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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If `x + y + z =a` and the minimum value of `a/x+a/y+a/z` is 81, then the value of . `lambda` isA. `(1)/(2)`B. 1C. `(1)/(4)`D. 2 |
| Answer» Correct Answer - C | |
| 52. |
The maximum distance of the centre of the ellipse `x^2/(81)+y^2/(25)=1` from a normal to the ellipse is |
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Answer» let us take a point as `(9cos theta , 5sin theta)` the tangent on the point will be ` (x*9cos theta)/81 + (y*5*sin theta)/25 = 1 ` `(x*Cos theta)/9 + (y*sin theta)/5 = 1` `y= (-5*cos theta)/(9*sin theta)* x + 5/ sin theta` so slope will be `(5Cos theta)/ (9 sin theta)` as we know that, slope of tangent * slope of normal = -1 so, slope of normal is `(9*Sin theta)/(5*cos theta)` for normal `y- sin theta = (9*sin theta)/(5* cos theta) * (x- 9*cos theta)` `5*y*cos theta - 25*sin theta*cos theta = 9*x*sin theta - 81* sin theta` `5*y*cosec theta - 25 = 9*x*sec theta - 81` `9*x*sec theta - 5*y*cosec theta - 56=0 ` ` now, for centre (0,0)` `|(0-0-56)/sqrt(81*cosec^2 theta +25*cosec^2 theta)| = 56/sqrt(81*sec^2 theta + 25*cosec^2 theta) ` sloving for denominator let `sqrt(81*sec^2 theta + 25*cosec^2 theta) = x` for maximum value the condition must be true that `(dx)/(d theta) = 0` So, differentiating we get, ` (81*sec^2 theta * tan theta - 25* 2*cosec^2 theta * cot theta)/ (2*sqrt(81*sec^2 theta + 25* cosec^2 theta)) = 0` `81*sec^2 theta * tan theta = 25* cosec^2 theta * cot theta` `(81*sin theta)/ cos ^ 3 theta = 25*cos theta / sin ^ 3 theta` `Sin^4 theta/ cos^4 theta = 25/81` `tan^2 theta = 5/9` `cot^2 theta = 9/5` now putting in main equation `56/sqrt(81*(1+tan^2 theta) + 25*(1+ cot^2 theta)) ` `= 56/ sqrt(81*(1+5/9) + 25*(1+9/5))` `= 56/sqrt(9*14+ 5*14)` `= 56/ 14 = 4` |
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| 53. |
Let `T_r` be the `r^(th)` term of an A.P whose first term is `a` and common difference is `d` IF for some integer m,n, `T_m=1/n` and `T_n=1/m` then `a-d=`A. `(1)/(mn)`B. `(1)/(m)+(1)/(n)`C. 1D. 0 |
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Answer» Correct Answer - C Let a be the term and d be the common difference of the given AP. Then, `T_(m)=(1)/(n)rArra=(m+1)d=(1)/(n)` . . . (i) `and,T_(n)=(1)/(m)rArra+(n-1)d=(1)/(m)` . . . (ii) Solving (i) and (ii), we get `d=(1)/(mn)anda=(1)/(mn)` `:." "T_(mn)=a+(mn-1)d=(1)/(mn-1)(1)/(mn)=1` |
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| 54. |
Let `V_r` denote the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is `(2r-1). Let `T_r=V_(r+1)-V_r-2 and Q_r =T_(r+1)-T_r for r=1,2` `T_r` is always (A) an odd number (B) an even number (C) a prime number (D) a composite num,berA. `Q_(1), Q_(2), Q_(3)`,.... are in an AP with common difference 5B. `Q_(1), Q_(2), Q_(3)`... are in an AP with common difference 6C. `Q_(1), Q_(2), Q_(3)`,.... are in an AP with common difference 11D. `Q_(1) = Q_(2) = Q_(3) = ...` |
