If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP.
Question: If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP. Solution: In the given AP, let the first term beaand the common difference bed.Then, Tn=a+ (n- 1)dNow, we have:T10=a+ (10- 1)d⇒a+9d = 52 ...(1)T13=a+ (13 - 1)d=a+12d ...(2)T17=a+ (17 - 1)d =a+16d ...(3) But, it is given that T17= 20 + T13i.e.,a+16d = 20 +a+12d⇒ 4d= 20 ⇒d= 5On substitutingd= 5 in (1), we get:a+ 9 ⨯ 5 = 52⇒a= 7 Thus,a= 7 andd= 5 The terms of the AP are 7, 12, 17, 22,......
Read More →Prove the following
Question: Let $\alpha$ and $\beta$ be the roots of $x^{2}-6 x-2=0$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_{8}}{3 a_{9}}$ is:(1) 4(2) 1(3) 2(4) 3Correct Option: , 3 Solution: and $\alpha^{2}-6 \alpha-2=0 \Rightarrow \alpha^{2}-2=6 \alpha$ Now $\beta^{2}-6 \beta-2=0 \Rightarrow \beta^{2}-2=6 \beta$ $\frac{a_{10}-2 a_{8}}{3 a_{9}}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)}{3\left(\alpha^{9}-\beta^{9}\right)}$ $=\frac{\...
Read More →An unsaturated hydrocarbon X on ozonolysis gives A.
Question: An unsaturated hydrocarbon $\mathrm{X}$ on ozonolysis gives $\mathrm{A}$. Compound $\mathrm{A}$ when warmed with ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon $X$ is:Correct Option: , 3 Solution: As (A) compound given positive tollen's test hence it may consist - $\mathrm{CHO}$ (aldehyde group). or it can be $\mathrm{HCOOH}$ So for the given option: (c) and for other compounds (options): (d) $\mathrm{CH}_{3}-\mathrm{...
Read More →Prove the following
Question: If $\alpha, \beta \in \mathrm{R}$ are such that $1-2 \mathrm{i}$ (here $\mathrm{i}^{2}=-1$ ) is a root of $\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$, then $(\alpha-\beta)$ is equal to:(1) 7(2) $-3$(3) 3(4) $-7$Correct Option: , 4 Solution: $(1-2 i)^{2}+\alpha(1-2 i)+\beta=0$ $1-4-4 i+\alpha-2 i \alpha+\beta=0$ $(\alpha+\beta-3)-i(4+2 \alpha)=0$ $\alpha+\beta-3=0 \quad \ 4+2 \alpha=0$ $\alpha=-2 \quad \beta=5$ $\alpha-\beta=-7$...
Read More →Which term of the AP 5, 15, 25, ...will be 130 more than its 31st term?
Question: Which term of the AP 5, 15, 25, ...will be 130 more than its 31st term? Solution: Here, a = 5and d = (15 - 5) = 10The 31stterm is given byT31=a+ (31 - 1)d =a+ 30d= 5 +30 ⨯10 = 305 Required term = (305 + 130) = 435Let this be be thenthterm.Then Tn= 435⇒ 5 + (n- 1)⨯ 10 = 435⇒ 10n= 440⇒n= 44Hence, the 44thterm will be 130 more than its 31st term....
Read More →The integer ' k ', for which the inequality
Question: The integer ' $k$ ', for which the inequality $x^{2}-2(3 k-1) x+8 k^{2}-70$ is valid for every $x$ in $R$ is :(1) 3(2) 2(3) 4(4) 0Correct Option: 1 Solution: $D0$ $(2(3 k-1))^{2}-4\left(8 k^{2}-7\right)0$ $4\left(9 k^{2}-6 k+1\right)-4\left(8 k^{2}-7\right)0$ $k^{2}-6 k+80$ $(k-4)(k-2)0$ $2k4$ then $k=3$...
