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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. |
An aeroplane flys around squares whose all sides are of length `100` miles. If the aeroplane covers at a speed of `100 mph` the first side, `200 mph` the second side `300 mph` the third side and `400 mph` the fourth side. The average speed of aeroplane around the square isA. `190mph`B. `195mph`C. `192mph`D. `200mph` |
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Answer» Correct Answer - C `(c )` `"Average speed"=H.M` `(4)/((1)/(100)+(1)/(200)+(1)/(300)+(1)/(400))=192` |
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| 52. |
In a G.P the sum of the first and last terms is 66, the product of the second and the last but one is 126, and the sum of the terms is 128 In any case, the difference of the least and greatest terms isA. 78B. 126C. 126D. none of these |
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Answer» Correct Answer - D Let a be the term and r the common ratio of the given G.P. Further, let there be n terms in the given G.P. Then `a_(1)+a_(n)=66rArra+ar^(n-1)=66` `a_(2)xxa_(n-1)=128` `rArrarxxar^(n-2)=128` or `a^(2)r^(n-1)=128` or `axx(ar^(n-1))=128` or `ar^(n-1)=128/a` Putting this value of `ar^(n-1)` in (1), we get `a+128/a=66` or `a^(2)-66a+128=0` or `(a-2)(a-64)=0` or a=2,64 Putting a=2 in (1), we get `2+2xxr^(n-1)=66` or `r^(n-1)=32` Putting a=64 in (1), we get `64+64r^(n-1)=66` or `r^(n-1)=1/32` For an increasing G.P., `rgt1`. Now, `S_(n)=126` `rArr2((r^(n)-1)/(r-1))=126` or `(r^(n)-1)/(r-1)=63` or `(r^(n-1)xxr-1)/(r-1)=63` or `(32r-1)/(r-1)=63` or r=2 `thereforer^(n-1)=32` `rArr2^(n-1)=32=2^(5)` `rArrn-1=5` `rArrn=6` For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128. For a=2,r=2, terms are 2,4,8,16,32,64. For a=64,r=1/2 terms are 64,32,16,8,4,2. Hence, difference is 62. |
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| 53. |
If `(1+x)(1+x^2)(1+x^4)….(1+x^(128))=Sigma_(r=0)^(n) x^r`, then n is equal isA. 256B. 255C. 254D. none of these |
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Answer» Correct Answer - B Degree of x on L.H.S is 1+2+4+…+128=`1+2+2^(2)+….+2^(7)` `=(2^(8)-1)/(2-1)=255` |
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| 54. |
In a n increasing G.P. , the sum of the firstand the last term is 66, the product of the second and the last but one is128 and the sum of the terms is 126. How many terms are there in theprogression?A. 64B. 128C. 256D. 729 |
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Answer» Correct Answer - B Let a be the term and r the common ratio of the given G.P. Further, let there be n terms in the given G.P. Then `a_(1)+a_(n)=66rArra+ar^(n-1)=66` `a_(2)xxa_(n-1)=128` `rArrarxxar^(n-2)=128` or `a^(2)r^(n-1)=128` or `axx(ar^(n-1))=128` or `ar^(n-1)=128/a` Putting this value of `ar^(n-1)` in (1), we get `a+128/a=66` or `a^(2)-66a+128=0` or `(a-2)(a-64)=0` or a=2,64 Putting a=2 in (1), we get `2+2xxr^(n-1)=66` or `r^(n-1)=32` Putting a=64 in (1), we get `64+64r^(n-1)=66` or `r^(n-1)=1/32` For an increasing G.P., `rgt1`. Now, `S_(n)=126` `rArr2((r^(n)-1)/(r-1))=126` or `(r^(n)-1)/(r-1)=63` or `(r^(n-1)xxr-1)/(r-1)=63` or `(32r-1)/(r-1)=63` or r=2 `thereforer^(n-1)=32` `rArr2^(n-1)=32=2^(5)` `rArrn-1=5` `rArrn=6` For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128. For a=2,r=2, terms are 2,4,8,16,32,64. For a=64,r=1/2 terms are 64,32,16,8,4,2. Hence, difference is 62. |