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Answer» Correct Answer - B Since, `T_(r) = 3r^(2) + 2r -1` and `T_(r+1) = 3 (r +1)^(2) + 2(r +1) -1` `:. Q_(r) = T_(r +1) - T_(r) = 3 [2r + 1] + 2 [1]` `rArr Q_(r) = 6r + 5` `rArr Q_(r +1) = 6(r +1) + 5` Common difference `= Q_(r +1) - Q_(r) = 6` |
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| 55. |
Two sequences `lta_(n)gtandltb_(n)gt` are defined by `a_(n)=log((5^(n+1))/(3^(n-1))),b_(n)={log((5)/(3))}^(n)`, thenA. `lta_(n)gt ` is an A.P. and `lta_(n)gt` is a G.PB. `lta_(n)gt` and `ltb_(n)gt` both are G.P.C. `lta_(n)gt` and `ltb_(n)gt` both are A.P.D. `lta_(n)gt` is a G.P. and `ltb_(n)gt` is neither an A.P. nor a G.P. |
| Answer» Correct Answer - A | |
| 56. |
If the arithmetic mean of `a_(1),a_(2),a_(3),"........"a_(n)` is a and `b_(1),b_(2),b_(3),"........"b_(n)` have the arithmetic mean b and `a_(i)+b_(i)=1` for `i=1,2,3,"……."n,` prove that `sum_(i=1)^(n)(a_(i)-a)^(2)+sum_(i=1)^(n)a_(i)b_(i)=nab`. |
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Answer» `suma_(i)b_(i)=suma_(i)(1-a_(i))=na-suma_(i)^(2)` `=na-sum(a_(i)-a+a)^(2)` `=na-sum[(a_(i)-a)^(2)+a^(2)+2a(a_(i)-a)]` `=na-sum(a_(i)-a)^(2)+suma^(2)+2asum(a_(i)-a)` `:.suma_(i)b_(i)+sum(a_(i)-a)^(2)=na-na^(2)-2a(na-na)` `=na(1-a)=nab [{:(sumb_(1)=sum1-suma_(i)),(therefore nb=n-na),(" or "a+b=1):}]`. |
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| 57. |
Let the harmonic mean of two positive real numbers a and b be 4, If q is a positive real number such that a, 5, q, b is an arithmetic progression, then the value(s) of |q -a| is (are) |
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Answer» `4=(2ab)/(a+b)` `(ab)/(a+b)=2-(1)` a,5,q,b->AP common diff. d `a=5-d-(2)` `q=a+2d` `b=a+3d` `b=5+2d-(3)` `((5-d)(5+2d))/(10+d)=2` `2d^2-3d-5=0` `(2d-5)(d+1)=0` `d=-1,5/2` `d=-1` `2|d|=2`. |
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| 58. |
If a,b,c and d are four positive real numbers such that abcd=1 , what is the minimum value of `(1+a)(1+b)(1+c)(1+d)`.A. 1B. 4C. 16D. 64 |
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Answer» Correct Answer - C `:.(1+a)(1+b)(1+c)(1+d)` ,brgt `=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+abd+cda+cdb+abcd " " [" 16 terms "]` `:.AMgeGM` `((1+a)(1+b)(1+c)(1+d))/(16)ge (a^(8)b^(8)c^(8)d^(8))^((1)/(16))` `=(abcd)^((1)/(2))=(1)^((1)/(2))=1 " " [:. Abcd=1]` `((1+a)(1+b)(1+c)(1+d))/(16)ge1` `(1+a)(1+b)(1+c)(1+d)ge16` `:.` Minimum value of `(1+a)(1+b)(1+c)(1+d)` is 16. |
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| 59. |
Let `a_(1),a_(2),a_(3), . . .,a_(100)` be an arithmetic progression with `a_(1)=3andS_(p)=sum_(i=1)^(p)a_(i),aleple100`. For any integer n with `1lenle20`, let m=5n. If `(S_(m))/(S_(n))` does not depend on n, then `a_(2)` isA. 9B. 8C. 7D. 5 |
