76)$ $=-540$...
Read More →The nth term of an A.P is given by
Question: The $n$th term of an A.P is given by $(-4 n+15)$, Find the sum of first 20 terms of this A.P. Solution: $a_{n}=-4 n+15$ $\Rightarrow a_{1}=-4+15=11$ Also, $a_{2}=-8+15=7$ Common difference, $d=a_{2}-a_{1}=7-11=-4$ Now, $S_{20}=\frac{20}{2}[2 \times 11+(20-1)(-4)]$ $=10(22-76)$ $=-540$...
Read More →Show that
Question: $\sin \left[\cot ^{-1}\left\{\tan \left(\cos ^{-1} x\right)\right\}\right]$ is equal to (a) $x$ (b) $\sqrt{1-x^{2}}$ (C) $\frac{1}{x}$ (d) none of these Solution: (a) $x$ Let $\cos ^{-1} x=y$ Then, $\sin \left[\cot ^{-1}\left\{\tan \left(\cos ^{-1} x\right)\right\}\right]=\sin \left[\cot ^{-1}\{\tan y\}\right]$ $=\sin \left[\cot ^{-1}\left\{\cot \left(\frac{\pi}{2}-y\right)\right\}\right]$ $=\sin \left(\frac{\pi}{2}-y\right)$ $=\cos y$ $=x \quad[\because \cos y=x]$...
Read More →The sum of first m terms of an A.P.
Question: The sum of first $m$ terms of an A.P. is $4 m^{2}-m$. If its nth term is 107 . find the value of $n$. Also, find the $21 s t$ term of this A.P. Solution: $S_{m}=4 m^{2}-m$ We know $a_{m}=S_{m}-S_{m-1}$ $\therefore a_{m}=4 m^{2}-m-4(m-1)^{2}+(m-1)$ $a_{m}=8 m-5$ Now, $a_{n}=107$ $\Rightarrow 8 n-5=107$ $\Rightarrow 8 n=112$ $\Rightarrow n=14$ $a_{21}=8(21)-5=163$...
Read More →If
Question: If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=$ (a) $\frac{1}{2}$ (b) $\frac{\sqrt{3}}{2}$ (c) $-\frac{1}{2}$ (d) none of these Solution: (b) $\frac{\sqrt{3}}{2}$ We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$. $\therefore \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ $\Rightarrow \frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ $\Rightarrow-2 \cos ^{-1} x=\frac{\pi}{6}-\frac{\pi}{2}$ $\Rightarrow-2 \cos ^{-1} x=-\frac{\pi}{3}$ $\Rightarrow \cos ^{-1} x=\frac{\pi}{6}$ ...
Read More →The sum of first m terms of an A.P. is 4m2 − m.
Question: The sum of first $m$ terms of an A.P. is $4 m^{2}-m$. If its nth term is 107 . find the value of $n$. Also, find the 21 st term of this A.P. Solution: $S_{m}=4 m^{2}-m$ We know $a_{m}=S_{m}-S_{m-1}$ $\therefore a_{m}=4 m^{2}-m-4(m-1)^{2}+(m-1)$ $a_{m}=8 m-5$ Now, $a_{n}=107$ $\Rightarrow 8 n-5=107$ $\Rightarrow 8 n=112$ $\Rightarrow n=14$ $a_{21}=8(21)-5=163$...
Read More →The sum of first q terms of an A.P. is 63q − 3q2.
Question: The sum of first q terms of an A.P. is $63 q-3 q^{2}$. If its pth term is $-60$, find the value of $p$. Also, find the 11 th term of this A.P. Solution: $S_{q}=63 q-3 q^{2}$ We know $a_{q}=S_{q}-S_{q-1}$ $\therefore a_{q}=63 q-3 q^{2}-63(q-1)+3(q-1)^{2}$ $a_{q}=66-6 q$ Now, $a_{p}=-60$ $\Rightarrow 66-6 p=-60$ $\Rightarrow 126=6 p$ $\Rightarrow p=21$ $a_{11}=66-6 \times 11=0$...