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| 55. |
If a,b and c also represent the sides of a triangle and a,b,c are in g.p then the complete set of `alpha^2 = (r^2 +r +1)/(r^2-r+1)`isA. `(1/3,3)`B. (2,3)C. `[1/3,2]`D. `(sqrt(5+3)/2,3)` |
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Answer» Correct Answer - D Let b=ar,c=`ar^(2)andrgt0` As the sum of two sides is more than the third side, we have `rin((sqrt5-1)/2,(sqrt5+1)/2)` - {1} `rArrr+1/r-1in(1,sqrt5-1)` As `alpha^(2)=(r^(2)+r+1)/(r^(2)-r+1)=1+2/(r+1/r-1)` `thereforealpha^(2)in((sqrt5+3)/2,3)` |
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| 56. |
There are two sets A and B each of which consists of three numbers in A.P.whose sum is 15 and where D and d are the common differences such that `D-d=1. If p/q=7/8`, where p and q are the product of the numbers ,respectively, and ` d gt 0` in the two sets . The sum of the product of the numbers in set B taken two at a time isA. 74B. 64C. 73D. 81 |
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Answer» Correct Answer - A Let numbers in set A be a-D,a,a+D and those in set B be b-d,b,b+d. Now, 3a=3b=15 or a=b=5 Set A = {5-D,5,5+D} set B = {5-d,5,5+d} where D=d+1 Also, `p/q=(5(25-D^(2)))/(5(25-d^(2)))=7/8` `rArr25(8-7)=8(d+1)^(2)-7d^(2)` `rArrd=-17,1` but `dgt0rArrd=1` So, the numbers in set A are 3,5,7 and the nnumbers in set are 4,5,6. Now, sum of product of numbers in set are 4,5,6. Now, sum of product of numbers in set A taken two at a time is `3xx5+3xx7+5xx7=71`. The sum of product of numbers in set B taken two at a time is `4xx5+5xx6+6xx4=74`. Also, `p=3xx5xx7=105 and q=4xx5xx6=120` `rArrq-p=15` |
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| 57. |
Four different integers form an increasing A.P .One of these numbers is equal to the sum of the squares of the other three numbers. Then The product of all numbers isA. `-2`B. 1C. 0D. 2 |
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Answer» Correct Answer - C Let the four integers be a-d,a,a+d, and a+2d, where a and d are integers and `dgt0`. Now, `a+2d=(a-d)^(2)+a^(2)+(a+d)^(2)` or `2d^(2)-2d+3a^(2)-a=0` (1) `therefored=1/2[1pmsqrt(1+2a-6a^(2))]` (2) Since d is a positive integer, so `1+2a-6a^(2)gt0` or `6a^(2)-2a-1lt0` or `(1-sqrt7)/6ltalt(1+sqrt7)/6` or a=0 (`because` a is an integer) Hence, from (2), d=1 or 0 But since `dgt0`, `therefore` d=1 Hence, the four numbers are -1,0,1,2. |
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| 58. |
In a n increasing G.P. , the sum of the firstand the last term is 66, the product of the second and the last but one is128 and the sum of the terms is 126. How many terms are there in theprogression?A. 9B. 8C. 12D. 6 |
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Answer» Correct Answer - D Let a be the term and r the common ratio of the given G.P. Further, let there be n terms in the given G.P. Then `a_(1)+a_(n)=66rArra+ar^(n-1)=66` `a_(2)xxa_(n-1)=128` `rArrarxxar^(n-2)=128` or `a^(2)r^(n-1)=128` or `axx(ar^(n-1))=128` or `ar^(n-1)=128/a` Putting this value of `ar^(n-1)` in (1), we get `a+128/a=66` or `a^(2)-66a+128=0` or `(a-2)(a-64)=0` or a=2,64 Putting a=2 in (1), we get `2+2xxr^(n-1)=66` or `r^(n-1)=32` Putting a=64 in (1), we get `64+64r^(n-1)=66` or `r^(n-1)=1/32` For an increasing G.P., `rgt1`. Now, `S_(n)=126` `rArr2((r^(n)-1)/(r-1))=126` or `(r^(n)-1)/(r-1)=63` or `(r^(n-1)xxr-1)/(r-1)=63` or `(32r-1)/(r-1)=63` or r=2 `thereforer^(n-1)=32` `rArr2^(n-1)=32=2^(5)` `rArrn-1=5` `rArrn=6` For decreasing G.P., a=64 and r=1/2. Hence, the sum of infiite terms is 64/{1-(1/2)}=128. For a=2,r=2, terms are 2,4,8,16,32,64. For a=64,r=1/2 terms are 64,32,16,8,4,2. Hence, difference is 62. |
