2 + 5 + 10 + 17 + 26 + ...
Question: 2 + 5 + 10 + 17 + 26 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ $0=2+\left[3+5+7+9+\ldots+\left(T_{n}-T_{n-1}\right)\right]-T_{n}$ The sequence of difference of s...
Read More →2 + 5 + 10 + 17 + 26 + ...
Question: 2 + 5 + 10 + 17 + 26 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ $0=2+\left[3+5+7+9+\ldots+\left(T_{n}-T_{n-1}\right)\right]-T_{n}$ The sequence of difference of s...
Read More →Write the ratio in which the line segment joining points
Question: Write the ratio in which the line segment joining points (2, 3) and (3, 2) is divided by X axis. Solution: Let $P(x, 0)$ be the point of intersection of $x$-axis with the line segment joining $A(2,3)$ and $B(3,-2)$ which divides the line segment AB in the ratio $\lambda: 1$. Now according to the section formula if point a point $\mathrm{P}$ divides a line segment joining $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ in the ratio $\mathrm{m}$ : $\mathrm...
Read More →The distance between an object and a screen is 100 cm.
Question: The distance between an object and a screen is $100 \mathrm{~cm}$. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is $40 \mathrm{~cm}$. If the power of the lens is close to $\left(\frac{N}{100}\right) D$ where $N$ is an integer, the value of $N$ is_________ Solution: $(476.19)$ Given, Distance between an object and screen, $D=100 \mathrm{~cm}$ Distance between the two posi...
Read More →Write the perimeter of the triangle formed
Question: Write the perimeter of the triangle formed by the pointsO(0, 0),A(a, 0) and B (0, b). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The perimeter of a triangle is the sum of lengths of its sides. The three vertices of the given triangle are O(0,0), A(a,0) and B(0, b). Let us now find the lengths of the sides of the triangle. $O A=\sqrt...
Read More →In a compound microscope,
Question: In a compound microscope, the magnified virtual image is formed at a distance of $25 \mathrm{~cm}$ from the eye-piece. The focal length of its objective lens is $1 \mathrm{~cm}$. If the magnification is 100 and the tube length of the microscope is $20 \mathrm{~cm}$, then the focal length of the eye-piece lens (in $\mathrm{cm}$ ) is Solution: $(4.48)$ According to question, final image i.e., $v_{2}=25 \mathrm{~cm}$, $f_{0}=1 \mathrm{~cm}$, magnification, $m=m_{1} m_{2}=100^{2}$ Using le...
Read More →Write the distance between the points
Question: Write the distance between the points A (10 cos , 0) and B (0, 10 sin ). Solution: We have to find the distance between $\mathrm{A}(10 \cos \theta, 0)$ and $\mathrm{B}(0,10 \sin \theta)$. In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ So, $\mathrm{AB}=\sqrt{(10 \cos \theta-0)^{2}+(0-10 \sin \theta)^{2}}$ $=\sqrt{10^{2}\left(\sin ^...
Read More →3 + 5 + 9 + 15 + 23 + ...
Question: 3 + 5 + 9 + 15 + 23 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum to $n$ terms of the given series. Thus, we have: $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n} \ldots(2)$ On subtracting (2) from (1), we get: $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$ $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$ $0=3+\left[2+4+6+8+\ldots+\left(T_{n}-T_{n-1}\right)\right]-T_{n}$ The sequence of difference of suc...
Read More →Find the area of a parallelogram ABCD if three of its vertices
Question: Find the area of a parallelogram $A B C D$ if three of its vertices are $A(2,4), B(2+\sqrt{3}, 5)$ and $C(2,6)$. Solution: It is given that $\mathrm{A}(2,4), \mathrm{B}(2+\sqrt{3}, 5)$ and $\mathrm{C}(2,6)$ are the vertices of the parallelogram $\mathrm{ABCD}$. We know that the diagonal of a parallelogram divides it into two triangles having equal area. $\therefore$ Area of the parallogram $\mathrm{ABCD}=2 \times$ Area of the $\triangle \mathrm{ABC}$ Now, $\operatorname{ar}(\Delta \mat...
Read More →Show that the following points are the vertices of a square.
