If ∑ n = 210,
Question: If n= 210, then n2= (a) 2870 (b) 2160 (c) 2970 (d) none of these Solution: (a) 2870 Given: n= 210 $\Rightarrow n\left(\frac{n+1}{2}\right)=210$ $\Rightarrow n^{2}+n-420=0$ $\Rightarrow(n-20)(n+21)=0$ $\Rightarrow n=20$$\quad(\because n0)$ Now, $\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$ $\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}$ $\Rightarrow \frac{n(n+1)}{2} \times \frac{(2 n+1)}{3}$ $\Rightarrow(210) \times\left(\frac{41}{3}\right)$ $\Rightarrow(70) \times(41)$ $\Rightarrow 2870$...
Read More →The value of
Question: The value of $\sum_{r=1}^{n}\left\{(2 r-1) a+\frac{1}{b^{r}}\right\}$ is equal to (a) $a n^{2}+\frac{b^{n-1}-1}{b^{n-1}(b-1)}$ (b) $a n^{2}+\frac{b^{n}-1}{b^{n}(b-1)}$ (c) $a n^{3}+\frac{b^{n-1}-1}{b^{n}(b-1)}$ (d) none of these Solution: (b) $a n^{2}+\frac{b^{n}-1}{b^{n}(b-1)}$ we have, $\sum_{r=1}^{n}\left\{(2 r-1) a+\frac{1}{b^{r}}\right\}$ $=\sum_{r=1}^{n}\left\{2 r a-a+\frac{1}{b^{r}}\right\}$ $=\sum_{r=1}^{n} 2 a r-\sum_{r=1}^{n} a+\sum_{r=1}^{n} \frac{1}{b^{r}}$ $=a n(n+1)-a n+\...
Read More →What is the distance between the points
Question: What is the distance between the pointsA(c, 0) andB(0, c)? Solution: We have to find the distance between $\mathrm{A}(c, 0)$ and $\mathrm{B}(0,-c)$. 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{(c-0)^{2}+(0-c)^{2}}$ $=c \sqrt{2}$...
Read More →If the coordinates of points A and B are (−2, −2) and (2, −4) respectively,
Question: If the coordinates of pointsAandBare (2, 2) and (2, 4) respectively, find the coordinates of the pointP such that $A P=\frac{3}{7} A B$, where $P$ lies on the line segment $A B$. Solution: The coordinates of the points $A$ and $B$ are $(-2,-2)$ and $(2,-4)$ respectively, where $A P=\frac{3}{7} A B$ and $P$ lies on the line segment $A B$. So $A P+B P=A B$ $\Rightarrow A P+B P=\frac{7 A P}{3} \quad \because A P=\frac{3}{7} A B$ $\Rightarrow B P=\frac{7 A P}{3}-A P$ $\Rightarrow \frac{A P...
Read More →The sum of the series $
Question: The sum of the series $\frac{1}{\log _{2} 4}+\frac{1}{\log _{4} 4}+\frac{1}{\log _{8} 4}+\ldots+\frac{1}{\log _{2}{ }^{n} 4}$ is (a) $\frac{n(n+1)}{2}$ (b) $\frac{n(n+1)(2 n+1)}{12}$ (c) $\frac{n(n+1)}{4}$ (d) none of these Solution: (c) $\frac{n(n+1)}{4}$ Let $S_{n}=\frac{1}{\log _{2} 4}+\frac{1}{\log _{4} 4}+\frac{1}{\log _{8} 4}+\ldots+\frac{1}{\log _{2}{ }_{2} 4}$ $\Rightarrow S_{n}=\frac{\log 2}{\log 4}+\frac{\log 4}{\log 4}+\frac{\log 8}{\log 4}+\ldots+\frac{\log 2^{n}}{\log 4}$ ...