Read More →The positive integral solution of the equation
Question: The positive integral solution of the equation $\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{1+y^{2}}}=\sin ^{-1} \frac{3}{\sqrt{10}} \mathrm{i}$ (a) $x=1, y=2$ (b) $x=2, y=1$ (c) $x=3, y=2$ (d) $x=-2, y=-1$. Solution: (a) $x=1, y=2$ We have, $\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{1+y^{2}}}=\sin ^{-1} \frac{3}{\sqrt{10}}$ $\Rightarrow \tan ^{-1} x+\tan ^{-1}\left[\frac{\sqrt{\sqrt{1-\left(\frac{y}{\sqrt{1+y^{2}}}\right)^{2}}}}{\frac{y}{\sqrt{1+y^{2}}}}\right]=\tan ^{-1}\left[\frac{\frac{3}{...
Read More →The sum of first n terms of an A.P is 5n2 + 3n.
Question: The sum of firstnterms of an A.P is 5n2+ 3n. If itsmth terms is 168, find the value ofm. Also, find the 20th term of this A.P. Solution: $S_{n}=5 n^{2}+3 n$ We know $a_{n}=S_{n}-S_{n-1}$ $\therefore a_{n}=5 n^{2}+3 n-5(n-1)^{2}-3(n-1)$ $a_{n}=10 n-2$ Now, $a_{m}=168$ $\Rightarrow 10 m-2=168$ $\Rightarrow 10 m=170$ $\Rightarrow m=17$ $a_{20}=10(20)-2=198$...
Read More →Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.
Question: Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25thterm. Solution: First term,a= 10 Sum of first 14 terms, $S_{14}=1505$ $\Rightarrow \frac{14}{2}[2 \times 10+(14-1) d]=1505$ $\Rightarrow 7 \times(20-13 d)=1505$ $\Rightarrow 20-13 d=\frac{1505}{7}=215$ $\Rightarrow 13 d=-195$ $\Rightarrow d=-15$ Now, $a_{25}=10+24(-15)=-350$...
Read More →A piece of equipment cost a certain factory Rs 60,000.
Question: A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost? Solution: In the given problem, Cost of the equipment = Rs 600,000 It depreciates by 15% in the first year. So, Depreciation in 1 year $=600000-495000$ $=105000$ $=90000$ It depreciates by 13.5% of the original cost in the 2 year. So, Depreciatio...
Read More →A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour
Question: A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount. Solution: In the given problem, the total amount = Rs 10710. For the first half and hour (30 minutes) he counts at a rate of Rs 180 per minute. So, The amount counted in 30 minutes So, amount left after half an hour After 30 minutes he counts at a rat...
Read More →In the given figure, side BC of ∆ABC is produced to D.
Question: In the given figure, sideBCof ∆ABCis produced toD. If ACD= 128 and ABC= 43, find BACand ACB. Solution: SideBCof triangleABCis produced toD. $\therefore \angle A C D=\angle A+\angle B \quad$ [Exterior angle property] $\Rightarrow 128^{\circ}=\angle A+43^{\circ}$ $\Rightarrow \angle A=(128-43)^{\circ}$ $\Rightarrow \angle A=85^{\circ}$ $\Rightarrow \angle B A C=85^{\circ}$ Also, in triangleABC, $\angle B A C+\angle A B C+\angle A C B=180^{\circ} \quad$ [Sum of the angles of a triangle] $...
Read More →In the expansion of
Question: In the expansion of (1 +x)nthe binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value ofn. Solution: Suppose the three consecutive terms are $T_{r-1}, T_{r}$ and $T_{r+1}$. Coefficients of these terms are ${ }^{n} C_{r-2},{ }^{n} C_{r-1}$ and ${ }^{n} C_{r}$, respectively. These coefficients are equal to 220,495 and 792 . $\therefore \frac{{ }^{n} C_{r-2}}{{ }^{n} C_{r-1}}=\frac{220}{495}$ $\Rightarrow \frac{r-1}{n-r+2}=\frac{4}{9}$ $\Rightar...
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