Question: Show that the following points are the vertices of a square.(i)A(3, 2),B(0, 5),C(3, 2) andD(0, 1) (ii)A(6, 2),B(2, 1),C(1, 5) andD(5, 6) (iii)P(0, 2),Q(3, 1),R(0, 4) andS(3, 1) Solution: (i) The given points areA(3, 2),B(0, 5),C(3, 2) andD(0, 1). $A B=\sqrt{(0-3)^{2}+(5-2)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units $B C=\sqrt{(-3-0)^{2}+(2-5)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units $C D=\sqrt{(0+3)^{2}+(-1-2)^{2}}=\sqrt{(3)^{2}+(-3)...
Read More →Find the 20th term and the sum of 20 terms of the series
Question: Find the 20thterm and the sum of 20 terms of the series 2 4 + 4 6 + 6 8 + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=2 n(2 n+2)=4 n^{2}+4 n$ Forn= 20, we have: $T_{20}=4 n^{2}+4 n$ $=4(20)^{2}+4(20)$ $=1600+80$ $=1680$ Therefore, the 20 th term of the given series is 1680 . Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+4 k\right)$ $\Righ...
Read More →When an object is kept at a distance of 30 cm from a concave mirror,
Question: When an object is kept at a distance of $30 \mathrm{~cm}$ from a concave mirror, the image is formed at a distance of $10 \mathrm{~cm}$ from the mirror. If the object is moved with a speed of $9 \mathrm{cms}^{-1}$, the speed (in $\mathrm{cms}^{-1}$ ) with which image moves at that instant is______ Solution: (1) Distance of object, $u=-30 \mathrm{~cm}$ Distance of image, $v=10 \mathrm{~cm}$ Magnification, $m=\frac{-v}{u}=\frac{(-10)}{-30}=\frac{1}{3}$ Speed of image $=m^{2} \times$ spee...
Read More →If the vertices of a triangle are (1, −3), (4, p) and (−9, 7)
Question: If the vertices of a triangle are (1, 3), (4,p) and (9, 7) and its area is 15 sq. units, find the value(s) ofp. Solution: Let A(1, 3), B(4,p) and C(9, 7) be the vertices of the ∆ABC.Here,x1= 1,y1= 3;x2= 4,y2=pandx3= 9,y3= 7ar(∆ABC) = 15 square units $\Rightarrow \frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|=15$ $\Rightarrow \frac{1}{2}|1(p-7)+4[7-(-3)]+(-9)(-3-p)|=15$ $\Rightarrow \frac{1}{2}|p-7+40+27+9 p|=15$ $\Right...
Read More →An observer can see through a small hole on the side of a jar (radius 15 cm )
Question: An observer can see through a small hole on the side of a jar (radius $15 \mathrm{~cm}$ ) at a point at height of $15 \mathrm{~cm}$ from the bottom (see figure). The hole is at a height of $45 \mathrm{~cm}$. When the jar is filled with a liquid up to a height of $30 \mathrm{~cm}$ the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is $N / 100$, where $N$ is an integer, the value of $N$ is_________ Solution: (158) From figure, $\sin i=\frac...
Read More →Find the sum of the series whose nth term is:
Question: Find the sum of the series whosenth term is: (i) 2n2 3n+ 5 (ii) 2n3+ 3n2 1 (iii)n3 3n (iv)n(n+ 1) (n+ 4) (v) (2n 1)2 Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: (i) $T_{n}=2 n^{2}-3 n+5$ Let $S_{n}$ be the sum of $n$ terms of the given series. Now, $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{2}-3 k+5\right)$ $\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{2}-3 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 5$ $\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)...
Read More →Find the value of a for which the area of the triangle
Question: Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(2, 6) and C(3, 1) is 10 square units. Solution: The formula for the area 'A' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ The three given points areA(a,2a), B(2,6) andC(3,1). It...
Read More →Show that the points O(0, 0) A
Question: Show that the points $O(0,0) A(3, \sqrt{3})$ and $B(3,-\sqrt{3})$ are the vertices of an equilateral triangle. Find the area of this triangle. Solution: The given points are $O(0,0) A(3, \sqrt{3})$ and $B(3,-\sqrt{3})$. $O A=\sqrt{(3-0)^{2}+\{(\sqrt{3})-0\}^{2}}=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{9+3}=\sqrt{12}=2 \sqrt{3}$ units $A B=\sqrt{(3-3)^{2}+(-\sqrt{3}-\sqrt{3})^{2}}=\sqrt{0+(2 \sqrt{3})^{2}}=\sqrt{4(3)}=\sqrt{12}=2 \sqrt{3}$ units $O B=\sqrt{(3-0)^{2}+(-\sqrt{3}-0)^{2}}=\sqrt...