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Answer» Correct Answer - A Since `a_(1),a_(2),a_(3), . . . .,a_(100)` is an A.P. with `a_(1)=3andS_(p)=underset(i=1)overset(p)suma_(i)`. `:." "(S_(m))/(S_(n))=(S_(5n))/(S_(n))=((5n)/(2){2xx3+(5n-1)xxd})/((n)/(2){2xx3+(n-1)xxd})=(5(5nd+6-d))/((nd+6-d))` Clearly, it will be independent of n if 6-d=0 i.e. d=6. `:." "a_(2)=a_(1)+d=3+6=9` |
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| 60. |
If `b-c,2b-lambda,b-a " are in HP, then " a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are isA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - B `(2b-lambda)=(2(b-c)(b-a))/((b-c)+(b-a))` `implies (2b-lambda)=(2b-(a+c))=2[b^(2)-(a+c)b+ac]` `implies 2b^(2)-2blambda+lambda (a+c)-2ac=0` `implies b^(2)-blambda+(lambda)/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)-lambda^(2)/(4)+lambda/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)=lambda^(2)/(4)-(lambda)/(2)(a+c)+ac` `implies (b-(lambda)/(2))^(2)=(a-lambda/(2))(c-(lambda)/(2))` Hence, `a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are in GP. |
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| 61. |
Suppose `alpha, beta` are roots of `ax^(2)+bx+c=0` and `gamma, delta` are roots of `Ax^(2)+Bx+C=0`. If `alpha,beta,gamma,delta` are in AP, then common difference of AP isA. `(1)/(4)((b)/(a)-(B)/(A))`B. `(1)/(3)((b)/(a)-(B)/(A))`C. `(1)/(2)((c)/(a)-(B)/(A))`D. `(1)/(3)((c)/(a)-(C)/(A))` |
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Answer» Correct Answer - A `:.alpha +beta=-(b)/(a),alphabeta=(c)/(a),alpha-beta=(sqrt(b^(2)-4ac))/(a)` and `gamma+delta=-(B)/(A),gammadelta=(C)/(A),gamma-delta=(sqrt(B^(2)-4AC))/(A)` Since, `alpha,beta, gamma` are in AP. Let `beta=alpha+D,gamma=alpha+2D" and "delta=alpha+3D` `:.alpha+ beta=(-b)/(a)" " implies alpha+alpha+D=-(b)/(a)` or `2alpha +D=-(b)/(a)" " ".......(i)"` and `gamma +delta=-(B)/(A)" " implies 2alpha +5D=-(B)/(A)" " "...........(ii)"` From Eqs. (i) and(ii), we get `4D=(-(B)/(A)+(b)/(a)) " or " D=(1)/(4)((b)/(a)-(B)/(A))`. |
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| 62. |
Let `(a_(1),b_(1))` and `(a_(2),b_(2))` are the pair of real numbers such that 10,a,b,ab constitute an arithmetic progression. Then, the value of `((2a_(1)a_(2)+b_(1)b_(2))/(10))` is |
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Answer» Let `a=10+D " " "……(i)"` `b=10+2D" " "…..(ii)"` `ab=10+3D" " "…….(iii)"` On substituting the values of a and b in Eq. (iii), we get `(10+D)(10+2D)=(10+3D)` `implies 2D^(2)+27D+90=0` `:.d=-6,D=-(15)/(2)` `:.a_(1)=10-6=4,a_(2)=10-(15)/(2)=(5)/(2)` and `b_(1)=10-12=-2, b_(2)=10-15=-5` Now, `((2a_(1)a_(2)+b_(1)b_(2))/(10))=((2xx10+10)/(10))=3`. |
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| 63. |
If a,b,c are in AP and `(a+2b-c)(2b+c-a)(c+a-b)=lambdaabc`, then `lambda` isA. 1B. 2C. 4D. None of these |
