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InterviewSolution
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.
| 1. |
Let `f(n)` denote the `n^(th)` terms of the seqence of `3,6,11,18,27,….` and `g(n)` denote the `n^(th)` terms of the seqence of `3,7,13,21,….` Let `F(n)` and `G(n)` denote the sum of `n` terms of the above sequences, respectively. Now answer the following: `lim_(ntooo)(f(n))/(g(n))=`A. `0`B. `1`C. `2`D. `oo` |
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Answer» Correct Answer - B `(b)` `S=3+6+11+18+...+t_(n)` `S=3+6+11+...+t_(n-1)+t_(n)` `thereforeoverline(0=3+(3+5+7+9+...(n-1)"terms"-t_(n))` `:.t_(n)=n^(2)+2` Similarly `nth` terms of `g(n)=n^(n)+n+1` `:.lim_(ntooo)(n^(2)+2)/(n^(2)+n+1)=1` `F(n)=sum(n^(2)+2)` `=(n(n+1)(2n+1))/(6)+2n` `=(n(2n^(2)+3n+13))/(6)` `G(n)=sum(n^(2)+n+1)=(n(n^(2)+3n+5))/(3)` `:.lim_(ntooo)((n(2n^(2)+3n+13))/(6))/((n(n^(2)+3n+5))/(3))=1` |
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| 2. |
Let `f(n)` denote the `n^(th)` terms of the seqence of `3,6,11,18,27,….` and `g(n)` denote the `n^(th)` terms of the seqence of `3,7,13,21,….` Let `F(n)` and `G(n)` denote the sum of `n` terms of the above sequences, respectively. Now answer the following: `lim_(ntooo)(F(n))/(G(n))=`A. `2`B. `1`C. `0`D. `oo` |
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Answer» Correct Answer - B `(b)` `S=3+6+11+18+...+t_(n)` `S=3+6+11+...+t_(n-1)+t_(n)` `thereforeoverline(0=3+(3+5+7+9+...(n-1)"terms"-t_(n))` `:.t_(n)=n^(2)+2` Similarly `nth` terms of `g(n)=n^(n)+n+1` `:.lim_(ntooo)(n^(2)+2)/(n^(2)+n+1)=1` `F(n)=sum(n^(2)+2)` `=(n(n+1)(2n+1))/(6)+2n` `=(n(2n^(2)+3n+13))/(6)` `G(n)=sum(n^(2)+n+1)=(n(n^(2)+3n+5))/(3)` `:.lim_(nto oo)((n(2n^(2)+3n+13))/(6))/((n(n^(2)+3n+5))/(3))=1` |
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| 3. |
Find the sum of the series `1^2+3^2+5^2+ ton`terms. |
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Answer» Correct Answer - `(n(2n-1)(2n+1))/(3)` `1^(2)+3^(2)+5^(2)+…=Sigma(2n-1)^(2)` `=Sigma(4n^(2)-4n+1)` `=4Sigman^(2)-4Sigman+n` `=4(n(n+1)(2n+1))/3-2n(n+1)+n` `=n/3(4n^(2)-1)` `=(n(2n-1)(2n+1))/3` |
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| 4. |
The 15th term of the series `2 1/2+1 7/(13)+1 1/9+(20)/(23)+..` isA. `10/39`B. `10/21`C. `10/23`D. none of these |
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Answer» Correct Answer - A Reciprocals of the terms of the series are 2/5,13/20,9/10,23/20,…. Or 8/20,13/20,… Its nth term is `(8+(n-1)5)/20=(5n+3)/20` Therefore, the 15th term is `20/78=10/39`. |
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| 5. |
If the first two terms of a H.P. are `2//5and12//23` respectively. Then, largest term is |
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Answer» Correct Answer - Second therm is the largest Let the H.P. be `1/a,1/(a+d),1/(a+2d),1/(a+3d),..` Then `1/a=2/5 and 1/(a+d)=12/13` `rArra=5/2and d=-17/12` Now, nth term of the H.P. is `1/(a+(n-1)d)=12/(47-17n)` So, the nth term is largest when 47-17n has the least positive value. Clearly, 12/(47-17n) is least for n=2. Hence, 2nd term is the largest term. |
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| 6. |
Let `a_(1),a_(2),a_(3), . . .` be a harmonic progression with `a_(1)=5anda_(20)=25`. The least positive integer n for which `a_(n)lt0`, isA. 22B. 23C. 24D. 25 |
