In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.
Question: In the given figure, AB || CD, BAE = 65 and OEC = 20. Find ECO. Solution: In the given figure,AB || CD and AE is the transversal.DOE =BAE (Pair of corresponding angles)⇒DOE =65In∆COE,DOE =OEC +ECO (Exterior angle of a triangle is equal to the sum of two opposite interior angles)⇒65 =20+ECO⇒ECO =6520 =45Thus, the measure of ECO is 45....
Read More →In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°,
Question: In the given figure, AB || CD and EF is a transversal. If AEF = 65, DFG = 30, EGF = 90 and GEF =x, find the value ofx. Solution: It is given that,AB || CD and EF is a transversal.EFD =AEF (Pair of alternate angles)⇒EFD =65⇒EFG + GFD=65⇒EFG +30=65⇒EFG =6530 =35In∆EFG,EFG +GEF +EGF = 180 (Angle sum property)⇒35 +x+90 =180⇒125 +x=180⇒x=180125 =55Thus, the value ofxis 55....
Read More →The first term of an A.P. is p and its common difference is q. Find its 10th term.
Question: The first term of an A.P. ispand its common difference isq. Find its 10th term. Solution: Here, we are given, First term (a) =p Common difference (d) =q We need to find the 10thterm (an). As we know, $a_{n}=a+(n-1) d$ So, for 10thterm (n= 10), we get, $a_{10}=p+(10-1) q$ $=p+9 q$ Therefore, $a_{10}=p+9 q$...
Read More →In the given figure, AB || DE and BD || FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.
Question: In the given figure, AB || DE and BD || FG such that ABC = 50 and FGH = 120. Find the values ofxandy. Solution: We have,FGE +FGH = 180 (Linear pair of angles)y+ 120= 180⇒y=180 120 = 60Now, AB || DF and BD is the transversal.ABD =BDF (Pair of alternate angles)⇒BDF =50Also, BD || FG and DF is the transversal.BDF =DFG (Pair of alternate angles)⇒DFG =50 .....(1)In∆EFG,FGH =EFG +FEG (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)⇒120 =50 +x [Using (1)]...
Read More →Write the expression of the common difference of an A.P. whose first term is a and nth term is b.
Question: Write the expression of the common difference of an A.P. whose first term isaandnth term isb. Solution: Here, we are given First term =a Last term =b Let us take the common difference asd Now, we know $a_{n}=a+(n-1) d$ So, For the last term (an), $b=a+(n-1) d$ $b-a=(n-1) d$ $d=\frac{b-a}{n-1}$ Therefore, common difference of the A.P. is $d=\frac{b-a}{n-1}$...
Read More →The coefficient of
Question: The coefficient of $x^{-17}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ is (a) 1365 (b) 1365 (c) 3003 (d) 3003 Solution: (b) 1365 Suppose the $(r+1)$ th term in the given expansion contains the coefficient of $x^{-17}$. Then, we have : $T_{r+1}={ }^{15} C_{r}\left(x^{4}\right)^{15-r}\left(\frac{-1}{x^{3}}\right)^{r}$ $=(-1)^{r}{ }^{15} C_{r} x^{60-4 r-3 r}$ For this term to contain $x^{-17}$, we must have : $60-7 r=-17$ $\Rightarrow 7 r=77$ $\Rightarrow r=11$...
Read More →In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1
Question: In the given figure,ABCis a triangle in which A: B: C= 3 : 2 : 1 andACCD. Find the measure of ECD. Solution: Let $\angle A=(3 x)^{\circ}, \angle B=(2 x)^{\circ}$ and $\angle C=x^{\circ}$ From $\triangle A B C$, we have: $\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow 3 x+2 x+x=180^{\circ}$ $\Rightarrow 6 x=180^{\circ}$ $\Rightarrow x=30^{\circ}$ $\begin{aligned} \therefore \angle A=3(30)^{\circ}=60^{\circ} \\ \angle B=2(30)^{\circ}=60^{\cir...