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Answer» Correct Answer - C `:.a,b,c` are in AP. `:. 2b=a+c" " "…..(i)"` Now, `(a+2b-c)(2b+c-a)(c+a-b)` `(a+a+c-c)(a+c+c-a)(2b-b)" " [" from Eq.(i) "]` `=(2a)(2c)(b)=4abc` `:. Lambda =4`. |
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| 64. |
If the roots of the cubic equation `ax^3+bx^2+cx+d=0` are in G.P thenA. `c^(3)a=b^(3)d`B. `ca^(2)=bd^(3)`C. `a^(3)b=c^(3)d`D. `ab^(3)=cd^(3)` |
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Answer» Correct Answer - A Let `(A)/(R),A,AR` be the roots of the equation `ax^(3)+bx^(2)+cx+d=0` `:." Product of the roots "=-(d)/(a)`. `rArr" "A^(3)=-(d)/(a)A=-((d)/(a))^(1//3)` Since A is a root of the given equation. `:." "aA^(3)+bA^(2)+cA+d=0` `rArr" "a(-(d)/(a))+b(-(d)/(a))^(2//3)+c(-(d)/(a))^(1//3)+d=0` `rArr" "b((d)/(a))^(2//3)=c((d)/(a))^(1//3)rArrb^(3)xx(d)/(a)rArrb^(3)d=c^(3)a` |
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| 65. |
If one root of `Ax^(3)+Bx^(2)+Cx+D=0,Dne0` is the arithmetic mean of the other two roots, then the relation `2B^(2)+lambdaABC+muA^(2)D=0` holds good. Then, the value of `2lambda+mu` is |
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Answer» Given equation, `Ax^(3)+Bx^(2)+Cx+D=0" " ".......(i)"` where, `Ane 0` Let roots are `alpha, beta, gamma` then `beta=(alpha+gamma)/(2)" " "…….(ii)"` Given relation, `2B^(3)+lambdaABC+muA^(2)D=0" " "....(iii)"` From Eq. (i), `alpha+beta+gamma=-(beta)/(A)` `implies 3beta=- (B)/(A)" " [" from Eq. (ii) "]` `implies beta=- (B)/(3A)` Now, `beta` satisfy Eq. (i), so `A((-B)/(3A))^(3)+B((-B)/(3A))^(2)+C((-B)/(3A))+D=0` `implies (-B^(3))/(27A^(2))+(B^(3))/(9A^(2))-(BC)/(3A)+D=0` `implies (2)/(27)(B^(3))/(A^(2))-(BC)/(3A)+D=0` `implies 2B^(3)-9ABC+27AD^(2)=0` Compare with Eq. (iii), we get `lambda=-9,mu=27` `2lambda +mu=-18+27=9`. |
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| 66. |
If the sum of the first n terms of an A.P. is pn+`qn^(2)` then its common difference isA. p-qB. p+qC. 2qD. 2p |
| Answer» Correct Answer - C | |
| 67. |
The sum of the series `a-(a+d)+(a+2d)-(a+3d)+`up to `(2n+1)`terms isA. `a^(2)+3nd^(2)`B. `a^(2)+2nad+n(n-1)d^(2)`C. `a^(2)+nad+n(n-1)d^(2)`D. `a^(2)+2nad+n(2n+1)d^(2)` |
| Answer» Correct Answer - D | |
| 68. |
The sum of the series `a-(a+d)+(a+2d)-(a+3d)+`up to `(2n+1)`terms is`-n d`b. `a+2n d`c. `a+n d`d. `2n d`A. `-nd`B. a+2ndC. a+ndD. 2nd |
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Answer» Correct Answer - C The given series is an AGP with common ratio-1. S=a-(a+b)+(a+2d)-(a+3d)+ . . . .+(a+2nd) `rArr" "-S=-a+(a+d)-(a+2d)+ . . . .+(a+(2n-1)d)-(a+2nd)` `:." "2S=a+{-d+d-d+d . . .` upto 2n terms}+(a+2nd) `rArr" "2S=2a+2ndrArrS=a+nd` |
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| 69. |
IF `a_(1),a_(2),a_(3),"...."a_(10)` be in AP and `h_(1),h_(2),h_(3),"...."h_(10)` be in HP. If `a_(1)=h_(1)=2` and `a_(10)=h_(10)=3`, then find value of `a_(4)h_(7)`. |