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Answer» Correct Answer - D `a_(1),a_(2),a_(3),…..` are in H.P. `rArr1/a_(1),1/a_(2),1/a_(3),`… are in A.P. `rArr1/a_(n)=1/a_(1)+(n-1)dlt0` where `(1/25-5/25)/19=d=((-4)/(19xx25)) ` `rArr1/5+(n-1)((-4)/(19xx25))lt0` or `(4(n-1))/(19xx5)gt1` or `n-1gt(19xx5)/4` or `ngt(19xx5)/4`+1 or `nge25` |
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| 7. |
If `Sigma_(r=1)^(50) (2)/(r^2+(11-r^2))`, then the value of n is __________ |
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Answer» Correct Answer - 50 `S=1/(2^(0)+2^(50))+1/(2^(1)+2^(50))+..+1/(2^(100)+2^(50))` `S=1/(2^(100)+2^(50))+1/(2^(99)+2^(50))+…+1/(2^(0)+2^(50))` `2S=1/(2^(50)+1/(2^(50))+..` 100 times `thereforeS=50/(2^(50))` |
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| 8. |
`Sigma_(r=1)^(50)(r^2)/(r^2+(11-r)^2)` is equal to ______. |
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Answer» Correct Answer - 25 Let I-`sum_(r=1)^(50)r^(2)/(r^(2)+(11-r)^(2))` ….(1) Writing the series in reverse order, we get `I=sum_(r=1)^(50)((11-r)^(2))/(r^(2)+(11-r)^(2))` …(2) Adding (1) and (2), we get 2I=50 `rArrI=25` |
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| 9. |
The value of the sum `Sigma_(i=1)^(20) i(1/i+1/(i+1)+1/(i+2)+.....+1/(2))` is ______________. |
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Answer» Correct Answer - 115 Given sum is `1(1/1+1/2+1/3+…+1/20)+2(1/2+1/3+..+1/20)+3(1/3+..+1/20)+20(1/20)` `=1/1(1)+1/2(1+2)+1/3(1+2+3)+…+1/20(1+2+3+….+20)` `=sum_(r=1)^(20)1/rcdot(r(r+1))/2` `=(1+1)/2+(2+1)/2+(3+1)/2+…+(20+1)/2` `=(20xx21)/4+10=115` |
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| 10. |
Find the sum of series upto n terms `((2n+1)/(2n-1))+3((2n+1)/(2n-1))^2+5((2n+1)/(2n-1))^3+...` |
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Answer» Correct Answer - 20 Let `(2n+1)/(2n-1)=r` Then the given series is `S=r+3r^(2)+5r^(3)+7r^(4)+…+(2n-1)r^(n)` `rS=r^(2)+3r^(3)+5r^(4)+..+(2n-3)r^(n)+(2n-1)r^(n+1)` ltbr. Subtracting, we get `(1-r)S=r+2r^(2)+2r^(3)+..+2r^(n)-(2n-1)r^(n+1)` `rArr820(1-r)=r(2r^(2)(1-r^(n-1)))/(1-r)-(2n-1)r^(n+1)` `rArr820(1-r)^(2)=r-r^(2)+2r^(2)-2r^(n+1)-(2n+1)r^(n+1)+(2n-1)r^(n+2)` `rArr820(1-r)^(2)=r+r^(2)-(2n+1)r^(n+1)+(2n-1)r^(n+2)` `rArr820(1-r)^(2)=r+r^(2)-(2n-1)[(2n+1)/(2n-1)r^(n+1)-r^(n+2)]` `rArr820(1-r)^(2)=r+r^(2)-(2n-1)[rcdotr^(n+1)-r^(n+2)]` `rArr820(1-r)^(2)=r(1+r)` `rArr820[1-(2n+1)/(2n-1)]^(2)=(2n+1)/(2n-1)[1+(2n+1)/(2n-1)]` `rArr[(-2)/(2n-1)]^(2)=(2n+1)/(2n-1)[(4n)/(2n-1)]` `rArr820=n(2n+1)` `rArrn=20` |
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| 11. |
Let `S=Sigma_(n=1)^(999) (1)/((sqrt(n)+sqrt(n+1))(4sqrt(n)+4sqrtn+1))` , then S equals ___________. |
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Answer» Correct Answer - 9 Given `S=sum_(n=1)^(9999)1/((sqrtn+sqrn(n+1)(root4n+root(4)(n+1)))` `=sum_(n=1)^(9999)1/((sqrtn+sqrt(n+1))(root4n+root(4)(n+1)))((root4n-root(4)(n+1))/(root4n-root(4)(n+1)))` `=sum_(n=1)^(9999)((n+1)^(1//4)-n^(1//4))` `=((2^(1/4)-1)+(3^(1/4)-2^(1/4))+(4^(1/4)-3^(1/4))+....+((9999+1)^(1/4)-(9999)^(1/4)))` `=(10^(4))^(1/4)-1` =9 |
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| 12. |
Find the sum `Sigma_(j=1)^(n) Sigma_(i=1)^(n) I xx 3^j` |
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Answer» Correct Answer - `(3n(3^n-1)(n+1))/(4)` `sum_(j=1)^(n)sum_(i=1)^(n)ixx3^(j)=(sum_(j=1)^(n)3^(j))(sum_(i=1)^(n)i)` `=(3+3^(2)+3^(3)+….+3^(n))xx(1+2+3+..+n)` `(3(3^(n)-1))/(3-1)xx(n(n+1))/2` `=(3n(3^(n)-1)(n+1))/4` |
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| 13. |