Read More →If the sum of n terms of an A.P. is Sn = 3n2 + 5n.
Question: If the sum ofnterms of an A.P. is Sn= 3n2+ 5n. Write its common difference. Solution: Here, we are given, $S_{n}=3 n^{2}+5 n$ Let us take the first term asaand the common difference asd. Now, as we know, $a_{n}=S_{n}-S_{n-1}$ So, we get, $a_{n}=\left(3 n^{2}+5 n\right)-\left[3(n-1)^{2}+5(n-1)\right]$ $=3 n^{2}+5 n-\left[3\left(n^{2}+1-2 n\right)+5 n-5\right]$$\left[\right.$ Using $\left.(a-b)^{2}=a^{2}+b^{2}-a b\right]$ $=3 n^{2}+5 n-\left(3 n^{2}+3-6 n+5 n-5\right)$ $=3 n^{2}+5 n-3 n^...
Read More →In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.
Question: In the given figure, BAD || EF, AEF = 55 and ACB = 25, find ABC. Solution: In the given figure, EF || BD and CE is the transversal.CAD =AEF (Pair of corresponding angles)⇒CAD =55In∆ABC,CAD =ABC +ACB (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)⇒ 55 =ABC +25⇒ABC =55 25 = 30Thus, the measure of ABC is 30....
Read More →The number of irrational terms in the expansion of
Question: The number of irrational terms in the expansion of $\left(4^{1 / 5}+7^{1 / 10}\right)^{45}$ is (a) 40 (b) 5 (c) 41 (d) none of these Solution: (c) 41 The general term $T_{r+1}$ in the given expansion is given by ${ }^{45} C_{r}\left(4^{1 / 5}\right)^{45-r}\left(7^{1 / 10}\right)^{r}$ For $T_{r+1}$ to be an integer, we must have $\frac{r}{5}$ and $\frac{r}{10}$ as integers i.e. $0 \leq r \leq 45$ $\therefore r=0,10,20,30$ and 40 Hence, there are 5 rational and 41 , i.e., $46-5$, irratio...
Read More →Write the sum of first n even natural numbers.
Question: Write the sum of firstneven natural numbers. Solution: In this problem, we need to find the sum of firstneven natural numbers. So, we know that the first odd natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2. So here, First term (a) = 2 Common difference (d) = 2 So, let us take the number of terms asn Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ So, fornterms, $S_{n}=\frac{n}{2}[2(2)+(n-1) 2]$ $=\frac{n}{2}[4+2 n-2]$ $=\frac{n}{2}(2+2 n...
Read More →In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?
Question: In the given figure,AMBCandANis the bisector of A. If ABC= 70 and ACB= 20, then MAN= ? Solution: In $\Delta A B C$, we have: $\angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle] $\Rightarrow \angle A+70^{\circ}+20^{\circ}=180^{\circ}$ $\Rightarrow \angle A=90^{\circ}$ $\Rightarrow \frac{1}{2} \angle A=45^{\circ}$ $\Rightarrow \angle B A N=45^{\circ}$ $\ln \triangle A B M$, we have: $\angle A B M+\angle A M B+\angle B A M=180^{\circ} \quad$ [Sum of the angles...
Read More →Write the sum of first n odd natural numbers.
Question: Write the sum of firstnodd natural numbers. Solution: In this problem, we need to find the sum of firstnodd natural numbers. So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2. So here, First term (a) = 1 Common difference (d) = 2 So, let us take the number of terms asn Now, as we know, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ So, fornterms, $S_{n}=\frac{n}{2}[2(1)+(n-1) 2]$ $=\frac{n}{2}[2+2 n-2]$ $=\frac{n}{2}(2 n)$ $...