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Answer» ` therefore a_(1),a_(2),a_(3),"...."a_(10)` are in AP. If d be the common difference, then `d=(a_(10)-a_(1))/(9)=(3-2)/(9)=(1)/(9)` `therefore a_(4)=a_(1)+3d=2+(3)/(9)=2+(1)/(3)=(7)/(3) "……..(i)"` and given ` h_(1),h_(2),h_(3),"...."h_(10)` are in HP. If D be common difference of corresponding AP. Then, ` D=((1)/h_(10)-(1)/h_(1))/(9)=((1)/(3)-(1)/(2))/(9)=-(1)/(54)` ` therefore (1)/h_(7)=(1)/h_(1)+6D=(1)/(2)-(6)/(54)=(1)/(2)-(1)/(9)=(7)/(18) implies h_(7)=(18)/(7)` Hence, ` a_(4)*h_(7)=(7)/(3)xx(18)/(7)=6` |
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| 70. |
If `a_(1),a_(2),a_(3),".....",a_(n)` are in HP, than prove that `a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+a_(n-1)a_(n)=(n-1)a_(1)a_(n)` |
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Answer» Given, `a_(1),a_(2),a_(3),"…",a_(n)` are in HP. ` therefore (1)/a_(1),(1)/a_(2),(1)/a_(3),"....."(1)/a_(n)` are in AP. Let D be the common difference of the AP, than `(1)/a_(2)-(1)/a_(1)=(1)/a_(3)-(1)/a_(2)=(1)/a_(4)-(1)/a_(3)="....."=(1)/a_(n)-(1)/a_(n-1)=D` ` implies (a_(1)-a_(2))/(a_(1)a_(2))=(a_(2)-a_(3))/(a_(2)a_(3))=(a_(3)-a_(4))/(a_(3)a_(4))="....."=(a_(n-1)-a_(n))/(a_(n-1)a_(n))=D` ` implies a_(1)a_(2)=(a_(1)-a_(2))/(D),a_(2)a_(3)=(a_(2)-a_(3))/(D),a_(3)a_(4)=(a_(3)-a_(4))/(D),".....",a_(n-1)-a_(n)=(a_(n-1)a_(n))/(D)` On adding all such expressions, we get ` a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+a_(n-1)-a_(n)=(a_(1)-a_(n))/(D)=(a_(1)a_(n))/(D)((1)/a_(n)-(1)/a_(1))` ` (a_(1)a_(n))/(D)[(1)/a_(1)+(n-1)D-(1)/(a_(1))]=(n-1)a_(1)a_(n)` Hence, ` a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+ a_(n-1)a_(n)=(n-1)a_(1)a_(n)` |
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| 71. |
if `S` is the sum , `P` the product and `R` the sum of reciprocals of `n` terms in `G.P.` prove that `P^2 R^n=S^n`A. PB. `P^(2)`C. `P^(3)`D. `P^(n)` |
| Answer» Correct Answer - B | |
| 72. |
The sum of three numbers in HP is 37 and the sum of their reciprocals is `(1)/(4)`. Find the numbers. |
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Answer» Three numbers in HP can be taken as ` (1)/(a-d),(1)/(a),(1)/(a+d)`. Then, ` (1)/(a-d)+(1)/(a)+(1)/(a+d)=37 "…….(i)"` and `a-d+a+a+d=(1)/(4)` `therefore " " a=(1)/(12)` From Eq.(i), `(12)/(1-12d)+12+(12)/(1+12d)=37` `implies (12)/(1-12d)+(12)/(1+12d)=25 implies (24)/(1-144d^(2))=25` `1-144d^(2)=(24)/(25)` or `d^(2)=(1)/(25xx144)` `therefore " " d= pm (1)/(60)` `therefore a-d,a,a+d, " are " (1)/(15),(1)/(12),(1)/(10), " or " (1)/(10),(1)/(12),(1)/(15).` Hence, three numbers in HP are `15,12,10, " or " 10,12,15`. |
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| 73. |
If the A.M. of the roots of a quadratic equation is `(8)/(5)` and the A.M. of their reciprocals is `(8)/(7)` then the equation isA. `5x^(2)-8x+7`=0B. `5x^(2)-16x+7`=0C. `7x^(2)-16x+5`=0D. `7x^(2)-16x+5`=0 |
| Answer» Correct Answer - B | |
| 74. |