The value of `Sigma_(i=1)^(n) Sigma_(j=1)^(i) underset(k=1)overset(j)` =220, then the value of n equalsA. 11B. 12C. 10D. 9 |
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Answer» Correct Answer - C The sum equals `(n(n+1)(n+2))/6=220` Which is true for n=10 |
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| 14. |
Let `T_r`denote the rth term of a G.P. for `r=1,2,3,`If for some positive integers `ma n dn ,`we have `T_m=1//n^2`and `T_n=1//m^2`, then find the value of `T_(m+n//2.)` |
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Answer» Correct Answer - 1/mn Given that `T_(m)=AR^(m-1)=1/n^(2)andT_(n)=AR^(n-1)=1/m^(2)` `rArrA^(2)R^(m+n-2)=1/(m^(2)n^(2))` `rArrAR^((m+n)/2-1)=1/(mn)` `rArrT_((m+n)/2)=1/(mn)` |
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| 15. |
The value of `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)=220` , then the value of `n` equalsA. `11`B. `12`C. `10`D. `9` |
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Answer» Correct Answer - C `(c )` `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)1` `=sum_(i=1)^(n)sum_(j=1)^(i)j` `=sum_(i=1)^(n)(i(i+1))/(2)` `=(1)/(2)sum_(i=1)^(n)(i^(2)+i)` `=(n(n+1)(n+2))/(6)=220` `:.n=10` |
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| 16. |
If three positive real numbers a,b,c are in AP such that abc=4, then the minimum value of b isA. `2^(1//3)`B. `2^(2//3)`C. `2^(1//2)`D. `2^(3//2)` |
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Answer» Correct Answer - B Since a,b,c are in A.P., therefore, b-a=d and c-b=d, where d is the common difference of the A.P. `therefore a=b-d and c=b+d` Now, abc=4 `rArr(b-d)b(b+d)=4` `rArrb(b^(2)-d^(2))=4` But, `b(b^(2)-d^(2))ltbxxb^(2)` `rArrb(b^(2)-d^(2))ltb^(3)` `rArr4ltb^(3)` `rArrb^(3)gt4` `rArrbgt2^(2//3)` Hence, the minimum value of b is `2^(2//3)`. |
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| 17. |
Let the sum of first three terms of G.P. with real terms be 13/12 andtheir product is -1. If the absolute value of the sum of their infinite termsis `S ,`then the value of `7S`is ______. |
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Answer» Correct Answer - 4 Let `a/r`,a,ar be the three tems in G.P. `therefore` Product of terms=`a^(3)`=-1 (Given) `rArra=-1` Now, sum of terms `=a/r+a+ar=13/12` (Given) `rArr(-1)/r-1-r=13/12` or `12r^(2)+25r+12=0` or `(3r+4)(4r+3)=0` or `r=(-4)/3,(-3)/4` But `rne(-4)/3` `thereforeabsS=abs(a/(1-r))=abs((-1)/(1-((-3)/4)))=abs((-1)/(1+3/4))` `=abs((-4)/7)=4/7` |
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| 18. |
`a`, `b`, `c` are positive integers formaing an incresing `G.P.` and `b-a` is a perfect cube and `log_(6)a+log_(6)b+log_(6)c=6`, then `a+b+c=`A. `100`B. `111`C. `122`D. `189` |
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Answer» Correct Answer - D `(d)` `log_(6)(abc)=6` `implies(abc)=6^(6)` Let `a=(b)/(r )` and `c=br` ,brgt `impliesb=36` and `a=(36)/(r )impliesr=2,3,4,6,9,12,18` Also `36(1-(1)/(r ))` is a perfect cube. `impliesr=4` `impliesa+b+c=9+36+144=189` |
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| 19. |
Given that `alpha`, `gamma` are roots of the equation `Ax^(2)-4x+1=0` and `beta`, `delta` are roots of the equation `Bx^(2)-6x+1=0`. If `alpha`, `beta`,`gamma` and `delta` are in `H.P.`, thenA. `A=5`B. `A=3`C. `B=8`D. `B=-8` |