Read More →The coefficients a, b and c
Question: The coefficients $a, b$ and $c$ of the quadratic equation, $a x^{2}+b x+c=0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is :(1) $\frac{1}{54}$(2) $\frac{1}{72}$(3) $\frac{1}{36}$(4) $\frac{5}{216}$Correct Option: , 4 Solution: $a x^{2}+b x+c=0$ $\mathrm{a}, \mathrm{b}, \mathrm{c} \in\{1,2,3,4,5,6\}$ $\mathrm{n}(\mathrm{s})=6 \times 6 \times 6=216$ $\mathrm{D}=0 \Rightarrow \mathrm{b}^{2}=4 \mathrm{ac}$ $\mathrm{ac}=\frac{\mathrm{b}^{...
Read More →Which term of the AP 8, 14, 20, 26, ... will be 72 more than its 41st term?
Question: Which term of the AP 8, 14, 20, 26, ... will be 72 more than its 41st term? Solution: In the given problem, let us first find the 41stterm of the given A.P. A.P. is 8, 14, 20, 26 Here, First term (a) = 8 Common difference of the A.P. $(d)=14-8=6$ Now, as we know, $a_{n}=a+(n-1) d$ So, for 41stterm (n= 41), $a_{41}=8+(41-1)(6)$ $=8+40(6)$ $=8+240$ $=248$ Let us take the term which is 72 more than the 41stterm asan. So, $a_{n}=72+a_{41}$ $=72+248$ $=320$ Also, $a_{n}=a+(n-1) d$ $320=8+(n...
Read More →Solve the following
Question: Identify the reagent(s) 'A' and condition(s) for the reaction:$\mathrm{A}=\mathrm{HCl}$; Anhydrous $\mathrm{AlCl}_{3}$$\mathrm{A}=\mathrm{HCl}, \mathrm{ZnCl}_{2}$$\mathrm{A}=\mathrm{Cl}_{2}$; UV light$\mathrm{A}=\mathrm{Cl}_{2}$; dark, Anhydrous $\mathrm{AlCl}_{3}$Correct Option: , 3 Solution: For substitution at allylic position in the given compound, the reagent used is $\mathrm{Cl}_{2} /$ uv light. The reaction is free radical halogenation....
Read More →The number of the real roots of the equation
Question: The number of the real roots of the equation $(x+1)^{2}+|x-5|=\frac{27}{4}$ is Solution: $x \geq 5$ $(x+1)^{2}+(x-5)=\frac{27}{4}$ $\Rightarrow x^{2}+3 x-4=\frac{27}{4}$ $\Rightarrow x^{2}+3 x-\frac{43}{4}=0$ $\Rightarrow 4 x^{2}+12 x-43=0$ $\boldsymbol{X}=\frac{-12 \pm \sqrt{144+688}}{8}$ $x=\frac{-12 \pm \sqrt{832}}{8}=\frac{-12 \pm 28.8}{8}$ $=\frac{-3 \pm 7.2}{2}$ $=\frac{-3+7.2}{2}, \frac{-3-7.2}{2}$ (Therefore no solution) For $x \leq 5$ $(x+1)^{2}-(x-5)=\frac{27}{4}$ $x^{2}+x+6-...
Read More →A small circular loop of conducting wire has radius a and carries current I.
Question: A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is $m$ then :(1) $\mathrm{T}=\sqrt{\frac{2 \mathrm{~m}}{\mathrm{IB}}}$(2) $\mathrm{T}=\sqrt{\frac{\pi \mathrm{m}}{2 \mathrm{IB}}}$(3) $\mathrm{T}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}$(4...
Read More →Which of the following is Lindlar catalyst?
Question: Which of the following is Lindlar catalyst?Zinc chloride and $\mathrm{HCl}$Cold dilute solution of $\mathrm{KMnO}_{4}$Sodium and Liquid $\mathrm{NH}_{3}$Partially deactivated palladised charcoalCorrect Option: Solution: Partially deactivated palladised charcoal $\left(\mathrm{H}_{2} / \mathrm{pd} / \mathrm{CaCO}_{3}\right)$ is lindlar catalyst....