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| 59. |
Consider three distinct real numbers a,b,c in a G.P with `a^2+b^2+c^2=t^2` and a+b+c =`alpha t` .The sum of the common ratio and its reciprocal is denoted by S. Complete set of `alpha^2` isA. `(-2,2)`B. `(-oo,-2) cup (2,oo)`C. `(-1,1)`D. `(-oo, -1)cup (1,oo)` |
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Answer» Correct Answer - B `S=r+1/r` `Sin(-oo,-2)cup(2,oo)` |
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| 60. |
There are two sets A and B each of which consists of three numbers in A.P.whose sum is 15 and where D and d are the common differences such that `D-d=1. If p/q=7/8`, where p and q are the product of the numbers ,respectively, and ` d gt 0` in the two sets . The sum of the products of the numbers is set A taken two at at time isA. 51B. 71C. 74D. 86 |
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Answer» Correct Answer - B Let numbers in set A be a-D,a,a+D and those in set B be b-d,b,b+d. Now, 3a=3b=15 or a=b=5 Set A = {5-D,5,5+D} set B = {5-d,5,5+d} where D=d+1 Also, `p/q=(5(25-D^(2)))/(5(25-d^(2)))=7/8` `rArr25(8-7)=8(d+1)^(2)-7d^(2)` `rArrd=-17,1` but `dgt0rArrd=1` So, the numbers in set A are 3,5,7 and the nnumbers in set are 4,5,6. Now, sum of product of numbers in set are 4,5,6. Now, sum of product of numbers in set A taken two at a time is `3xx5+3xx7+5xx7=71`. The sum of product of numbers in set B taken two at a time is `4xx5+5xx6+6xx4=74`. Also, `p=3xx5xx7=105 and q=4xx5xx6=120` `rArrq-p=15` |
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| 61. |
The sum of all the four numbers isA. 3B. 0C. 4D. 2 |
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Answer» Correct Answer - D Let the four integers be a-d,a,a+d, and a+2d, where a and d are integers and `dgt0`. Now, `a+2d=(a-d)^(2)+a^(2)+(a+d)^(2)` or `2d^(2)-2d+3a^(2)-a=0` (1) `therefored=1/2[1pmsqrt(1+2a-6a^(2))]` (2) Since d is a positive integer, so `1+2a-6a^(2)gt0` or `6a^(2)-2a-1lt0` or `(1-sqrt7)/6ltalt(1+sqrt7)/6` or a=0 (`because` a is an integer) Hence, from (2), d=1 or 0 But since `dgt0`, `therefore` d=1 Hence, the four numbers are -1,0,1,2. |
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| 62. |
If a, b and c are roots of the equation `x^3+qx^2+rx +s=0 ` then the value of r isA. 184B. 196C. 224D. none of these |
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Answer» Correct Answer - B We have, a+b+c=25 2a=b+2 `c^(2)=18b` Elminating a from (1) and (2), we have `b=16-(2c)/3` Then from (3), `c^(2)=18(16-(2c)/3)` or `c^(2)+12c-18xx16=0` or `(c-12)(c+24)=0` Now, c=-24 is not possible since it does not lie between 2 and 18. Hence, c=12. Then from (3), b=8 and finally from (2),a = 5. Thus, a=5,b=8 and c=12. Hence, abc`=5xx8xx12=480`. Also,equation `ax^(2)+bx+c=0 is 5x^(2)+8x+12=0`, which has imaginary roots. If a,b,c are roots of the equation `x^(3)+qx^(2)+rx+s=0`, then the sum of product of roots taken two at a time is `r=5xx8+5xx12+8xx12=196`. |
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| 63. |