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Answer» Correct Answer - B::C `(b,c)` `alpha,beta,gamma,delta` are in `H.P.` `implies(1)/(alpha)`, `(1)/(beta)`, `(1)/(gamma)`, `(1)/(delta)` are in `A.P.` Let `d` be the common difference of the `A.P.` Since `alpha`, `gamma` are roots of `Ax^(2)-4x+1=0` `:.(alpha+gamma)/(alphagamma)=(4//A)/(1//A)=4` or `(1)/(alpha)+(1)/(gamma)=4` `implies(1)/(alpha)+(1)/(alpha)+2d=4` or `(1)/(alpha)+d=2`.......`(i)` Also `beta`, `delta` are roots of `Bx^(2)-6x+1=0` `:.(beta+delta)/(betadelta)=(1)/(beta)+(1)/(delta)=(6//B)/(1//B)=6` or `(1)/(alpha)+d+(1)/(alpha)+3d=6` `implies(1)/(alpha)+2d=3`..........`(ii)` Solving `(i)` and `(ii)`, we get `(1)/(alpha)=1` and `d=1` `:. (1)/(alpha)=1`, `(1)/(beta)=2`, `(1)/(gamma)=3` and `(1)/(delta)=4` since `(1)/(alphagamma)=AimpliesA=3` Also, `(betagamma)=BimpliesB=8` |
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| 20. |
If `log_(2)(5.2^(x)+1),log_(4)(2^(1-x)+1)` and 1 are in A.P,then x equalsA. `log_(2)5`B. `1- log_5 2`C. `log_(5)2`D. `1-log_(2)5` |
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Answer» Correct Answer - D The given numbers are in A.P. Therefore, `2log_(4)(2^(1-x)+1)=log_(2)(5xx2^(x)+1)+1` or `2log_(2^(2))(2/(2^(x))+1)=log_(2)(5xx2^(x)+1)+log_(2)2` or `2/2loglog_(2)(2/(2^(x))+1)=log_(2)(10xx2^(x)+2)` or `2/(2^(x))+1=10xx2^(x)+2` `rArr2/y+1=10y+2`, where `2^(x)=y` or `10y^(2)+y-2=0` or `(5y-2)(2y+1)=0` `rArry=2//5` or `y=-1//2` `rArr2^(x)=2//5` or `2^(x)=-1//2` `rArr x=log_(2)2-log_(2)5` `[because2^(x)` cannot be negative`]` `rArrx=log_(2)2-log_(2)5` `rArrx=1-log_(2)5` |
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| 21. |
The terms `a_1, a_2, a_3`from an arithmetic sequence whose sum s 18. The terms `a_1+1,a_2, a_3,+2,`in that order, form a geometric sequence. Then the absolute value ofthe sum of all possible common difference of the A.P. is ________. |
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Answer» Correct Answer - (-1) Let `a_(1)=a-d,a_(2)=a,a_(3)=a+d` `therefore3a=18` or a=6 Hence, the number in A.P. `6-d,d,6+d` `a_(1)+1,a_(2),a_(3)+2` in G.P. i.e., 7-d,6,8+d in G.P. `therefore` 36=(7-d)(8+d) `36=56-d-d^(2)` `d^(2)+d-20=0` Hence, the sum of all possible common different is -1. |
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| 22. |
If a,x,b are in A.P.,a,y,b are in G.P. and a,z,b are in H.P. such that x=9z and `gt0,bgt0`, thenA. `|y|=3z`B. `x=3|y|`C. `2y=x+z`D. none of these |
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Answer» Correct Answer - B x is A.M. of a and b, y is G.M. of a and b ,z is H.M. of and b. `y^(2)=xz` Also given, x=9z `rArrx=9y^(2)//x` `rArrx=3absy` |
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| 23. |
If `x,y,z` be three numbers in `G.P.` such that `4` is the `A.M.` between `x` and `y` and `9` is the `H.M.` between `y` and `z` , then `y` isA. `4`B. `6`C. `8`D. `12` |
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Answer» Correct Answer - B `(b)` Let `x=(k)/(r )`, `y=k`, `z=kr` `implies((k)/(r )+k)/(2)=4` ……..`(i)` and `(2k*kr)/(k+kr)=9` ………`(ii)` Solving we get `k=6` |
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| 24. |
If `x,y,z` are in `G.P. (x,y,z gt 1)` , then `(1)/(2x+log_(e)x)`, `(1)/(4x+log_(e)y)`, `(1)/(6x+log_(ez)z)` are inA. `A.P.`B. `G.P.`C. `H.P.`D. none of these |
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Answer» Correct Answer - C `(c )` As `x,y,z` are in `G.P.` `impliese^(2x)x,e^(4x)y,e^(6x)z` are in `G.P.` `implieslog_(e)(e^(2x)x),log_(e)(e^(4x)y),log_(e)(e^(6x)z)` are in `A.P.` `implies(1)/(2x+log_(e)x),(1)/(4x+log_(e)y),(1)/(6x+log_(e)z)` are in `H.P.` |
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| 25. |