Read More →Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Question: Show that the pointsA(5, 6)B(3, 0) andC(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area. Solution: Let the given points beA(5, 6)B(3, 0) andC(9, 8). $A B=\sqrt{(3-(-5))^{2}+(0-6)^{2}}=\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}=\sqrt{100}=10$ units $B C=\sqrt{(9-3)^{2}+(8-0)^{2}}=\sqrt{(6)^{2}+(8)^{2}}=\sqrt{36+64}=\sqrt{100}=10$ units $A C=\sqrt{(9-(-5))^{2}+(8-6)^{2}}=\sqrt{(14)^{2}+(2)^{2}}=\sqrt{196+4}=\sqrt{200}=10 \sqrt{2}$ units Therefore, $A B=B C=10$ ...
Read More →Find the value of k, if the points A (8, 1) B(3, −4)
Question: Find the value of k, if the points A (8, 1) B(3, 4) and C(2, k) are collinear. Solution: The formula for the area 'A' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal to 0. The three given points areA(...
Read More →A light ray enters a solid glass sphere of refractive index
Question: A light ray enters a solid glass sphere of refractive index $\mu=\sqrt{3}$ at an angle of incidence $60^{\circ}$. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is________ Solution: $(\mathbf{9 0 . 0 0})$ In the figure, $Q R$ is the reflected ray and $Q S$ is refracted ray. $C Q$ is normal. Apply Snell's law at $P$ $1 \sin 60^{\circ}=\sqrt{3} \sin r$ $\Rightarrow \sin r=\frac{...
Read More →If R (x, y) is a point on the line segment joining the points
Question: If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove thatx+y=a+b. Solution: The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula, $\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$ If three points are collinear the area encompassed by them is equal ...
Read More →Show that the points (− 3, − 3), (3, 3) and
Question: Show that the points $(-3,-3),(3,3)$ and $(-3 \sqrt{3}, 3 \sqrt{3})$ are the vertices of an equilateral triangle. Solution: Let the given points be $A(-3,-3), B(3,3)$ and $C(-3 \sqrt{3}, 3 \sqrt{3})$. Now $A B=\sqrt{(-3-3)^{2}+(-3-3)^{2}}=\sqrt{(-6)^{2}+(-6)^{2}}$ $=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}$ $B C=\sqrt{(3+3 \sqrt{3})^{2}+(3-3 \sqrt{3})^{2}}$ $=\sqrt{9+27+18 \sqrt{3}+9+27-18 \sqrt{3}}=\sqrt{72}=6 \sqrt{2}$ $A C=\sqrt{(-3+3 \sqrt{3})^{2}+(-3-3 \sqrt{3})^{2}}=\sqrt{(3-3 \sqrt{3})...
Read More →Solve the following
Question: 3 12+ 5 22+ 7 32+ ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=(2 n+1) n^{2}=2 n^{3}+n^{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)$ $\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}$ $\Rightarrow S_{n}=\left[\frac{2 n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2 n+1)}{6}\right]$ $\Rightarrow S_{n}=\left[\frac{n^{2}(n+1)...
Read More →A spherical mirror is obtained as shown in the figure from a hollow glass sphere.
Question: A spherical mirror is obtained as shown in the figure from a hollow glass sphere. If an object is positioned in front of the mirror, what will be the nature and magnification of the image of the object? (Figure drawn as schematic and not to scale)Inverted, real and magnifiedErect, virtual and magnifiedErect, virtual and unmagnifiedInverted, real and unmagnifiedCorrect Option: 4 Solution: (4) Object is placed beyond radius of curvature $(R)$ of concave mirror hence image formed is real,...
Read More →1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...
Question: 1 2 + 2 3 + 3 4 + 4 5 + ... Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=n(n+1)=n^{2}+n$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} T_{k}$ $\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{2}+k\right)$ $\Rightarrow S_{n}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $\Rightarrow S_{n}=\left(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right)$ $\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{2 n+1}{3}+1\right)$ $\R...
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