Consider three distinct real numbers a,b,c in a G.P with `a^2+b^2+c^2=t^2` and a+b+c =`alpha t` .The sum of the common ratio and its reciprocal is denoted by S. Complete set of `alpha^2` isA. `(1/3,3)`B. `[1/3,3]`C. `(1/3,3)-{1}`D. `(-oo,1/3)cap (3,oo)` |
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Answer» Correct Answer - C a,b,c are in G.P. Hence,a,ar,`ar^(2)` are in G.P. So, `(a^(2)+a^(2)r^(2)+a^(2)r^(4))/((a+ar+ar^(2))^(2))=t^(2)/(alpha^(2)t^(2))=1/alpha^(2)` `alpha^(2)=(r^(2)+r+1)/(r^(2)-r+1)` Let `alpha^(2)=y`, `thereforey=(r^(2)+r+1)/(r^(2)-r+1)` `(y-1)r^(2)-r(y+1)+(y-1)=0` For real r, `(y+1)^(2)-4(y-1)^(2)ge0` `rArr1/3leyle3` But `ne1//3,1,3` `(becauserne1,-1,0)` `therefore1/3ltylt3andyne1` `alpha^(2)in(1/3,3)-{1}` |
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| 64. |
`a`, `b`, `c`, `d` are in increasing `G.P.` If the `AM` between `a` and `b` is `6` and the `AM` between `c` and `d` is `54`, then the `AM` of `a` and `b` isA. `15`B. `48`C. `44`D. `42` |
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Answer» Correct Answer - D `(d)` Let `r` is the common ratio. `implies(a+ar)/(2)=6` and `(ar^(2)+ar^(3))/(2)=54` `impliesr^(2)=9impliesr=+-3impliesr=3(r ne -3)` When `r=3`, `a=3, `AM` of `a` and `d=(a+ar^(3))/(2)=42` |
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| 65. |
Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..)A. 1088B. 1008C. 1040D. none of these |
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Answer» Correct Answer - B 1,2,2,3,3,3,4,4,4,4,.. Let us write the terms in the groups as follows: 1,(2,2),(3,3,3),(4,4,4,4),… consisting of 1,2,3,4,.. Terms. Let 2000th term fall in nth group. Then, `((n-1)n)/2lt2000le(n(n+1))/2` or n(n-1)`lt4000len(n+1)` Let us consider, `n(n-1)lt4000` or `n^(2)-n-4000lt0` or `nlt(1+sqrt(16001))/2` r `nlt64` We have `n(n+1)ge4000` or `n^(2)+n-4000ge0` or `nge63` That means 2000th term falls in 63rd group, which means that the 2000th term is 63. Now, the total number of terms up to 62nd group is `(62xx63)//2=1953`. Hence, the sum of first 2000 terms is `1^(2)+2^(2)+..+62^(2)+63(2000-1953)` `=(62(63)125)/6+63xx47=84336` Sum of the remaining terms=`63xx16=1008`. |
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| 66. |
Let `a ,b ,c ,d`be four distinct real numbers in A.P. Then half of the smallestpositive valueof `k`satisfying `a(a-b)+k(b-c)^2=(c-a)^3=2(a-x)+(b-d)^2+(c-d)^3`is __________. |
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Answer» Correct Answer - 16 Since a,b,c,d are in A.P., we have b-a=c-b=d-c=D (let common difference) or d=a+3D `rArra-d=-3D and d=b+2D` or b-d=-2D Also c=a+2D or c-a=2D `therefore` Given equation 2(a-b)`+k(b-c)^(2)+(c-a)^(3)` `=2(a-d)+(b-d)^(2)+(c-d)^(3)` becomes `-2D+kD^(2)+(2D)^(3)=-6D+4D^(2)-D^(3)` `rArr9D^(2)+(k-4)D+4=0` Since D is real, we have `(k-4)^(2)-4(4)(9)ge0` or `k^(2)-8k-128ge0` or `(k-16)(k+8)ge0` `thereforekin(-oo,-8]cup[16,oo)` Hence, the smallest positive value of k = 16. |
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| 67. |
The common difference of the divisible byA. 1B. 3C. 2D. `-2` |
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Answer» Correct Answer - A Let the four integers be a-d,a,a+d, and a+2d, where a and d are integers and `dgt0`. Now, `a+2d=(a-d)^(2)+a^(2)+(a+d)^(2)` or `2d^(2)-2d+3a^(2)-a=0` (1) `therefored=1/2[1pmsqrt(1+2a-6a^(2))]` (2) Since d is a positive integer, so `1+2a-6a^(2)gt0` or `6a^(2)-2a-1lt0` or `(1-sqrt7)/6ltalt(1+sqrt7)/6` or a=0 (`because` a is an integer) Hence, from (2), d=1 or 0 But since `dgt0`, `therefore` d=1 Hence, the four numbers are -1,0,1,2. |