Let `n in N ,n > 25.`Let `A ,G ,H`deonote te arithmetic mean, geometric man, and harmonic mean of 25 and `ndot`The least value of `n`for which `A ,G ,H in {25 , 26 , n}`isa. 49 b.81 c.169 d. 225A. 49B. 81C. 169D. 225 |
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Answer» Correct Answer - D `A=(25+n)/2,G=5sqrtn,H=(50n)/(25+n)` As A,G,H are natural numbers, n must be odd perfect square. Now, H will be a natural number, if we take n =225. |
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| 26. |
The arithmetic mean of two positive numbers is `6` and their geometric mean `G` and harmonic mean `H` satisfy the relation `G^(2)+3H=48`. Then the product of the two numbers isA. `24`B. `32`C. `48`D. `54` |
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Answer» Correct Answer - B `(b)` `a+b=12` `ab+(6ab)/(a+b)=48` `:.ab+(ab)/(2)=48` `:.ab=32` |
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| 27. |
If `1/(1^2)+1/(2^2)+1/(3^2)+ tooo=(pi^2)/6,t h e n1/(1^2)+1/(3^2)+1/(5^2)+`equals`pi^2//8`b. `pi^2//12`c. `pi^2//3`d. `pi^2//2`A. `pi^2//8`B. `pi^2//8`C. `pi//3`D. `pi^2//2` |
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Answer» Correct Answer - A `1/1^(2)+1/3^(2)+1/5^(2)+1/7^(2)+..` `=(1/1^(2)+1/2^(2)+1/3^(2)+1/4^(2)+1/5^(2)+1/6^(2)+1/7^(2)+..)` `-(1/2^(2)+1/4^(2)+1/6^(2)+..)` `=pi^(2)/6-1/4(1/1^(2)+1/2^(2)+1/3^(2)+..)` `=pi^(2)/6-1/4(pi^(2)/6)` `=pi^(2)/8` |
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| 28. |
Find the sum of the series `1+4/5+7/(5^2)+10/(5^3)+...`A. `7//16`B. `5//16`C. `105//64`D. `35//16` |
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Answer» Correct Answer - D Let ` S=1+4/5+7/5^(2)+10/5^(3)+..oo` Then `1/5S=1/5+4/5^(2)+7/5^(3)+..oo` or `S(1-1/5)=1+3[1/5+1/5^(2)+1/5^(3)+..oo]` or `4/5S=1+3xx(1//5)/(1-(1//5))=1+3/4=7/4` or `S=35/16` |
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| 29. |
If `S_p`denotes the sum of the series `1+r^p+r^(2p)+ toooa n ds_p`the sum of the series `1-r^(2p)r^(3p)+ tooo,|r| |
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Answer» Correct Answer - A `S_(p)=1/(1-r^(P)),s_(P)=1/(1+r^(p)),S_(2p)=1/(1-r^(2p))` Clearly `S_(p)+s_(p)=2/(1-r^(2p))=2S_(2p)` |
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| 30. |
If the sum to infinity of the series `1+2r+3r^2+4r^3+`is 9/4, then value of `r`is`1//2`b. `1//3`c. `1//4`d. none of theseA. `1//2`B. `1//3`C. `1//4`D. none of these |
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Answer» Correct Answer - B `S=1+2r+3r^(2)+4r^(3)+..` `thereforerS=r+2r^(2)+3r^(3)+4r^(4)+..` `rArr(1-r)S=1+r+r^(2)+r^(3)+..` `=1/(1-r)` `rArrS=1/((1-)^(2))` Given `S=9/4` `rArr1/((1-r)^(2))=9/4` `rArr1-r=pm2/3` `rArrr=1/3or5/3` Hence, r=1/3 as `0ltabsrlt1` |
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| 31. |
The sum to infinityof the series `1+2/3+6/(3^2)+(10)/(3^3)+(14)/(3^4). . . . . .`is(1) 2 (2) 3(3) 4 (4) 6A. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - B Let `S=1+2/3+6/3^(2)+10/3^(3)+14/3^(4)`+…. `therefore1/3S=1/3+2/3^(2)+6/3^(3)+10/3^(4)+….` subtracting (2) from (1), we get `S(1-1/3)=1+1/3+4/3^(2)+4/3^(3)+4/3^(4)+…..` `rArr2/3S=4/3+4/3^(2)(1+1/3+1/3^(2)+…)` `=4/3+4/3^(2)(1/(1-1/3)=4/3+4/3^(2)3/2=4/3+2/3=6/3` `rArrS=3` |
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| 32. |
The sum of `n` terms of series `ab+(a+1)(b+1)+(a+2)(b+2)+…+(a+(n-1)(b+(n-1))` if `ab=(1)/(6)` and `(1+b)=(1)/(3)` isA. `(n)/(6)(1-2n)^(2)`B. `(n)/(6)(1+n-2n^(2))`C. `(n)/(6)(1-2n+2n^(2))`D. None of these |
