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If 5+9+13+... to n terms7+9+11+...to (n+1) terms=1716, then n =

Question: If $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$, then $n=$ (a) 8(b) 7(c) 10(d) 11 Solution: Here, we are given, $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$ .......(1) We need to findn. So, first let us find out the sum ofnterms of the A.P. given in the numerator. Here we use the following formula for the sum ofnterms of an A.P., $S_{n}=\frac{n...

If a and b are the coefficients of

Question: If $a$ and $b$ are the coefficients of $x^{n}$ in the expansion of $(1+x)^{2 n}$ and $(1+x)^{2 n-1}$ respectively, find $\frac{a}{b}$. Solution: Coefficients of $x^{n}$ in the expansion of $(1+x)^{2 n}$ is ${ }^{2 n} C_{n}=a$. Coefficients of $x^{n}$ in the expansion of $(1+x)^{2 n-1}$ is ${ }^{2 n-1} C_{n}=b$. Now, $\frac{a}{b}=\frac{{ }^{2 n} C_{n}}{{ }^{2 n-1} C_{n}}$ $=\frac{\frac{(2 n) !}{n ! n !}}{\frac{(2 n-1) !}{n !(n-1) !}}$ $=\frac{2 n}{n}$ $=2$ Hence, $\frac{a}{b}=2$....

if

Question: If $3 \sin ^{-1} x=\pi-\cos ^{-1} x$, then $x=$__________________. Solution: $3 \sin ^{-1} x=\pi-\cos ^{-1} x$ $\Rightarrow 3\left(\frac{\pi}{2}-\cos ^{-1} x\right)=\pi-\cos ^{-1} x$ $\left(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)$ $\Rightarrow \frac{3 \pi}{2}-3 \cos ^{-1} x=\pi-\cos ^{-1} x$ $\Rightarrow 2 \cos ^{-1} x=\frac{3 \pi}{2}-\pi=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1} x=\frac{\pi}{4}$ $\Rightarrow x=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ If $3 \sin ^{-1} x=\pi-\cos ^{-1...

Find the number of terms in the expansion of

Question: Find the number of terms in the expansion of $(a+b+c)^{n}$. Solution: We have: $(a+b+c)^{n}=[a+(b+c)]^{n}$ $=a^{n}+{ }^{n} C_{1} a^{n-1}(b+c)^{1}+{ }^{n} C_{2} a^{n-2}(b+c)^{2}+\ldots+{ }^{n} C_{n}(b+c)^{n}$ Further, expanding each term of R.H.S., we note that First term consists of 1 term. Second term on simplification gives 2 terms. Third term on expansion gives 3 terms. Similarly, fourth term on expansion gives 4 terms and so on. $\therefore$ The total number of terms $=1+2+3+\ldots...

In ∆ABC, side AB is produced to D such that BD = BC. If ∠B = 60° and ∠A = 70°,

Question: In∆ABC, sideABis produced toDsuch thatBD=BC. IfB= 60 and A= 70, prove that (i)ADCDand (ii)ADAC. Solution: In triangle CBA, CBD is an exterior angle. i. e., $\angle C B A+\angle C B D=180^{\circ}$ $\Rightarrow 60^{\circ}+\angle C B D=180^{\circ}$ $\Rightarrow \angle C B D=120^{\circ}$ Triangle BCD is isosceles and BC = BD. Let $\angle B C D=\angle B D C=x^{\circ}$. In $\triangle C B D$, we have : $\Rightarrow \angle B C D+\angle C B D+\angle C D B=180^{\circ}$ $\Rightarrow x+120^{\circ}...

Write last two digits of the number

Question: Write last two digits of the number 3400. Solution: $3^{400}=(9)^{200}$' $=(10-1)^{200}$ $={ }^{200} C_{0}(10)^{200}+{ }^{200} C_{1}(10)^{199}(-1)^{1}+\ldots .+{ }^{200} C_{198}(10)^{2}(-1)^{198}+{ }^{200} C_{199}(10)^{1}(-1)^{199}+{ }^{200} C_{200}(-1)^{200}$ $=100\left[(10)^{198}+{ }^{200} C_{1}(10)^{197}(-1)^{1}+\ldots . .+{ }^{200} C_{198}(-1)^{198}\right]+200(10)^{1}(-1)^{199}+(-1)^{200}$ $=100\left[(10)^{198}-{ }^{200} C_{1}(10)^{197}+\ldots \ldots+{ }^{200} C_{198}-2(10)\right]+...

if

Question: If $x0, y0, x y1$, then $\tan ^{-1} x+\tan ^{-1} y=$ Solution: We know $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$, if $x0, y0$ and $x y1$ If $x0, y0, x y1$, then $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$...

