If the fifth term of a G.P. is 2, then write the product of its 9 terms.
Question: If the fifth term of a G.P. is 2, then write the product of its 9 terms. Solution: Here, $a_{5}=2$ $\Rightarrow a r^{4}=2$ Product of the nine terms, i.e. $a, a r, a r^{2}, a r^{3}, a r^{4}, a r^{5}, a r^{6}, a r^{7}$ and $a r^{8}$ : $\left(a \times a r^{8}\right)\left(a r \times a r^{7}\right)\left(a r^{2} \times a r^{6}\right)\left(a r^{3} \times a r^{5}\right)\left(a r^{4}\right)=\left(a r^{4}\right)^{9}$ $\because a r^{4}=2$ Required product $=2^{9}=512$...
Read More →In the given figure, ABCD and ABEF are two cyclic quadrilaterals.
Question: In the given figure,ABCDandABEFare two cyclic quadrilaterals. IfBCD= 110, then BEF= ?(a) 55(b) 70(c) 90(d) 110 Solution: (d) 110Since ABCD is a cyclic quadrilateral, we have:BAD + BCD = 180⇒BAD + 110 = 180⇒BAD= (180 - 110) = 70Similarly inABEF, we have:BAD + BEF = 180⇒ 70 + BEF = 180⇒BEF = (180 - 70) = 110...
Read More →The third term of a G.P. is 4, the product of the first five terms
Question: The third term of a G.P. is 4, the product of the first five terms is ____________. Solution: Given third term of G.P is 4 Letabe the first term of G.P andrdenote the common ratio. Then second term isar Third term isar2 fourth term isar3and fifth term isar4. Sincear2= 4 Product of first 5 terms isa.ar.ar2ar3ar4 =a5r10 = (ar2)5 Hence, product of first 5 terms = 45...
Read More →In the given figure, O is the centre of a circle and ∠AOB = 130°.
Question: In the given figure,Ois the centre of a circle andAOB= 130. Then, ACB= ?(a) 50(b) 65(c) 115(d) 155 Solution: (c) 115Join AB.Then chord AB subtends AOB at the centre and ADB at a point D of the remaining parts of a circle. AOB= 2ADB $\Rightarrow \angle \mathrm{ADB}=\frac{1}{2} \angle \mathrm{AOB}=\left(\frac{1}{2} \times 130^{\circ}\right)=65^{\circ}$ In cyclic quadrilateral, we have:ADB + ACB = 180⇒ 65 + ACB = 180ACB= (180 - 65) = 115...
Read More →If a, b, c are in G.P. then the value of
Question: If $a, b, c$ are in G.P. then the value of $\frac{a-b}{b-c}$ is equal to __________________ Solution: Leta, bandcbe in g.p i.eb=ar c=ar2 i. e $\frac{a-b}{b-c}=\frac{a-a r}{a r-a r^{2}}$ $=\frac{a(1-r)}{a\left(r-r^{2}\right)}$ $=\frac{1-r}{r(1-r)}$ $\therefore \frac{a-b}{b-c}=\frac{1}{r}$...
Read More →In the given figure, O is the centre of a circle and ∠AOB = 140°.
Question: In the given figure,Ois the centre of a circle andAOB= 140. Then, ACB= ?(a) 70(b) 80(c) 110(d) 40 Solution: (c) 110Join AB.Then chord AB subtends AOB at the centre and ADB at a point D of the remaining parts of a circle. AOB= 2ADB $\Rightarrow \angle \mathrm{ADB}=\frac{1}{2} \angle \mathrm{AOB}=\left(\frac{1}{2} \times 140^{\circ}\right)=70^{\circ}$ In the cyclic quadrilateral, we have:ADB + ACB = 180⇒ 70 + ACB = 180ACB= (180 - 70) = 110...