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Answer» Correct Answer - C `(c )` `S=ab+[ab+(a+b)+1]+[ab+2(a+b)+2^(2)]+…+[ab+(n-1)(a+b)+(n-1)^(2)]` `=nab+(a+b)sum_(r=1)^(n-1)r+sum_(r=1)^(n-1)r^(2)` `=nab+(a+b)(n(n-1))/(2)+((n-1)(n)(2n-1))/(6)` `=(n)/(6)[1+(n-1){1+2n-1}]` `=(n)/(6)[1+2n(n-1)]=(n)/(6)(1-2n+2n^(2))` |
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| 33. |
The sum `(7)/(2^2xx5^2)+13/(5^2xx8^2)+19/(8^2xx11^2)+…10` terms is S, then the value of 1024(S) is __________. |
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Answer» Correct Answer - 85 `S=7/(2^(2)xx5^(2))+13/(5^(2)xx8^(2))+19/(8^(2)xx11^(2))+….10` terms `=sum_(r=1)^(10)(6r+1)/((3r-1)^(2)(3r+2)^(2))` ltbr gt`=1/3sum_(r=1)^(10)((3r+2)^(2)-(3r-1)^(2))/((3r-1)^(2)(3r+2)^(2))` `=1/3sum_(r=1)^(10)(1/((3r-1)^(2))-1/((3r+2)^(2)))` `=1/3(1/2^(2)-1/2^(10))` `1/3((2^(8)-1)/2^(10))` `=85/1024` |
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| 34. |
The sum `2xx5+5xx9+8xx13+…10` terms isA. `4500`B. `4555`C. `5454`D. None of these |
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Answer» Correct Answer - B `(b)` `S=2xx5+5xx9+8xx13+…10`terms `t_(r )=(3r-1)(4r+1)` `=12r^(2)-r-1` `:.S=sum_(r=1)^(10)(12r^(2)-r-1)` `=12(10.11.21)/(6)-(10.11)/(2)-10` `=10(462-(11)/(2)-1)` `=4555` |
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| 35. |
A person drops a ball from an 80 m tall building and each time the ball bounces, it rebounds to p% of its previous height. If the ball travels a total distance of 320 m, then the value of p is |
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Answer» Correct Answer - 60 The ball is dropped from the heights, h=80 m. So, initially it travels the distance of 80 m. Then rebounds to x% of its previous height. So, it rebounds by the distance of `(80x)/100`m and from the same height it drps again. It continous like this till it comes to rest. So, toatal distance travelled by the ball is `320=80+(2xx(80x)/(100)+2xx(80x^(2))/(100^(2))+...)` `therefore4=1+((2x)/(100)+(2x^(2))/(100^(2))+..)` `rArr3=((2x)/100)/(1-x/100)` `rArr3=(2x)/(100-x)` `rArrx=60` |
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| 36. |
If `(10)^9 + 2(11)^1 (10)^8 + 3(11)^2 (10)^7+...........+10 (11)^9= k (10)^9` , then k is equal to :A. `121/10`B. `441/100`C. 100D. 110 |
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Answer» Correct Answer - C `S=10^(9)+2cdot11^(1)cdot10^(8)+…+10cdot11^(9)` `therefore11/10cdotS=11^(1)cdot10^(8)+..+9cdot11^(9)+11^(10)` Subtracting `rArr-1/10cdotS=10^(9)+11^(1)cdot10^(8)+11^(2)cdot10^(7)+…+11^(9)-11^(10)` `rArr-1/10S=10^(9)(((11/10)^(10)-1)/(11/10-1))-11^(10)` `rArr-1/10S=11^(10)-10^(10)-11^(10)` `rArrS=10^(11)` `rArrS=100cdot10^(9)` `rArrk=100` |
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| 37. |
The sum of first 20terms of the sequence 0.7, 0.77, 0.777, .. , is(1) `7/9(99-10^(-20))`(2)`7/(81)(179+10^(-20))`(3) `7/9(99+10^(-20))`(3)`7/(81)(179-10^(-20))`A. `7/81(179-10)^(20))`B. `7/9(99 -10^(20))`C. `7/81(179 +10^(-20))`D. `7/9(99+10^(-20))` |
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Answer» Correct Answer - C 0.7+0.77+0.777+….+0.7777….7 `=7/9[0.9+0.99+0.999+….+0.999…9]` `=7/9[(1-0.1)+(1-0.01)+(1-0.001)+….+(1-0.000…1)]` `=7/9[20-1/10cdot(1-1/10^(20))/(1-1/10)]=7/9[20-1/9cdot((10^(20)-1)/10^(20))]` `=7/81[180-(1-1/10^(20))]=7/81[179+10^(-20)]` |
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| 38. |
Divide 28 into four parts in an A.P. so that the ratio of the productof first and third with the product of second and fourth is 8:15. |