Read More →Find the value of a so that the point (3, a)
Question: Find the value ofaso that the point (3,a) lies on the line represented by 2x 3y+ 5 = 0 Solution: If a point $\left(x_{1}, y_{1}\right)$ is said lie on a line represented by $a x+b y+c=0$, then the given equation of the line should hold true when the values of the co-ordinates of the points are substituted in it. Here it is said that the point $(3, a)$ lies on the line represented by the equation $2 x-3 y+5=0$. Substituting the co-ordinates of the values in the equation of the line we h...
Read More →Find the distance between the points
Question: Find the distance between the points $\left(-\frac{8}{5}, 2\right)$ and $\left(\frac{2}{5}, 2\right)$ Solution: We have to find the distance between $\mathrm{A}\left(-\frac{8}{5}, 2\right)$ and $\mathrm{B}\left(\frac{2}{5}, 2\right)$. 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, $A B=\sqrt{\left(\frac{2}{5}+\frac{8}{5}\right...
Read More →The sum to n terms of the series
Question: The sum to $n$ terms of the series $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ldots+\ldots$ is (a) $\sqrt{2 n+1}$ (b) $\frac{1}{2} \sqrt{2 n+1}$ (c) $\sqrt{2 n+1}-1$ (d) $\frac{1}{2}|\sqrt{2 n+1}-1|$ Solution: (d) $\frac{1}{2}|\sqrt{2 n+1}-1|$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\frac{1}{\sqrt{2 n-1}+\sqrt{2 n+1}}=\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}$ Now, Let $S_{n}$ be the sum of $n$ terms of the given ser...
Read More →Find the coordinates of the points of trisection of the line segment joining the points A(7, –2) and B(1, –5).
Question: Find the coordinates of the points of trisection of the line segment joining the pointsA(7, 2) andB(1, 5). Solution: Consider the figure.Here pointsPandQ trisectAB.Therefore,Pdivides AB into 1 : 2 andQdivides AB into 2 : 1.Using section formula, coordinates of P are; $\mathrm{P}(x, y)=\left(\frac{1 \times 1+2 \times 7}{3}, \frac{1 \times-5+2 \times-2}{3}\right)$ $\mathrm{P}(x, y)=\left(\frac{15}{3}, \frac{-9}{3}\right)=(5,-3)$ Similarly, coordinates of Q are; $\mathrm{Q}(a, b)=\left(\f...
Read More →Write the ratio in which the line segment doining the points A (3, −6),
Question: Write the ratio in which the line segment doining the points A (3, 6), and B (5, 3) is divided by X-axis. Solution: Let $\mathrm{P}(x, 0)$ be the point of intersection of $x$-axis with the line segment joining $\mathrm{A}(3,-6)$ and $\mathrm{B}(5,3)$ which divides the line segment $\mathrm{AB}$ in the ratio $\lambda: 1$. Now according to the section formula if point a point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ in the ratio $m$...
Read More →If we need a magnification of 375 from a compound microscope of tube length 150 mm
Question: If we need a magnification of 375 from a compound microscope of tube length $150 \mathrm{~mm}$ and an objective of focal length $5 \mathrm{~mm}$, the focal length of the eye-piece, should be close to:$22 \mathrm{~mm}$$12 \mathrm{~mm}$$2 \mathrm{~mm}$$33 \mathrm{~mm}$Correct Option: 1 Solution: (1) According question, $M=375$ $L=150 \mathrm{~mm}, f_{0}=5 \mathrm{~mm}$ and $f_{e}=$ ? Using, magnification, $M \simeq \frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)$ $\Rightarrow 375=\frac{150...
Read More →Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.
Question: (i) Find the coordinates of the point that divides the join ofA(1, 7) andB(4, 3) in the ratio 2 : 3.(ii) Find the coordinates of the point which divides the join ofA(5, 11) andB(4, 7) in the ratio 7 : 2. Solution: (i) The end points ofABareA(1, 7) andB(4, 3).Therefore, (x1= 1,y1= 7) and (x2= 4,y2= 3)Also,m= 2 andn= 3Let the required point beP(x,y).By section formula, we get: $x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)}$ $\Rightarrow x=\fra...