Read More →The minimum value of the expression
Question: The minimum value of the expression 3x+ 31 x,xR, is __________. Solution: Minimum value of 3x+ 31 x;xR Since arithmetic mean geometric mean of3xand 31 x $\therefore \frac{3 x+3^{1-x}}{2} \geq \sqrt{3^{x} 3^{1-x}}$ i. e $\frac{3^{x}+3^{1-x}}{2} \geq \sqrt{3^{x} 33^{-x}}$ i. e $\frac{3^{x}+3^{1-x}}{2}=\sqrt{3}$ i. e $3^{x}+3^{1-x} \geq 2 \sqrt{3}$ i.e minimum possible value of $3^{x}+3^{1-x}$ is $2 \sqrt{3}$....
Read More →The product of n geometric means between a and b
Question: The product ofngeometric means betweenaandbis _________. Solution: To find:- Product ofngeometric means between Let us supposeg1,g2, ............,gnrepresent geometric means betweenaandb. a = a i.eg1=ar g2=ar2 gn=arn b=arn+1 i. e $r^{n+1}=\frac{b}{a}$ $\Rightarrow g_{1} g_{2} g_{3-----} g_{n}=\left(a r^{\cdot}\right)\left(a r^{2}\right)\left(a r^{3}\right)_{---_{-}} a r^{n}$ $=a^{n} r^{1+2+3+_{--+}}$ $=a^{n} r^{\frac{n(n+1)}{2}} \quad\left[\because 1+2+_{---}+n=\frac{n(n+1)}{2}\right]$...
Read More →In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If ∠CBE = 100°, then ∠CDE = ?
Question: In the given figure, sidesABandADof quad.ABCDare produced toEandFrespectively. IfCBE= 100, then CDE= ?(a) 100(b) 80(c) 130(d) 90 Solution: (b) 80In a cyclic quadrilateral ABCD, we have:Interior opposite angle,ADC = exterior CBE = 100CDF = (180 - ADC) = (180 - 100) = 80 (Linear pair)...
Read More →The product of n geometric means between a and b
Question: The product ofngeometric means betweenaandbis _________. Solution: To find:- Product ofngeometric means between Let us supposeg1,g2, ............,gnrepresent geometric means betweenaandb. a = a i.eg1=ar g2=ar2 gn=arn b=arn+1 i. e $r^{n+1}=\frac{b}{a}$ $\Rightarrow g_{1} g_{2} g_{3-----} g_{n}=\left(a r^{\cdot}\right)\left(a r^{2}\right)\left(a r^{3}\right)_{---_{-}} a r^{n}$ $=a^{n} r^{1+2+3+_{--+}}$ $=a^{n} r^{\frac{n(n+1)}{2}} \quad\left[\because 1+2+_{---}+n=\frac{n(n+1)}{2}\right]$...
Read More →If the points A (a, −11), B (5, b), C(2, 15) and D (1, 1)
Question: If the points A (a, 11), B (5, b), C(2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values ofaandb. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (a,11); B (5,b); C (2, 15) and D (1, 1). Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\ri...
Read More →In the given figure, equilateral ∆ABC is inscribed in a circle and ABCD is a quadrilateral,
Question: In the given figure, equilateral∆ABCis inscribed in a circle andABCDis a quadrilateral, as shown. Then, BDC= ?(a) 90(b) 60(c) 120(d) 150 Solution: (c) 120Since ΔABC is an equilateral triangle, each of its angle is 60.BAC = 60In a cyclic quadrilateral ABCD, we have:BAC +BDC = 180⇒ 60 +BDC = 180⇒BDC = (180 - 60) = 120...
Read More →If the sum of the series 3 + 3x + 3x
Question: If the sum of the series $3+3 x+3 x^{2}+$ ____________ to $\infty$ is $\frac{45}{8}$, than $x=$ _______________ Solution: Sum of series $3+3 x+3 x^{2}+\ldots \quad=\frac{45}{8} \quad$ (given) Sincea =3 r = x and sum of infinite g.p is $\frac{a}{1-r}$ $\therefore \frac{a}{1-r}=\frac{45}{8}$ $\Rightarrow \frac{3}{1-x}=\frac{45}{8}$ $8 \times 3=45(1-x)$ i.e $8=15(1-x)$ $8=15-15 x$ i. e $15 x=7$ i. e $x=\frac{7}{15}$...