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Answer» Correct Answer - 4, 6, 8, 10 Let four member be a-3d,a-d,a+b,a+3d. Here, 4a=28 or a=7 Also, `((a-3d)(a+d))/((a-d)(a+3d))=8/15` or `15[a^(2)-3d^(2)-2ad]=8[a^(2)-3d^(2)+2ad]` or `7(a^(2)-3d^(2))=46ad` or `7(49-3d^(2))=46xx7xxd` or `49-3d^(2)=46d` `3d^(2)+46d-49=0` or (d-1)(3d+49)=0 or d=1 Therefore, the required numbers are 4, 6, 8, 10. |
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| 39. |
एक समांतर श्रेणी के प्रथम चार पदों का योगफल 56 है | अंतिम चार पदों का योगफल 112 है | यदि इसका प्रथम पद 11 है, तो पदों की संख्या ज्ञात किजिए | |
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Answer» Correct Answer - 11 terms Let the A.P. be 11,11+d,11+2d,11+3d,…,11+(n-2)d,11+(n-1)d. Sum of first four terms=11+(11+d)+(11+2d)+(11+3d) =44+6d Sum of last four terms=[11+(n-4)d]+[11+(n-3)d]+[11+(n-2)d]+[11+(n-1)d] =44+(4n-10)d According to the given condition, 44+6d=56`rArr`d=2 and 44+(4n-10)d=112 or 44+2(4n-10)=112 or n=11 Thus, the number of terms of the A.P is 11 |
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| 40. |
Let `a_1,a_2,a_3…… ,a_n ` be in G.P such that `3a_1+7a_2 +3a_3-4a_5=0` Then common ratio of G.P can beA. 2B. `3/2`C. `5/2`D. `-1/2` |
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Answer» Correct Answer - B::D Given `3a_(1)+7a_(2)+3a_(3)-4a_(5)=0` `rArr7(a_(1)+a_(2)+a_(3))=4(a_(1)+a_(3)+a_(5))` `rArr7(1+r+r^(2))=4(1+r^(2)+r^(4))` `rArr7=4(r^(2)-r+1)` `rArr4r^(2)-4r+1=4` `rArr(2r-1)^(2)=4` `rArr2r-1=pm2` `rArrr=3//2,-1//2` |
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| 41. |
The sum `1+3+7+15+31+...to100`terms is`2^(100)-102 b`b. `2^(99)-101`c. `2^(101)-102`d. none of theseA. `2^(100)-102`B. `2^(99)-101`C. `2^(101)-102`D. none of these |
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Answer» Correct Answer - C Here the successive differences are 2,4,8,16,… which are in G.P. `S=1+3+7+15+31+…+T_(100)` `=(2^(1)-1)+(2^(2)-1)+(2^(3)-1)+..+(2^(100)-1)` `=(2+2^(2)+2^(3)+….+2^(100))-100` `=2((2^(100)-1)/(2-1))-100` `=2^(101)-102` |
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| 42. |
Suppose that `F(n +1) =( 2f(n)+1)/2` for n = 1, 2, 3,.....and f(1)= 2 Then F(101) equals = ?A. 50B. 52C. 54D. none of these |
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Answer» Correct Answer - B Given, `F(n+1)=(2F(n)+1)/2` `rArrF(n+1)-F(n)=1//2` Hence, the given series is an A.P. with common difference 1/2 and first term being 2. F(101) is 101st term of A.P. given by 2+(101-1)(1/2)=52 |
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| 43. |
If pth, qth , rth and sth terms of an AP are in GP then show that (p-q), (q-r), (r-s) are also in GPA. A.PB. G.PC. H.PD. none of these |
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Answer» Correct Answer - B Given that a+(p-1)d=A a+(q-1)d=AR `a+(r-1)d=AR^(2)` `a+(s-1)d=AR^(3)` where R is common ratio of G.P. Now, `p-q=(A-AR)/d,q-r=R((A-AR)/d)` Clearly,p-q,q-r,r-s are in G.P. |
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| 44. |
If the pth term of an A.P. is q and the qth term isp, then find its rth term. |
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Answer» Correct Answer - p+q-r a+(p-1)d=q and a+(q-1)d=p `rArr(p-q)d=q-p` `rArr d=-1` `T_(r)=a+(r-1)d=a+(p-1)d+(r-p)d` `=q-(r-p)=p+q-r` |
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| 45. |