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| 68. |
Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..) The sum of first 2000 terms isA. 84336B. 96324C. 78466D. none of these |
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Answer» Correct Answer - A 1,2,2,3,3,3,4,4,4,4,.. Let us write the terms in the groups as follows: 1,(2,2),(3,3,3),(4,4,4,4),… consisting of 1,2,3,4,.. Terms. Let 2000th term fall in nth group. Then, `((n-1)n)/2lt2000le(n(n+1))/2` or n(n-1)`lt4000len(n+1)` Let us consider, `n(n-1)lt4000` or `n^(2)-n-4000lt0` or `nlt(1+sqrt(16001))/2` r `nlt64` We have `n(n+1)ge4000` or `n^(2)+n-4000ge0` or `nge63` That means 2000th term falls in 63rd group, which means that the 2000th term is 63. Now, the total number of terms up to 62nd group is `(62xx63)//2=1953`. Hence, the sum of first 2000 terms is `1^(2)+2^(2)+..+62^(2)+63(2000-1953)` `=(62(63)125)/6+63xx47=84336` Sum of the remaining terms=`63xx16=1008`. |
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| 69. |
Consider the sequence in the form of group (1),(2,2)(3,3,3),(4,4,4,4),(5,5,5,5,5…..) The `2000^(th)` term of the sequence is not divisible byA. 3B. 9C. 7D. none of these |
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Answer» Correct Answer - D 1,2,2,3,3,3,4,4,4,4,.. Let us write the terms in the groups as follows: 1,(2,2),(3,3,3),(4,4,4,4),… consisting of 1,2,3,4,.. Terms. Let 2000th term fall in nth group. Then, `((n-1)n)/2lt2000le(n(n+1))/2` or n(n-1)`lt4000len(n+1)` Let us consider, `n(n-1)lt4000` or `n^(2)-n-4000lt0` or `nlt(1+sqrt(16001))/2` r `nlt64` We have `n(n+1)ge4000` or `n^(2)+n-4000ge0` or `nge63` That means 2000th term falls in 63rd group, which means that the 2000th term is 63. Now, the total number of terms up to 62nd group is `(62xx63)//2=1953`. Hence, the sum of first 2000 terms is `1^(2)+2^(2)+..+62^(2)+63(2000-1953)` `=(62(63)125)/6+63xx47=84336` Sum of the remaining terms=`63xx16=1008`. |
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| 70. |
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6: 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is |
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Answer» Correct Answer - 9 Given `s_1/S_11=6/11` `rArr (7/2[2a+6d])/(11/2[2a+10d])=6/11` `rArr (7(a+3d))/(11(a+5d))=6/11` `rArr 7a+21d= 6a+30d ` `rArr a = 9d` `rArr T_7=a+6d=15d` Given `130 lt 15 d lt 140` `rArr d=9` |
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| 71. |
The value of `Sigma_(r=1)^(n) (a+r+ar)(-a)^r` is equal toA. `(-1)^n[n+1)a^(n+1)-a]`B. `(-1)^n(n+1)a^(n+1)`C. `(-1)^n((n+2)a^(n+1))/2`D. `(-1)^n(na^n)/(2)` |
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Answer» Correct Answer - B `T_(r)=r(-a)^(r )+(r+1)a(-a)^(r )` `=r(-a)^(r )-(r+1)(-a)^(r+1)` `=v(r )-v(r+1)` So, `sum_(r=0)^(n)T_(r )=sum_(r=0)^(n)(v(r )-v(r+1))` `=v_(0)-v_(n+1)` `=-(n+1)(-a)^(n+1)` |
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| 72. |
Value of `(1+1/3)(1+1/(3^2))(1+1/(3^4))(1+1/(3^8))oo`is equal to`3`b. `6/5`c. `3/2`d. none of theseA. 3B. `6/5`C. `3/2`D. none of these |