Read More →Solve the following
Question: $\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.14}+\frac{1}{14.19}+\ldots+\frac{1}{(5 n-4)(5 n+1)}$ Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\frac{1}{(5 n-4)(5 n+1)}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} \frac{1}{(5 k-4)(5 k+1)}$ $=\frac{1}{5} \sum_{k=1}^{n}\left(\frac{1}{(5 k-4)}-\frac{1}{(5 k+1)}\right)$ $=\frac{1}{5} \sum_{k=1}^{n} \frac{1}{(5 k-4)}-\frac{1}{5} \sum_{k=1}^{n} \frac{1}{...
Read More →Find the values of x for which the distance between
Question: Find the values of x for which the distance between the point P(2, 3), and Q (x, 5) is 10. Solution: It is given that distance between $P(2,-3)$ and $Q(x, 5)$ is 10 . 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, $\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$ So, $10^{2}=(x-2)^{2}+(5+3)^{2}$ On further simplification, $(x-2)^{2}=36$ $x=2 \pm 6$ $=8,-4$...
Read More →Solve the following
Question: $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots$ Solution: Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\frac{1}{(3 n-2)(3 n+1)}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n} \frac{1}{(3 k-2)(3 k+1)}$ $=\frac{1}{3} \sum_{k=1}^{n}\left(\frac{1}{(3 k-2)}-\frac{1}{(3 k+1)}\right)$ $=\frac{1}{3} \sum_{k=1}^{n} \frac{1}{(3 k-2)}-\frac{1}{3} \sum_{k=1}^{n} \frac{1}{(3 k+1)}$ $=\frac{1}{3}\left[\left(1+\frac{...
Read More →Write the condition of collinearity of points
Question: Write the condition of collinearity of points (x1, y1), (x2, y2) and (x3, y3). Solution: The condition for co linearity of three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is that the area enclosed by them should be equal to 0 . The formula for the area 'A' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{1}, y_{3}\right)$ is given by the formula, $\mathrm{A}=\frac{1}{2}\left|\begin{array}{...
Read More →2 + 4 + 7 + 11 + 16 + ...
Question: 2 + 4 + 7 + 11 + 16 + ... Solution: Let $S_{n}$ be the sum of $n$ terms and $T_{n}$ be the $n$th term of the given series. Thus, we have: $S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$ ....(1) Equation (1) can be rewritten as $S_{n}=2+4+7+11+16+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: $\Rightarrow 2+\left[\frac{(n-1)}{2}(4+(n-2) 1)\right]-T_{n}=0$ $\Rightarrow 2+\left[\frac{(n-1)}{2}(n+2)\right]-T_{n}=0$ $\Rightarrow 2+\left[\frac{n^{2}+n}{2}-1\right]-T_{n}=0$ $\Righ...
Read More →Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4).
Question: Show that ΔABCwith verticesA(2, 0),B(0, 2) andC(2, 0) is similar to ΔDEFwith verticesD(4, 0),F(4, 0) andE(0, 4). Solution: In ΔABC, the coordinates of the vertices are A(2, 0), B(0,2), C(2,0). $\mathrm{AB}=\sqrt{(0+2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$ $\mathrm{CB}=\sqrt{(0-2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$ $\mathrm{AC}=\sqrt{(2+2)^{2}+(0-0)^{2}}=4$ In ΔDEF, the coordinates of the vertices are D(4, 0), E(4, 0), F(0, 4). $\mathrm{DF}=\sqrt{(4+4)^{2}+(0-0)^{2}}=8$ $\mathrm{FE}=\sqr...