For what value of b, will the roots of the equation cos x=b, `-1legle1` when arranged in ascending order of their magnitudes, form an A.P. ?A. -1B. `(sqrt(3))/(2)`C. `(1)/(sqrt(2))`D. `1//2` |
| Answer» Correct Answer - A | |
| 75. |
The interior angles of a polygon are in AP The smallest angle is `120` and the common difference is 5. Find the number of sides of the polygon.A. 9 or 16B. 9C. 16D. 13 |
| Answer» Correct Answer - B | |
| 76. |
The interior angles of a polygon are in arithmetic progression. The smallest angle is `120^@` and the common difference is `5^@` Find the number of sides of the polygon |
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Answer» Correct Answer - 9 Since, angle of polygon are in an AP. `:.` Sum of all angles `= (n -2) xx 180^(@) = (n)/(2) {2(120^(@)) + (n _1) 5^(@)}` `rArr 5n^(2) - 125 n + 720 = 0` `rArr n^(2) - 25n + 144 = 0` `rArr (n -9) (n-16) = 0` `rArr n = 9, 16` If n = 9, then largest angle `= a + 8d = 160^(@)` Again, if n = 16, then n largest angle `= a + 15d = 120^(@) + 75 = 195^(@)` Which is not possible. [since, any angle of polygon cannot be `gt 180^(@)`] Hence, n = 9 [neglecting n = 16] |
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| 77. |
If the `n^(th)` term of a G.P. is `3 (4^(n+1))` then its first term and common ratio are respectivelyA. 3,nB. n,4C. 3,4D. 4,3 |
| Answer» Correct Answer - C | |
| 78. |
If 2+x, 3+x, 9+x are in a G.P., then : x =A. `-(9)/(5)`B. `(9)/(5)`C. `-(5)/(9)`D. `(5)/(9)` |
| Answer» Correct Answer - A | |
| 79. |
If the positive numbers 3,x,5,y are in a G.P., then : `(x,y)=`A. `(3sqrt(15),2sqrt(5))`B. `(3sqrt(5),15sqrt(5))`C. `(2sqrt(15),3sqrt(5))`D. none of these |
| Answer» Correct Answer - D | |
| 80. |
Three given numbers whose sum is 24 are in an A.P. If the first is decreased by 1 the second is decreased by 2 and the third is left unchanged the resulting numbers are in a G.P. Then the given numbers areA. 3,13,8B. 4,8,12C. 13,3,8D. none of these |
| Answer» Correct Answer - B | |
| 81. |
Four number in G.P. such that the product of their extremes is 108 and the sum of the middle two is 24 areA. 9,10,11,12B. 2,18,6,54C. 2,6,8,54D. none of these |
| Answer» Correct Answer - C | |
| 82. |
If three numbers are in H.P., then the numbers obtained by subtracting half of the middle number from each of them are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 83. |
Prove that `9^(1//3)xx9^(1//9)xx9^(1//27)xx...oo=3`.A. 9B. 1C. 3D. none of these |
| Answer» Correct Answer - C | |
| 84. |
The minimum number of terms from the beginning of the series `20+22(2)/(3)+25(1)/(3)+ . . . .`, so that the sum may exceed 1568, isA. 25B. 27C. 28D. 29 |
| Answer» Correct Answer - D | |
| 85. |
Theaverage marks of boys in a class is 52 and that of girls is 42. The averagemarks of boys and girls combined is 50. The percentage of boys in the classis(1)40 (2) 20(3) 80 (4) 60 |