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Answer» Correct Answer - C Consider the first product, `P=(1+1/3)(1+1/3^(2))(1+1/3^(4))(1+1/3^(8))..(1+1/(3^(2^(n))))` `=((1+1/3)(1+1/3^(2))(1+1/3^(4))(1+1/3^(8))...(1+1/3^(2^(n))))/((1-1/3))` `=((1+1/3^(2))(1+1/3^(2))(1+1/3^(4))(1+1/3^(8))...(1+1/3^(2^(n))))/((1-1/3))` `=((1+1/3^(4))(1+1/3^(4))(1+1/3^(8))...(1+1/3^(2^(n))))/((1-1/3))` `1/((1-1/3))(1-(1/3)^(2^(n-1)))` `=3/2(1-(1/3)^(2^(n+1)))` `rArr(1+1/3)(1+1/3^(2))(1+1/3^(4))(1+1/3^(8))...` infinity `=lim_(ntoo)3/2(1-(1/3)^(2^(n+1)))=3/2` |
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| 73. |
Find the sum `11^2-1^2+12^2-2^2+13^2-3^2+……+20^2-10^2` |
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Answer» Correct Answer - 1400 `11^(2)-1^(2)+12^(2)-2^(2)+13^(2)-3^(2)+…+20^(2)-10^(2)` `=(1^(2)+2^(2)+3^(2)+….+20^(2))-2(1^(2)+2^(2)+3^(2)+…+10^(2))` `=(20xx21xx41)/6-2xx(10xx11xx21)/6` `=(20xx21)/6(41-21)` =1400 |
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| 74. |
Given the sequence of numbers `x_(1),x_(2),x_(3),x_(4),….,x_(2005)`, `(x_(1))/(x_(1)+1)=(x_(2))/(x_(2)+3)=(x_(3))/(x_(3)+5)=...=(x_(2005))/(x_(2005)+4009)`, the nature of the sequence isA. `A.P.`B. `G.P.`C. `H.P.`D. None of these |
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Answer» Correct Answer - A `(a)` Given `(x_(1))/(x_(1)+1)=(x_(2))/(x_(2)+3)=(x_(3))/(x_(3)+5)=...=(x_(2005))/(x_(2005)+4009)` `impliesx_(1)=(lambda)/(1-lambda)`, `x_(2)=(3lambda)/(1-lambda)`, `x_(3)=(5lambda)/(1-lambda)`,…… Hence , `x_(1),x_(2),x_(3),…..,x_(2005)` are in arithmetic progression. |
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| 75. |
The sum of `25` terms of an `A.P.`, whose all the terms are natural numbers, lies between `1900` and `2000` and its `9^(th)` term is `55`. Then the first term of the `A.P.` isA. `5`B. `6`C. `7`D. `8` |
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Answer» Correct Answer - C `(c )` Let the first term be `a` and the common difference `d`. Then it is given that `a+8d=55` and ……..`(i)` `1900 lt (25)/(2)(2a+24d) lt 2000` `:.1900 lt 25(a+12d) lt 2000` `implies1900 lt 25(a+8d)+25xx4d lt 2000` `implies1900 lt 25xx55+100d lt 2000` `implies(525)/(100) lt d lt (625)/(100)` `impliesd=6` as `d` is an integar `:. a=55-8xx6=7` |
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| 76. |
If `x_1,x_2 …,x_(20)` are in H.P and `x_1,2,x_(20)` are in G.P then `Sigma_(r=1)^(19)x_rr_(x+1)`A. 76B. 80C. 84D. none of these |
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Answer» Correct Answer - A Clearly, `1/x_(1),1/x_(2),…1/x_(20)` will be in A.P. Hence, `1/x_(2)-1/x_(1)=1/x_(3)-1/x_(2)=..=1/x_(r+1)-1/x_(r)=..=lamda` (say) `rArr(x_(r)-x_(r+1))/(x_(r)x_(r+1))=lamda` or `x_(r)x_(r+1)=-1/lamda(x_(r+1)-x_(r ))` `rArrsum_(r=1)^(19)x_(r )x_(r+1)=-1/lamdasum_(r=1)^(19)(x_(r+1)-x_(r ))=-1/lamda(x_(20)-x_(1))` Now, `1/x_(20)=1/x_(1)+19lamda` or `(x_(1)-x_(20))/(x_(1)x_(20))=19lamda` `rArrsum_(r=1)^(19)x_(r )x_(r+1)=19x_(1)x_(20)=19xx4=76` `(becausex_(1),2,x_(20)` are in G.P., then `x_(1)x_(20)=4)` |
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| 77. |
An infinite `G.P.` has `2^(nd)` term `x` and its sum is `4`. Then `x` belongs toA. `(0,2]`B. `(1,8)`C. `(-8,1]`D. none of these |
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Answer» Correct Answer - C `(c )` `"Sum"=((x)/(r ))/(1-r)=4` (where `r` is common ratio) `x=4r(1-r)` `=4(r-r^(2))` For `r in (-1,1)` `r-r^(2)in(-2,(1)/(4)]` `impliesx in (-8,1]` |
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| 78. |