Read More →In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and ∠CBD = 35°.
Question: In the give figure,ABCDis a cyclic quadrilateral in whichBC=CDandCBD= 35. Then, BAD= ?(a) 65(b) 70(c) 110(d) 90 Solution: (b) 70BC = CD (given)⇒BDC =CBD = 35InΔ BCD, we have:BCD + BDC + CBD = 180 (Angle sum property of a triangle)⇒BCD + 35 + 35 = 180⇒BCD = (180 - 70) = 110In cyclic quadrilateral ABCD, we have:BAD + BCD = 180⇒BAD + 110 = 180BAD = (180 - 110) = 70...
Read More →If three consecutive vertices of a parallelogram are
Question: If three consecutive vertices of a parallelogram are (1, 2), (3, 6) and (5, 10), find its fourth vertex. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1,2); B (3, 6) and C(5, 10). We have to find the co-ordinates of the forth vertex. Let the forth vertex be Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. Now to find the mid-pointof two pointsandwe use se...
Read More →In the given figure, O is the centre of a circle and ∠OAB = 50°.
Question: In the given figure,Ois the centre of a circle andOAB= 50. Then, BOD= ?(a) 130(b) 50(c) 100(d) 80 Solution: (c) 100OA = OB (Radii of a circle)⇒OBA = OAB = 50InΔ OAB, we have: OAB + OBA + AOB = 180 (Angle sum property of a triangle)⇒50 + 50 + AOB = 180⇒AOB = (180 - 100) = 80Since AOB + BOD = 180 (Linear pair)BOD = (180 - 80) = 100...
Read More →The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4)
Question: The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not. Solution: Let A (2, 0); B (9, 1); C (11, 6) and D (4, 4) be the vertices of a quadrilateral. We have to check if the quadrilateral ABCD is a rhombus or not. So we should find the lengths of sides of quadrilateral ABCD. $\mathrm{AB}=\sqrt{(9-2)^{2}+(1-0)^{2}}$ $=\sqrt{49+1}$ $=\sqrt{50}$ $B C=\sqrt{(11-9)^{2}+(6-1)^{2}}$ $=\sqrt{4+25}$ $=\sqrt{29}$ $\mat...
Read More →The value of the product (32) × (32)
Question: The value of the product (32) (32)1/6 (32)1/36 _________ to , is _________. Solution: $(32) \times(32)^{\frac{1}{6}} \times(32)^{\frac{1}{36}} \times \ldots \ldots \ldots \infty$ $=(32)^{\left(1+\frac{1}{6}+\frac{1}{36}+\ldots \ldots\right)}$ Since $1+\frac{1}{6}+\frac{1}{6^{2}}+\ldots \ldots$ from an g. $\mathrm{p}$ $=32^{\left(\frac{1}{1-\frac{1}{6}}\right)}$ $=(32)^{\left(\frac{6}{5}\right)}$ $=\left(2^{5}\right)^{\frac{6}{5}}$ $=2^{5 \times \frac{6}{5}}$ $=2^{6}=64$ i. e $(32) \tim...
Read More →if
Question: If $A=\left[\begin{array}{rrr}2 4 -1 \\ -1 0 2\end{array}\right], B=\left[\begin{array}{rr}3 4 \\ -1 2 \\ 2 1\end{array}\right]$, find $(A B)^{T}$ Solution: Here, $A B=\left[\begin{array}{ccc}2 4 -1 \\ -1 0 2\end{array}\right]\left[\begin{array}{cc}3 4 \\ -1 2 \\ 2 1\end{array}\right]$ $\Rightarrow A B=\left[\begin{array}{cc}6-4-2 8+8-1 \\ -3-0+4 -4+0+2\end{array}\right]$ $\Rightarrow A B=\left[\begin{array}{cc}0 15 \\ 1 -2\end{array}\right]$ $\Rightarrow(A B)^{T}=\left[\begin{array}{c...