Read More →If A and B are the sums of odd and even terms respectively in the expansion of
Question: IfAandBare the sums of odd and even terms respectively in the expansion of (x+a)n, then (x+a)2n (xa)2nis equal to (a) 4 (A+B) (b) 4 (AB) (c)AB (d) 4AB Solution: (d) 4AB If $A$ and $B$ denote respectively the sums of odd terms and even terms in the expansion $(x+a)^{n}$ Then,$(x+a)^{n}=A+B \quad \ldots(1)$ $(x-a)^{n}=A-B \quad \ldots(2)$ Squaring and subtraction equation (2) from (1) we get $(x+a)^{2 n}-(x-a)^{2 n}=(A+B)^{2}-(A-B)^{2}$ $\Rightarrow(x+a)^{2 n}-(x-a)^{2 n}=4 A B$...
Read More →Write the nth term of an A.P. the sum of whose n terms is Sn.
Question: Write thenth term of an A.P. the sum of whosenterms is Sn. Solution: We are given an A.P. the sum of whosenterms isSn. So, to calculate thenthterm of the A.P. we use following formula, $a_{n}=S_{n}-S_{n-1}$ So, the $n^{\text {th }}$ term of the A.P. is given by $a_{n}=S_{n}-S_{n-1}$....
Read More →In the adjoining figure, show that
Question: In the adjoining figure, show that $\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^{\circ} .$ Solution: In $\Delta A C E$, we have : $\angle A+\angle C+\angle E=180^{\circ} \ldots(i)$ [Sum of the angles of a triangle] In $\Delta B D F$, we have: $\angle B+\angle D+\angle F=180^{\circ} \ldots(i i)$ [Sum of the angles of a triangle] Adding $(i)$ and $(i i)$, we get: $\angle A+\angle C+\angle E+\angle B+\angle D+\angle F=(180+180)^{\circ}$ $\Rightarrow \angle A+\angle B+\angle ...
Read More →If in the expansion of
Question: If in the expansion of (a+b)nand (a+b)n+ 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, thennis (a) 3 (b) 4 (c) 5 (d) 6 Solution: (c)n= 5 Coefficients of the 2 nd and 3 rd terms in $(a+b)^{n}$ are ${ }^{n} C_{1}$ and ${ }^{n} C_{2}$ Coefficients of the 3 rd and 4 th terms in $(a+b)^{n+3}$ are ${ }^{n+3} C_{2}$ and ${ }^{n+3} C_{3}$ Thus, we have $\frac{{ }^{n} C_{1}}{{ }^{n} C_{2}}=\frac{{ }^{n+3} C_{2}}{{ }^{n+3} C_{3}}$ ...
Read More →Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.
Question: Write the value of x for which 2x,x+ 10 and 3x+ 2 are in A.P. Solution: Here, we are given three terms, First term $\left(a_{1}\right)=2 x$ Second term $\left(a_{2}\right)=x+10$ Third term $\left(a_{3}\right)=3 x+2$ We need to find the value ofxfor which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get, $d=a_{2}-a_{1}$ $d=(x+10)-(2 x)$ $d=x+10-2 x$ $d=10-x$ $\ldots(1)$ Also, $d=a_{3}-a_{2}$ $d=(3 x+2)-(x+10)$ $d=3 x+2-x-10$ $d=2...
Read More →If rth term in the expansion of
Question: If $r$ th term in the expansion of $\left(2 x^{2}-\frac{1}{x}\right)^{12}$ is without $x$, then $r$ is equal to (a) 8 (b) 7 (c) 9 (d) 10 Solution: (c) 9 $r$ th term in the given expansion is ${ }^{12} C_{r-1}\left(2 x^{2}\right)^{12-r+1}\left(\frac{-1}{x}\right)^{r-1}$ $=(-1)^{r-1}{ }^{12} C_{r-1} 2^{13-r} x^{26-2 r-r+1}$ For this term to be independent of $x$, we must have : $27-3 r=0$ $\Rightarrow r=9$ Hence, the 9 th term in the expansion is independent of $\mathrm{x}$....
Read More →Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.