The sequence `{x_(k)}` is defined by `x_(k+1)=x_(k)^(2)+x_(k)` and `x_(1)=(1)/(2)`. Then `[(1)/(x_(1)+1)+(1)/(x_(2)+1)+...+(1)/(x_(100)+1)]` (where `[.]` denotes the greatest integer function) is equal toA. `0`B. `2`C. `4`D. `1` |
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Answer» Correct Answer - D `(d)` `(1)/(x_(k+1))=(1)/(x_(k)(x_(k)+1))=(1)/(x_(k))-(1)/(x_(k)+1)` `implies(1)/(x_(k)+1)=(1)/(x_(k))-(1)/(x_(k-1))` `:. (1)/(x_(1)+1)+(1)/(x_(2)+1)+...+(1)/(x_(100)+1)=(1)/(x_(1))=(1)/(x_(101))` As `0 lt (1)/(x_(101)) lt 1` `:.[(1)/(x_(1)+1)+(1)/(x_(2)+1)+...+(1)/(x_(100)+1)]=1` |
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| 79. |
If, for a positive integer `n ,`the quadratic equation,`x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n`has two consecutive integral solutions, then `n`is equal to :`10`(2) `11`(3) `12`(4) `9`A. 11B. 12C. 9D. 10 |
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Answer» Correct Answer - A we have `x(x+1)+(x+1)(x+2)+…+(x+bar(n-1))(x+n)=10n ` Roots of above equation are consecutive integers . Let roots be `alpha " and " alpha +1` `rArr alpha(alpha +1)+(alpha +1)(alpha+2)+…..+(alpha +n )(alpha +n+1)=10 n ` and `(alpha +1)(alpha+2)+(alpha+2)(alpha+3)+.....+(alpha +n)(alpha+n+1)=10n ` Subractiong (2) from (1) we get `alpha(alpha+1)-(alpha +n)(alpha +n+1)=0` `rArr alph^2+alpha-alpha^2(2n +1)alpha-n(n+1)=0` `rArr alpha = (n+1)/(2)` Putting this vlaue in the original equation i.e `nx^2+x underset(r=1)overset(n)Sigma (2r-1)+underset(r=1)overset(n)Sigma(r-1)r= 10n` ,we get `rArr 3(n+1)^2-6n(n+1)+2(n+1)-6(n+1)=120` `rArr n^2=121` `rArr n=11` |
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| 80. |
If `sum_(r=1)^(r=n)(r^(4)+r^(2)+1)/(r^(4)+r)=(675)/(26)`, then `n` equal toA. `10`B. `15`C. `25`D. `30` |
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Answer» Correct Answer - C `(c )` `T_(e)=(r^(4)+r^(2)+1)/(r^(4)+r)=((r^(2)+r+1)(r^(2)-r+1))/(r(r+1)(r^(2)-r+1))=(r^(2)+r+1)/(r(r+1))` `T_(r)=1+(1)/(r )-(1)/(r+1)` `T_(1)=1+(1)/(1)-(1)/(2)` `T_(2)=1+(1)/(2)-(1)/(3)` `T_(3)=1+(1)/(3)-(1)/(4)` ……………….. `T_(n)=1+(1)/(n)-(1)/(n+1)` `:.S_(n)=n+1-(1)/(n+1)=(675)/(26)` `:.26(n+1)^(2)-26=675(n+1)` `implies26(n+1)^(2)-675(n+1)-26=0` `implies26(n+1)[n+1-26]+[(n+1)-26]=0` `implies(n-25)(26n-27)=0` `:. n=25` |
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| 81. |
If `n >1`, the value of the positive integer `m`for which `n^m+1`divides `a=1+n+n^2+ddot+n^(63)`is/are`8`b. `16`c. `32`d. `64`A. 8B. 16C. 32D. 64 |
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Answer» Correct Answer - A::B::C `a=(n^(64)-1)/(n-1)` `=(n+1)(n^(2)+1)(n^(4)+1)(n^(8)+1)(n^(16)+1)(n^(32)+1)` |
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| 82. |
If `a,b,c` are in `H.P`, `b,c,d` are in `G.P` and `c,d,e` are in `A.P.` , then the value of `e` isA. `(ab^(2))/((2a-b)^(2))`B. `(a^(2)b)/((2a-b)^(2))`C. `(a^(2)b^(2))/((2a-b)^(2))`D. None of these |
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Answer» Correct Answer - A `(a)` `a,b,c` are in `H.P.impliesb=(2ac)/(a+c)`….`(i)` `b,c,d` are in `G.P.impliesc^(2)=bd`..`(ii)` `c,d,e` are in `A.P.impliesd=(c+e)/(2)`…..`(iii)` From `(i)`, `ab+bc=2ac` `impliesc=(ab)/(2a-b)`….`(iv)` From `(iii)` and `(iv)` `d=(1)/(2)[(ab)/(2a-b)+e]` ......`(v)` From `(ii)`, `(iv)` and `(v)` `(a^(2)b^(2))/((2a-b)^(2))=(b)/(2)[(ab)/(2a-b)+e]` `impliese=(ab^(2))/((2a-b)^(2))` |
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| 83. |