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Answer» `B:G=x:y` `Mavg=(Mb.x+My.y)/(x+y)` `50=(52x+42y)/(x+y)` `50x+50y=52x+42y` `2x=8y` `x/y=4` `%b=x/(x+y)xx100` `=x/(5x/4)xx100` `=100xx4/5` `=80%` option`(c)=80` |
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| 86. |
If the sum of a certain number of terms of the A.P. 25, 22, 19.... is 116. Find the last term. |
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Answer» Here, first term, `a = 25` Common difference, `d = -3` Sum,`S_n = 116` So, `n/2(2a+(n-1)d ) = 116` `=>n/2(2**25+(n-1)(-3) ) = 116` `=>n(50+3-3n) = 232` `=>3n^2-53n +232 = 0` `=>3n^2-29n-24n+232 = 0` `=>n(3n-29)-8(3n-29) = 0` `=>(n-8)(3n-29) = 0` `=>n = 8 and n = 29/3` As n is a whole number. So, it can not be `29/3` `:.` Number of terms,`n = 8`. `:.` Last term `= a+(n-1)d = 25+7(-3) = 4` |
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| 87. |
What is the geometric mean of the sequence `1,2, 4, 8,..., 2^n` |
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Answer» GM=`(1*2*4*8....2^n)^(1/(n+1))` GM=`(2^(0+1+2+3+...+n))^(1/(n+1))` GM=`(2^(n(n+1)/2))^(1/(n+1))` GM=`(2)^(n/2)`. |
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| 88. |
The sum of the first 5 terms and the sum of the first 10 terms of an AP are same.Which one of the following is the correct statement? (a) The first term must be negative (b) The common difference must be negative (c) Either the first term or the common difference is negative but not both (d) Both the first term and the common difference are negative |
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Answer» `S_5=5/2(2a+(5-1)d)=5/2(2a+4d)-(1)` `S_10=10/2[2a+(10-1)d]=5(2a+9d)-(2)` `S_5=S_10` `5/2(2a+4d)=5(2a+9d)` `2a+4d=4a+18d` `2a+14d=0` `a=-7d` option C is correct. |
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| 89. |
If `S_(n)` denotes the sum of first n terms of an A.P., then `(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))` is equal toA. 2n-1B. 2n+1C. 4n+1D. 2n+3 |
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Answer» Correct Answer - B We have, `(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))=((3n)/(2){2a+(3n-1)d}-(n-1)/(2){2a+(n-2)d})/((2n)/(2){2a+(2n-1)d}-(2n-1)/(2){2a+(2n-2)d})` `=(2a(3n-n+1)+{3n(3n-1)-(n-1)(n-2)}d)/(2a(2n-2n+1)+{2n(2n-1)-(2n-1)(2n-2)}d)` `=(2a(2n+1)+{9n^(2)-3n-n^(2)+3n-2}d)/(2a+{4n^(2)-2n-4n^(2)+6n-2}d)` `=(2a(2n+1)+(8n^(2)-2)d)/(2a+(4n-2)d)` `=(2(2n+1){a+(2n-1)d})/(2{a+(2n-1)d})=2n+1` |
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| 90. |
Let `S_n`denote the sum of first `n`terms of an A.P. If `S_(2n)=3S_n ,`then find the ratio `S_(3n)//S_ndot`A. 4B. 6C. 8D. 10 |
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Answer» Correct Answer - B Let a be the first term and d be the common difference of the A.P. Then, `S_(2n)=3S_(n)` `rArr" "(2n)/(2){2a+(2n-1)d}=(3n)/(2){2a+(n-1)d}rArr2a=(n+1)d`. `:." "(S_(3)n)/(S_(n))=((3n)/(2)[2a+(3n-1)d])/((n)/(2)[2a+(n-1)d])=(3[(n+1)d+(3n-1)d])/([(n+1)d+(n-1)d])=6` |