Read More →Write the formula for the area of the triangle having its vertices
Question: Write the formula for the area of the triangle having its vertices at (x1, y1), (x2, y2) and (x3, y3). 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, $\mathrm{A}=\frac{1}{2}\left|\begin{array}{ll}x_{1}-x_{2} y_{1}-y_{2} \\ x_{2}-x_{3} y_{2}-y_{3}\end{array}\right|$ $\mathrm{A}=\frac{1}{2}\left|\left(x_{1}-x_{2}\right)\left(y_{2}-y_{3}\right)-\left(x_{2}-x_{...
Read More →If points Q and reflections of point P (−3, 4)
Question: If points Q and reflections of point P (3, 4) in X and Y axes respectively, what is QR? Solution: We have to find the reflection of (3, 4) alongx-axis andy-axis. Reflection of any point $P(a, b)$ along $x$-axis is $(a,-b)$ So reflection of $(-3,4)$ along $x$-axis is $Q(-3,-4)$ Similarly, reflection of any point $\mathrm{P}(\mathrm{a}, \mathrm{b})$ along $y$-axis is $(-a, b)$ So, reflection of $(-3,4)$ along $y$-axis is $R(3,4)$ Therefore, $Q R=\sqrt{(3+3)^{2}+(4+4)^{2}}$ $=\sqrt{36+64}...
Read More →Write the coordinates the reflections of points
Question: Write the coordinates the reflections of points (3, 5) in X and Y -axes. Solution: We have to find the reflection of (3, 5) alongx-axis andy-axis. Reflection of any point $P(a, b)$ along $x$-axis is $(a,-b)$ So reflection of $(3,5)$ along $x$-axis is $(3,-5)$ Similarly, reflection of any point $\mathrm{P}(\mathrm{a}, \mathrm{b})$ along $y$-axis is $(-a, b)$ So, reflection of $(3,5)$ along $y$-axis is $(-3,5)$...
Read More →4 + 6 + 9 + 13 + 18 + ...
Question: 4 + 6 + 9 + 13 + 18 + ... 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: $S_{n}=4+6+9+13+18+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=4+6+9+13+18+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: The sequence of difference between successive terms is 2, 3, 4, 5,... We observe that it is an AP with common difference 1 and first term 2. Now, $4+\left[\frac{(n-1)}{2}\{4+(n-2) 1\}\r...
Read More →Show that the following points are the vertices of a rectangle.
Question: Show that the following points are the vertices of a rectangle.(i)A(4, 1),B(2, 4)C(4, 0) andD(2, 3) (ii)A(2, 2),B(14, 10)C(11, 13) andD(1, 1)(iii)A(0, 4),B(6, 2)C(3, 5) andD(3, 1) Solution: (i) Given : A $(-4,-1)$, B $(-2,-4)$, C $(4,0)$ and D $(2,3)$ In rectangle the opposite sides are equal Using $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ $\mathrm{AB}=\sqrt{(-4+2)^{2}+(-1+4)^{2}}$ $\mathrm{AB}=\sqrt{13}$ $\mathrm{BC}=\sqrt{(4+2)^{2}+(0+4)^{2}}$ $\mathrm{BC}=...
Read More →Two vertices of a triangle have coordinates (−8, 7) and
Question: Two vertices of a triangle have coordinates (8, 7) and (9, 4) . If the centroid of the triangle is at the origin, what are the coordinates of the third vertex? Solution: We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be $(x, y)$. The co-ordinates of other two vertices are (8, 7) and (9, 4) The co-ordinate of the centroid is (0, 0) We know that the co-ordinates of the centroid of a triangle whose vertices are $\left(x...
Read More →If the mid-point of the segment joining A (x, y + 1)
Question: If the mid-point of the segment joining $A(x, y+1)$ and $B(x+1, y+2)$ is $C\left(\frac{3}{2}, \frac{5}{2}\right)$, find $x, y .$ Solution: It is given that mid-point of line segment joining $\mathrm{A}(x, y+1)$ and $\mathrm{B}(x+1, y+2)$ is $\mathrm{C}\left(\frac{3}{2}, \frac{5}{2}\right)$ In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\left(\f...
Read More →