The sum of n terms of two A.P.'s are in the ratio

Question: The sum ofnterms of two A.P.'s are in the ratio 5n+ 9 : 9n+ 6. Then, the ratio of their 18th term is (a) $\frac{179}{321}$ (b) $\frac{178}{321}$ (C) $\frac{175}{321}$ (d) $\frac{176}{321}$ Solution: In the given problem, the ratio of the sum ofnterms of two A.Ps is given by the expression, $\frac{S_{n}}{S_{n}^{1}}=\frac{5 n+9}{9 n+6}$.......(1) We need to find the ratio of their18thterms. Here we use the following formula for the sum of $n$ terms of an A.P., $S_{n}=\frac{n}{2}[2 a+(n-1...

Find the ratio of the coefficients of

Question: Find the ratio of the coefficients of $x^{\rho}$ and $x^{q}$ in the expansion of $(1+x)^{p+q}$. Solution: Coefficient of $x^{p}$ in the expansion of $(1+x)^{p+q}$ is $^{p+q} C_{p}$. Coefficient of $x^{q}$ in the expansion of $(1+x)^{p+q}$ is $^{p+q} C_{q}$. Now, $\frac{p+q_{p}}{p+q_{q}}=\frac{\frac{(p+q) !}{p l q !}}{\frac{(p+q) !}{q ! p !}}=1$ Hence, the ratio of the coefficients of $x^{p}$ and $x^{q}$ in the expansion of $(1+x)^{p+q}$ is $1: 1$....

In ∆ABC, ∠A = 90°. Which is its longest side?

Question: (i) In∆ABC,A= 90. Which is its longest side?(ii)In∆ABC,A= B =45. Which is its longest side?(iii) In∆ABC,A= 100 and C =50. Which is its shortest side? Solution: (i) Given: In∆ABC,A= 90So, sum of the other two angles in triangle B+ C= 90i.e. B, C 90Since,Ais the greatest angle.So, the longest side isBC.(ii) Given: A= B= 45Using angle sum property of triangle,C= 90Since,Cis the greatest angle.So, the longest side isAB.(iii) Given: A= 100 and C= 50Using angle sum property of triangle,B= 30...

if

Question: If $\cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{3}$, then $\sin ^{-1} x+\sin ^{-1} y=$____________________. Solution: We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$, for all $x \in \mathrm{R}$ ...(1) Also, $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$, for all $y \in \mathrm{R}$ ...(2) Adding (1) and (2), we get $\sin ^{-1} x+\cos ^{-1} x+\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2}$ $\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\cos ^{-1} x+\cos ^{-1} y=\pi$ $\Rightarrow \sin ^{-1} x+\sin ^...

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In ΔABC, ∠A = 50° and ∠B = 60°.

Question: In ΔABC, A= 50 and B= 60. Determine the longest and shortest sides of the triangle. Solution: Given: In ΔABC, A= 50 and B= 60InΔABC,A+ B+ C= 180 (Angle sum property of a triangle)⇒⇒50 +60 +C= 180⇒⇒110 +C= 180⇒⇒C= 180-110⇒⇒C= 70Hence, the longest side will be opposite to the largest angle (C= 70) i.e.AB.And, the shortest side will be opposite to the smallest angle (A= 50 ) i.e.BC....

Find the sum of the coefficients of two middle terms in the binomial expansion of

Question: Find the sum of the coefficients of two middle terms in the binomial expansion of $(1+x)^{2 n-1}$. Solution: $(1+x)^{2 n-1}$ Here, $n$ is an odd number. Therefore, the middle terms are $\left(\frac{2 n-1+1}{2}\right)^{\text {th }}$ and $\left(\frac{2 n-1+1}{2}+1\right)^{\text {th }}$, i. e.,$n^{\text {th }}$ and $(n+1)^{\text {th }}$ terms. Now, we have $T_{n}=T_{n-1+1}$ $={ }^{2 n-1} C_{n-1}(x)^{n-1}$ And, $T_{n+1}=T_{n+1}$ $={ }^{2 n-1} C_{n}(x)^{n}$ $\therefore$ the coefficients of ...

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.

Question: Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.(i) 5 cm, 4 cm, 9 cm(ii) 8 cm, 7 cm, 4 cm(iii) 10 cm, 5 cm, 6 cm(iv) 2.5 cm, 5 cm, 7 cm(v) 3 cm, 4 cm, 8 cm Solution: (i) No, because the sum of two sides of a triangle is not greater than the third side.5 + 4 = 9(ii) Yes, because the sum of two sides of a triangle is greater than the third side.7 + 4 8; 8 + 7 4; 8 + 4 7(iii) Yes, because the sum of two sides of a triangle is gr...

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.