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| 91. |
Let `S_n` denote the sum of first n terms of an AP and `3S_n=S_(2n)`What is `S_(3n):S_n` equal to?What is `S_(3n):S_(2n)` equal to? |
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Answer» `S_n=n/2[2a+(n-1)d]` `3S_n=S_(2n)` `3(n/2)[2a+(n-1)d]=(2n)/2[2a+(2n-1)d]` `3/2[2a+(n-1)d]=2a+(2n-1)d` `3a+3/2nd-3/2d=2a+2nd-d` `a=(nd)/2+d/2` `a=(n+1)d/2` `2a=(n+1)d` `S_(3n)/S_n=((3n/2)[2a+(3n-1)d])/(n/2[2a+(n-1)d]` `=(3[nd+d+3nd-d])/(nd+d+nd-d)` `(12nd)/(2nd)` `S_(3n):S_n=6:1` option b is correcct. |
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| 92. |
If `S_(n)` is the sum of the first n terms of an A.P. then : (a) `S_(3n)=3(S_(2n)-S_n)` (b) `S_(3n)=S_n+S_(2n)` (c) `S_(3n)=2(S_(2n)-S_(n)` (d) none of theseA. `S_(3n)=3(S_(2n)-S_(n))`B. `2.S_(3n)=S_(n)+S_(2n)`C. `S_(3n)=2(S_(2n)-S_(n))`D. none of these |
| Answer» Correct Answer - A | |
| 93. |
If `S_(n)` denotes the sum of first n terms of an A.P., then `(S_(3n)-S_(n-1))/(S_(2n)-S_(n-1))` is equal toA. 21B. 15C. 16D. 19 |
| Answer» Correct Answer - B | |
| 94. |
If `a ,1/b ,a n d1/p ,q ,1/r`from two arithmetic progressions of the common difference, then `a ,q ,c`are in A.P. if`p ,b ,r`are in A.P. b. `1/p ,1/b ,1/r`are in A.P.c. `p ,b ,r`are in G.P. d. none oftheseA. p,b,r, in A.P.B. `(1)/(p),(1)/(b),(1)/( r)` in A.P.C. p,b,r in G.P.D. none of these |
| Answer» Correct Answer - B | |
| 95. |
IF in an A.P. `{a_(n)} a_(1)+a_(4)+a_(7)+cdots+a_(16)=147,` then : `a_(1)+a_(6)+a_(11)+a_(16)= cdots `. (a) 49 (b) 98 (c)147 (d) 196A. 49B. 98C. 147D. 196 |
| Answer» Correct Answer - B | |
| 96. |
If in an A.P. `{a_(n)}`, `a_(1)+a_(5)+a_(10)+a_(15)+_(20)+a_(24)=`225 then : `S_(24)= cdots`A. 550B. 900C. 1150D. 1400 |
| Answer» Correct Answer - B | |
| 97. |
The middle term of the progression `20,16,12, cdots , -176,-180` is (a)-46 (b)-76 (c)-80 (d) None of theseA. -46B. -76C. -80D. none of these |
| Answer» Correct Answer - C | |
| 98. |
If in an A.P. `{t_(n)}`, it is given that p.`t_(p)=q.t_(q)` then : `t_(p+q)= cdots`A. -1B. 1C. 0D. `-(p+q)` |
| Answer» Correct Answer - C | |
| 99. |
If in an A.P. `{t_(n)}`, it is given that `t_(p)`=q and `t_(q)`=p then : `t_(p+q)= cdots` |
| Answer» Correct Answer - A | |
| 100. |
Let `S_(k)=lim_(n to oo) sum_(i=0)^(n) (1)/((k+1)^(i))." Then " sum_(k=1)^(n) kS_(k)` equalsA. `(n(n+1))/(2)`B. `(n(n-1))/(2)`C. `(n(n+2))/(2)`D. `(n(n+3))/(2)` |
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Answer» Correct Answer - D We have, `S_(k)=underset(ntooo)limunderset(i=0)overset(n)sum(1)/((k+1)^(i))` `rArr" "S_(k)=1+(1)/((k+1)^(2))+(1)/((k+1)^(3))+ . . . .. ` `rArr" "S_(k)=(1)/(1-(1)/(k+1))=(k+1)/(k)` `:." "underset(k=1)overset(n)sumkS_(k)=underset(k=1)overset(n)sum(k+1)=2+3+ . . . +(n+1)` `rArr" "underset(k=1)overset(n)sumkS_(k)=(n)/(2)(2+n+1)=(n(n+3))/(2)` |
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