Read More →Which term of the AP 21, 18, 15, ... is −81?
Question: Which term of the AP 21, 18, 15, ... is 81? Solution: The given AP is 21, 18, 15, ... .First term,a= 21Common difference,d= 18 21 =3Supposenthterm of the given AP is 81. Then, $a_{n}=-81$ $\Rightarrow 21+(n-1) \times(-3)=-81 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow-3(n-1)=-81-21=-102$ $\Rightarrow n-1=\frac{102}{3}=34$ $\Rightarrow n=34+1=35$ Hence, the 35th term of the given AP is 81....
Read More →A charged particle of mass ' m ' and charge ' q '
Question: A charged particle of mass ' $m$ ' and charge ' $q$ ' moving under the influence of uniform electric field $E \hat{i}$ and a uniform magnetic field $B \vec{k}$ follows a trajectory from point $\mathrm{P}$ to $\mathrm{Q}$ as shown in figure. The velocities at $\mathrm{P}$ and $\mathrm{Q}$ are respectively, $v \vec{i}$ and $-2 v \vec{j}$. Then which of the following statements $(A, B, C, D)$ are the correct? (Trajectory shown is schematic and not to scale) (A) $\mathrm{E}=\frac{3}{4}\lef...
Read More →Which term of the AP
Question: Which term of the AP $\frac{5}{6}, 1,1 \frac{1}{6}, 1 \frac{1}{3}, \ldots .$ is 3 ? Solution: In the given $\mathrm{AP}$, first term $=\frac{5}{6}$ and common difference, $\mathrm{d}=\left(1-\frac{5}{6}=\frac{1}{6}\right)$. Let itsnth term be 3. Now, $T_{n}=3$ $\Rightarrow a+(n-1) d=3$ $\Rightarrow \frac{5}{6}+(n-1) \times \frac{1}{6}=3$ $\Rightarrow \frac{2}{3}+\frac{n}{6}=3$ $\Rightarrow \frac{n}{6}=\frac{7}{3}$ $\Rightarrow n=14$ Hence, the 14th term of the given AP is 3....
Read More →A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic
Question: A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2 a$, respectively from the axis of the wire is:(1) $\frac{2}{3}$(2) 2(3) $\frac{1}{2}$(4) $\frac{3}{2}$Correct Option: 1 Solution: (1) Let $a$ be the radius of the wire Magnetic field at point $A$ (inside) $B_{A}=\frac{\mu_{0} i r}{2 \pi a^{2}}=\frac{\mu_{0} i \frac{a}{3}}{2 \pi a^{2}}=\frac{\mu_{0} i}{\pi a^{...
Read More →Which term of the AP 72, 68, 64, 60,... is 0?
Question: Which term of the AP 72, 68, 64, 60,... is 0? Solution: In the given AP, first term, a = 72 and common difference, d = (68 - 72) = -4.Let itsnthterm be 0.Then, Tn= 0 ⇒a+ (n- 1)d = 0⇒ 72 + (n- 1) ⨯ (-4) = 0⇒76 - 4n= 0⇒4n= 76⇒n= 19Hence, the 19th term of the given AP is 0....
Read More →Let a, b, c be in arithmetic progression.
Question: Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c),(2, b)$ and $(a, b)$ be $\left(\frac{10}{3}, \frac{7}{3}\right) .$ If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+1=0$, then the value of $\alpha^{2}+\beta^{2}-\alpha \beta$ is:(1) $\frac{71}{256}$(2) $-\frac{69}{256}$(3) $\frac{69}{256}$(4) $-\frac{71}{256}$Correct Option: , 4 Solution: $2 b=a+c$ $\frac{2 a+2}{3}=\frac{10}{3}$ and $\frac{2 b+c}{3}=\frac{7}{3}$ $\left.a=4, ...
Read More →Which term of the AP 3, 8, 13, 18,... is 88?