Question: Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201. Solution: In the given problem, we need to find the 5thterm from the end for the given A.P. 3, 5, 7, 9 201 Here, to find the 5thterm from the end let us first find the common difference of the A.P. So, First term (a) = 3 Last term (an) = 201 Common difference (d) = Now, as we know, thenthterm from the end can be given by the formula, So, the 5thterm from the end, $a_{5}=201-(5-1) 2$ $=201-(4) 2$ $=201-8$ $=193$ Therefore, th...
Read More →If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.
Question: If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles. Solution: SideBCof triangleABCis produced toD. $\angle A C D=\angle B+\angle A \ldots(i)$ SideACof triangleABCis produced toE. $\angle B A C=\angle B+\angle C \quad \ldots(i)$ And sideABof triangleABCis produced toF. $\angle C B F=\angle C+\angle A \quad \ldots($ iii $)$ Adding $(i),(i i)$ and $(i i i)$, we get: $\angle A C D+\angle B A E+\angle C B F=2(\...
Read More →The term without x in the expansion of
Question: The term without $x$ in the expansion of $\left(2 x-\frac{1}{2 x^{2}}\right)^{12}$ is (a) 495 (b) 495 (c) 7920 (d) 7920 Solution: (d) 7920 Suppose the $(\mathrm{r}+1)$ th term in the given expansion is independent of $x$. Then, we have : $T_{r+1}={ }^{12} C_{r}(2 x)^{12-r}\left(\frac{-1}{2 x^{2}}\right)^{r}$ $=(-1)^{r}{ }^{12} C_{r} \quad 2^{12-2 r} \quad x^{12-r-2 r}$ For this term to be independent of $x$, we must have: $12-3 r=0$ $\Rightarrow r=4$ $\therefore$ Required term : $(-1)^...
Read More →Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
Question: Write the value of $a_{30}-a_{10}$ for the A.P. $4,9,14,19, \ldots .$ Solution: In this problem, we are given an A.P. and we need to find $a_{30}-a_{10}$. A.P. is $4,9,14,19 \ldots$ Here, First term (a) = 4 Common difference of the A.P. (d) Now, as we know, $a_{n}=a+(n-1) d$ Here, we find $a_{30}$ and $a_{20}$. So, for $30^{\text {th }}$ term, $a_{30}=a+(30-1) d$ $=4+(29)(5)$ $=4+145$ $=149$ Also, for $10^{\text {th }}$ term, $a_{20}=a+(10-1) d$ $=4+(9)(5)$ $=4+45$ $=49$ So, $a_{30}-a_...
Read More →If in the expansion of
Question: If in the expansion of (1 +x)20, the coefficients ofrth and (r+ 4)th terms are equal, thenris equal to (a) 7 (b) 8 (c) 9 (d) 10 Solution: (c) 9 Coefficients of the $r$ th and $(r+4)$ th terms in the given expansion are ${ }^{20} C_{r-1}$ and ${ }^{20} C_{r+3}$. Here, ${ }^{20} C_{r-1}={ }^{20} C_{r+3}$ $\Rightarrow r-1+r+3=20 \quad\left[\because\right.$ if ${ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x=y$ or $\left.x+y=n\right]$ $\Rightarrow r=2$ or $2 r=18$ $\Rightarrow r=9$...
Read More →Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
Question: Write the value of $a_{30}-a_{10}$ for the A.P. $4,9,14,19, \ldots .$ Solution: In this problem, we are given an A.P. and we need to find. A.P. is Here, First term (a) = 4 Common difference of the A.P. (d) Now, as we know, $a_{n}=a+(n-1) d$ Here, we finda30anda20. So, for 30thterm, $a_{30}=a+(30-1) d$ $=4+(29)(5)$ $=4+145$ $=149$ Also, for 10thterm, $a_{20}=a+(10-1) d$ $=4+(9)(5)$ $=4+45$ $=49$ So, $a_{30}-a_{10}=149-49$ $=100$ Therefore, for the given A.P $a_{30}-a_{10}=100$....
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