Read More →In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad.
Question: In the given figure,Ois the centre of a circle in whichAOC= 100. SideABof quad.OABChas been produced toD. Then, CBD= ?(a) 50(b) 40(c) 25(d) 80 Solution: (a) 50Take a point E on the remaining part of the circumference.Join AE and CE. Then $\angle \mathrm{AEC}=\frac{1}{2} \angle \mathrm{AOC}=\left(\frac{1}{2} \times 100^{\circ}\right)=50^{\circ}$ Now, side AB of the cyclic quadrilateral ABCE has been produced to D. ExteriorCBD = AEC = 50...
Read More →If the points (−2, −1), (1, 0), (x, 3) and
Question: If the points (2, 1), (1, 0), (x, 3) and (1,y) form a parallelogram, find the values ofxandy. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (2,1); B (1, 0); C (x, 3) and D (1,y). Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}...
Read More →The sum of infinity of the series
Question: The sum of infinity of the series $9-3+1-\frac{1}{3}+$ , is Solution: $9-3+1-\frac{1}{3}+\ldots \ldots \ldots \ldots$ Here $a=9$ $r=\frac{a_{2}}{a_{1}}=\frac{-3}{9}=\frac{-1}{3}$ $r=\frac{a_{3}}{a_{2}}=\frac{1}{-3}=\frac{-1}{3}$ Sum of infinite terms of G.P is $S_{n}=\frac{a}{1-r}$ $S_{n}=\frac{9}{1-\left(\frac{-1}{3}\right)}$ $=\frac{9}{\frac{3+1}{3}}$ $=\frac{9 \times 3}{4}$ $S_{n}=\frac{27}{4}$ i.e the sum of infinity of the series...
Read More →The sum of infinity of the series
Question: The sum of infinity of the series $9-3+1-\frac{1}{3}+ ___________$ is_____________ Solution: $9-3+1-\frac{1}{3}+\ldots \ldots \ldots \ldots$ Here $a=9$ $r=\frac{a_{2}}{a_{1}}=\frac{-3}{9}=\frac{-1}{3}$ $r=\frac{a_{3}}{a_{2}}=\frac{1}{-3}=\frac{-1}{3}$ Sum of infinite terms of G.P is $S_{n}=\frac{a}{1-r}$ $S_{n}=\frac{9}{1-\left(\frac{-1}{3}\right)}$ $=\frac{9}{\frac{3+1}{3}}$ $=\frac{9 \times 3}{4}$ $S_{n}=\frac{27}{4}$ i.e the sum of infinity of the series...
Read More →The line segment joining the points P(3, 3) and Q(6, −6) is trisected
Question: The line segment joining the pointsP(3, 3) andQ(6, 6) is trisected at the pointsAandBsuch thatAis nearer toP. If A also lies on the line given by 2x+y+k= 0, find the value ofk. Solution: We have two points P (3, 3) and Q (6,6). There are two points A and B which trisect the line segment joining P and Q. Let the co-ordinate of $\mathrm{A}$ be $\mathrm{A}\left(x_{1}, y_{1}\right)$ Now according to the section formula if any point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\ri...
Read More →In the given figure, AOB is a diameter of a circle and CD || AB.
Question: In the given figure,AOBis a diameter of a circle andCD||AB. IfBAD= 30, then CAD= ?(a) 30(b) 60(c) 45(d) 50 Solution: (a) 30ADC = BAD = 30 (Alternate angles)ADB = 90 (Angle in semicircle) CDB = (90 + 30) = 120But ABCD being a cyclic quadrilateral, we have:BAC + CDB = 180⇒BAD +CAD +CDB = 180⇒ 30 +CAD + 120 = 180⇒ CAD = (180 - 150) = 30...
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