Let `A_1,A_2,A_3,….,A_m` be the arithmetic means between -2 and 1027 and `G_1,G_2,G_3,…., G_n` be the gemetric means between 1 and 1024 .The product of gerometric means is `2^(45)` and sum of arithmetic means is `1024 xx 171` The value of `Sigma_(r=1)^(n) G_r ` is |
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Answer» Correct Answer - C `G_(1)G_(2)…G_(n)=(sqrt(1xx1024))^(n)=2^(5n)` Given, `2^(5n)=2^(45)rArrn=9` Hence, `r=(1024)^(1/(9+1))=2` `rArrG_(1)=2,r=2` `rArrG_(1)+G_(2)+..+G_(9)=(2xx(2^(9)-1))/(2-1)=1024-2=1022` |
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| 46. |
The sum to 50 terms of the series `3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is `A. `100/17`B. `150/17`C. `200/51`D. `50/17` |
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Answer» Correct Answer - A Let `T_(r)` be the rth term of the given series. Then, `T_(r)=(2r+1)/(1^(2)+2^(2)+…+r^(2))=(6(2r+1))/((r )(r+1)(2r+1))` `=6(1/r-1/(r+1))` So, sum is given by `sum_(r=1)^(50)T_(r)=6sum_(r=1)^(50)(1/r-1/(r+1))` `=6[(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/50-1/51)]` `=6[1-1/51]=100/17` |
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| 47. |
Let S=`4/19+44/(19)^2+444/(19)^3+...oo` then find the value of SA. `40//9`B. `38//81`C. `36//171`D. none of these |
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Answer» Correct Answer - B `S=4/19+44/19^(2)+444/19^(3)+…` (1) `rArr1/19S=4/19^(2)+44/19^(3)+…` (2) Subtracting (2) from (1), we get `18/19S=4/19+40/19^(2)+400/19^(3)+…` `=(4/19)/(1-10/19)=4/9` `rArrS=38/81` |
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| 48. |
If `a_(1), a_(2), …..,a_(n)` are in A.P. with common difference `d ne 0,` then the sum of the series sin `d[sec a_(1)sec a_(2) +..... sec a_(n-1) sec a_(n)]` isA. `cosec a_(n)-cosec a`B. `cot a_(n)-cot a`C. `sec a_(n)- sec a_(1)`D. `tan a_(n)- tan a_(1)` |
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Answer» Correct Answer - D As `a_(1),a_(2),a_(3),…,a_(n-1),a_(n)` are in A.P., hence `d=a_(2)-a_(1)=a_(3)-a_(2)=…=a_(n)-a_(n-1)` `sind[seca_(1)seca_(2)+seca_(2)seca_(3)+…+seca_(n-1)seca_(n)]` `=(sin(a_(2)-a_(1)))/(cosa_(1)cosa_(2))+(sin(a_(3)-a_(2)))/(cosa_(2)cosa_(3))+...+(sin(a_(n)-a_(n-1)))/(cosa_(n-1)cosa_(n))` `=(tana_(2)-tana_(1))+(tana_(3)-tana_(2))+...+(tana_(n)-tana_(n-1))` `=tana_(n)-tana_(1)` |
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| 49. |
Let `a,x,b` be in `A.P`, `a,y,b` be in `G.P` and `a,z,b` be in `H.P`. If `x=y+2` and `a=5z`, thenA. `y^(2)=xz`B. `x gt y gt z`C. `a=9`, `b=1`D. `a=1//4`, `b=9//4` |
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Answer» Correct Answer - A::B::C `(a,b,c)` `2x=a+b` `y^(2)=ab` `z=(2ab)/(a+b)` `x=y+2` and `a=5z` `z=(2y^(2))/(2x)` `impliesy^(2)=xz` `x=y+2` `:.(a+b)/(2)=sqrt(ab)+2`………`(i)` and `a=5((2ab))/(a+b)` `implies(a+b)=10b` `impliesa=9b` `:.` from `(i) (9b+b)/(2)=sqrt(9b^(2))+2` `implies5b=3b+2` `impliesb=1` So `a=9` `implies x gt y gt z` |
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| 50. |
If the sum of n terms of an A.P is cn (n-1)where `c ne 0` then the sum of the squares of these terms isA. `c^2n(n+1)^2`B. `2/3c^2n(n-1)(2n-1)`C. `(2c^2)/(3)n(n+1)(2n+1)`D. none of these |
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Answer» Correct Answer - B If `t_(r)` be the rth term of the A.P., then `t_(r)=S_(r)-S_(r-1)` `=cr(r-1)-c(r-1)(r-2)` `=c(r-1)(r-r+2)-2c(r-1)` We have, `t_(1)^(2)+t_(2)^(2)+…+t_(n)^(2)=4c^(2)(0^(2)+1^(2)+2^(2)+…+(n-1)^(2))` `=4c^(2)((n-1)n(2n-1))/6` `=2/3c^2n(n-1)(2n-1)` |
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