Question: Which term of the AP 3, 8, 13, 18,... is 88? Solution: In the given AP, first term, a = 3 and common difference, d = (8 - 3) = 5.Let's itsnthterm be 88.Then Tn= 88⇒a+ (n- 1)d = 88⇒ 3 + (n- 1) ⨯ 5 = 88⇒ 5n- 2= 88⇒ 5n= 90⇒n= 18Hence, the 18thterm of the given AP is 88....
Read More →How many terms are there in the AP 18,
Question: How many terms are there in the AP $18,15 \frac{1}{2}, 13, \ldots,-47 ?$ Solution: The given $\mathrm{AP}$ is $18,15 \frac{1}{2}, 13, \ldots,-47$. First term,a= 18 Common difference, $d=15 \frac{1}{2}-18=\frac{31}{2}-18=\frac{31-36}{2}=-\frac{5}{2}$ Suppose there arenterms in the given AP. Then, $a_{n}=-47$ $\Rightarrow 18+(n-1) \times\left(-\frac{5}{2}\right)=-47 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow-\frac{5}{2}(n-1)=-47-18=-65$ $\Rightarrow n-1=-65 \times\left(-\frac{2}{5}\...
Read More →A very long wire A B D M N D C is shown in figure carrying current I.
Question: A very long wire $A B D M N D C$ is shown in figure carrying current $I$. $A B$ and $B C$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $D M N D$ of radius $R$. $A B, B C$ parts are tangential to circular turn at $N$ and $D$. Magnetic field at the centre of circle is: (1) $\frac{\mu_{0} I}{2 \pi R}\left(\pi+\frac{1}{\sqrt{2}}\right)$(2) $\frac{\mu_{0} I}{2 \pi R}\left(\pi-\frac{1}{\sqrt{2}}\right)$(3) $\frac{\mu_{0} I}{2 \pi R}(\pi+1)$(4) $\frac{\mu_{0}...
Read More →Let p and q be two positive number such that
Question: Let $p$ and $q$ be two positive number such that $p+q=2$ and $p^{4}+q^{4}=272 .$ Then $p$ and $q$ are roots of the equation :(1) $x^{2}-2 x+2=0$(2) $x^{2}-2 x+8=0$(3) $x^{2}-2 x+136=0$(4) $x^{2}-2 x+16=0$Correct Option: , 4 Solution: $\left(p^{2}+q^{2}\right)^{2}-2 p^{2} q^{2}=272$ $\left((p+q)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=272$ $16+16 p q+2 p^{2} q^{2}=272$ $(p q)^{2}-8 p q-128=0$ $\mathrm{pq}=\frac{8 \pm 24}{2}=16,-8$ $\mathrm{pq}=16$ Now $x^{2}-(p+q) x+p q=0$ $x^{2}-2 x+16=0$...
Read More →How many terms are there in the AP 41, 38, 35,...., 8 ?
Question: How many terms are there in the AP 41, 38, 35,...., 8 ? Solution: In the given AP,a= 41 andd= (38 - 41) = -3Suppose that there arenterms in the given AP.Then Tn= 8⇒a+ (n- 1) d = 8⇒ 41 + (n- 1) ⨯ (-3) = 8⇒ 44 - 3n= 8⇒ 3n= 36⇒n= 12Hence, there are 12 terms in the given AP....
Read More →Increasing order of reactivity of the following compounds for
Question: Increasing order of reactivity of the following compounds for $S_{N} 1$ substitution is: (\mathrm{B})(\mathrm{C})(\mathrm{D})(\mathrm{A})$(B)(C)(A)(D)$$(B)(A)(D)(C)$$(\mathrm{A})(\mathrm{B})(\mathrm{D})(\mathrm{C})$Correct Option: , 3 Solution: In $\mathrm{S}_{\mathrm{N}} 1$ reaction carbocation acts as an intermediate. Carbocation produced by $(\mathrm{C})$ is more stable than carbocation produced by (D) due to + I effect of $-\mathrm{OCH}_{3}$ group. Further in (A) there is formation...
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