Question: If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent. Is the statement true? Why? Solution: Yes,the statement is true because the two triangles can be congruent by either AAS or ASA congruence criteria.Disclaimer: If corresponding angles of two triangles are equal, then the by angle sum property, the third corresponding angle will be equal. So, if we have two corresponding angles and a corresponding sid...

Write the number of terms in the expansion of

Question: Write the number of terms in the expansion of $\left[\left(2 x+y^{3}\right)^{4}\right]^{7}$. Solution: In the binomial expansion of $(a+b)^{n}$, total number of terms will be $(n+1)$. Now, $\left[\left(2 x+y^{3}\right)^{4}\right]^{7}=\left(2 x+y^{3}\right)^{28}$ Therefore, in the expansion of $\left[\left(2 x+y^{3}\right)^{4}\right]^{7}$, total number of terms will be $28+1=29$....

If 18, a, b, −3 are in A.P., the a + b =

Question: If 18,a,b, 3 are in A.P., thea+b=(a) 19(b) 7(c) 11(d) 15 Solution: Here, we are given four terms which are in A.P., First term (a1) = Second term (a2) = Third term (a3) = Fourth term (a4)= So, in an A.P. the difference of two adjacent terms is always constant. So, we get, $d=a_{2}-a_{1}$ $d=a-18$$\ldots \ldots$ (1) Also, $d=a_{4}-a_{3}$ $d=-3-b$.......$(2)$ Now, on equating (1) and (2), we get, $a-18=-3-b$ $a+b=18-3$ $a+b=15$ Therefore, $a+b=15$ Hence the correct option is(d)....

The value of

Question: The value of $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ for all $x \in R$, is____________________ Solution: We know $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$, for all $x \in \mathrm{R}$ $\therefore \cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=\cot \frac{\pi}{2}$ $\Rightarrow \cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=0$ The value of $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ for all $x \in \mathrm{R}$, is__0__....

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.”

Question: If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.Is the statement true? Why? Solution: No, the statement is not true because the two triangles are congruent only by SAS congruence condition but the statement contains ASS or SSA condition as well which are not any condition for congruence of triangles....

Write the coefficient of the middle term in the expansion of

Question: Write the coefficient of the middle term in the expansion of $(1+x)^{2 n}$. Solution: Here, $n$, i. e., $2 n$, is an even number. $\therefore$ Middle term $=\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th term Thus, we have : $T_{n+1}={ }^{2 n} C_{n}(1)^{2 n-n}(x)^{n}$ $={ }^{2 n} C_{n} x^{n}$ Hence, the coefficient of the middle term is ${ }^{2 n} C_{n}$...

In a ΔABC, D is the midpoint of side AC such that BD

Question: In a $\triangle A B C, D$ is the midpoint of side $A C$ such that $B D=\frac{1}{2} A C$. Show that $\angle A B C$ is a right angle. Solution: Given: In $\triangle A B C, D$ is the midpoint of side $A C$ such that $B D=\frac{1}{2} A C$. To prove: $\angle A B C$ is a right angle. Proof: In $\Delta A D B$, $A D=B D \quad\left(\right.$ Given, $\left.B D=\frac{1}{2} A C\right)$ $\Rightarrow \angle D A B=\angle D B A=x$ (Let) $\quad$ (Angles opposite to equal sides are equal) Similarly, in $...

Two A.P.'s have the same common difference.

Question: Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is(a) 11(b) 3(c) 8(d) 5 Solution: Here, we are given two A.P.s with same common difference. Let us take the common difference asd. Given, First term of first A.P. (a) = 8 First term of second A.P. (a) = 3 We need to find the difference between their 30thterms. So, let us first find the 30thterm of first A.P. $a_{30}=a+(30-1) d$ $=3+29 d$ $\l...

If a and b denote the sum of the coefficients in the expansions of

Question: If $a$ and $b$ denote the sum of the coefficients in the expansions of $\left(1-3 x+10 x^{2}\right)^{n}$ and $\left(1+x^{2}\right)^{n}$ respectively, then write the relation between $a$ and $b$. Solution: Here, $a=1-3+10=8=2^{3}$ $b=1+1=2$ $\Rightarrow a=b^{3}$...

Write the middle term in the expansion of

Question: Write the middle term in the expansion of $\left(x+\frac{1}{x}\right)^{10}$. Solution: Here, $n$, i. e., 10 , is an even number. $\therefore$ Middle term $=\left(\frac{10}{2}+1\right)$ th term $=6$ th term Thus, we have: $T_{6}=T_{5+1}$ $={ }^{10} C_{5}(x)^{10-5} \times\left(\frac{1}{x}\right)^{5}$ $={ }^{10} C_{5}$...

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