If `a+c`, `a+b`, `b+c` are in `G.P` and `a,c,b` are in `H.P.` where `a`,`b`,`c gt 0`, then the value of `(a+b)/(c )` isA. `3`B. `2`C. `(3)/(2)`D. `4` |
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Answer» Correct Answer - B `(b)` `a+c`, `a+b`, `a+c` are in `G.P.` `:. (a+b)^(2)=(a+c)(b+c)` `(a+b)^(2)=ab+c(a+b)+c^(2)`………`(1)` `a,c,b` are in `H.P.` `:.c=(2ab)/(a+b)`………`(2)` From `(1)` and `(2)` `(a+b)^(2)=(3)/(2)c(a+b)+c^(2)` ,brgt `:.2(a+b)^(2)-3c(a+b)-c^(2)=0` `:.a+b=-(c )/(2)`(rejected) or `a+b=2c` `:.(a+b)/(c )=2` |
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| 84. |
Find the sum of the infinte series `(1)/(9)+(1)/(18)+(1)/(30)+(1)/(45)+(1)/(63)+…`A. `(1)/(3)`B. `(1)/(4)`C. `(1)/(5)`D. `(2)/(3)` |
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Answer» Correct Answer - A `(a)` `T_(n)=(1)/(3)[(1)/(3)+(1)/(6)+(1)/(10)+(1)/(15)+(1)/(21)+…]` `=(2)/(3)[(1)/(2*3)+(1)/(3*4)+(1)/(4*5)+(1)/(5*6)+…]` Hence `T_(n)=(2)/(3)(1)/((n+1)(n+2))` `=(2)/(3)[(1)/(n+1)-(1)/(n+2)]` `:.S_(oo)=(2)/(3)*(1)/(2)=(1)/(3)` |
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| 85. |
Find the sum `(1^2)/(2)+(3^2)/(2^2)+(5^2)/(2^3)+(7^2)/(2^4)+….oo` |
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Answer» Correct Answer - 17 `S=1^(2)/2+3^(2)/2^(2)+5^(2)/2^(3)+7^(2)/2^(4)+….oo` (1) `rArr1/2s=1^(2)/2^(2)+3^(2)/2^(3)+5^(2)/2^(4)+…oo` (2) `rArr1/2S=1^(2)/2^(2)+3^(2)/2^(3)+5^(2)/2^(4)+…oo` (2) Subtracting (2) from (1), `1/2S=1^(2)/2+8/2^(2)+16/2^(3)+24/2^(4)+32/2^(5)+..oo` (3) Let `S_(1)=8/2^(2)+16/2^(3)+24/2^(4)+32/2^(5)+..oo` (4) `rArr1/2S_(1)=8/2^(3)+16/2^(4)+24/2^(5)+...oo` (5) Subtracting (5) from (4), `1/2S_(1)=8/2^(2)+8/2^(3)+8/2^(4)+8/2^(5)+...oo` Subtracting (5) from (4), `1/2S_(1)=8/2^(2)+8/2^(3)+8/2^(4)+8/2^(5)+...oo` `=(8/2^(2))/(1-1/2)` =4 or `S_(1)=8` Hence, from (3), `1/2S=1/2+8` `rArrS=17` |
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| 86. |
If the rootsof `10 x^3-n x^2-54 x-27=0`are inharmonic oprogresion, then `n`eqauls_________. |
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Answer» Correct Answer - 9 `10x^(3)-nx^(2)-54x-27`=0 has roots in H.P. put x=1/t `27t^(3)+54t^(2)+nt-10=0` This equation has roots in A.P. Let the roots be a-d,a, and a+d `therefore3a=-(54)/(27)` or `a=-2/3` Also `(a-d)a(a+d)=10/27` `therefore2/3(4/9-d^(2))=-10/27` or `(4/9-d^(2))=-5/9` `therefored^(2)=1` or `d=pm1` So, roots are `1/3,-2/3,-5/3` `thereforen/27=10/9-5/9-2/9` or `n/27=3/9` or n=9 |
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| 87. |
The number of ordered pairs `(x,y)` , where `x`, `y in N` for which `4`, `x`, `y` are in `H.P.` , is equal toA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - C `(c )` `4,x,y` are in `H.P.` ,brgt `(2)/(x)=(1)/(4)+(1)/(y)` `implies (2)/(x)-(1)/(4)=(1)/(y)` `implies(8-x)/(4x)=(1)/(y)` `impliesy=(4x)/(8-x)=(4(8-(8-x)))/(8-x)=(32)/(8-x)-4` `8-x` must be a factor of `32` `8-x=1impliesx=7`, `y=28` `8-x=2impliesx=6`, `y=12` `8-x=4impliesx=4`, `y=4` `8-x=8impliesx=0`, `y=0` (Not possible) `:.` Number of ordered pairs of `(x,y)` is `3`. |
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| 88. |
Find the sum `1^2+(1^2+2^2)+(1^2+2^2+3^2)+`up to 22nd term. |
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Answer» Correct Answer - 23276 `T_(n)=1^(2)+2^(2)+3^(2)+…+n^(2)` `=(n(n+1)(2n+1))/6` `=n/6(2n^(2)+3n+1)` `=(2n^(3)+3n^(2)+n)/6` `S_(n)=sumT_(n)` `=1/6[2sumn^(3)+3sumn^(2)+sumn]` `=1/6{2[(n(n+1))/2]^(2)+3xx(n(n+1)(2n+1))/6+(n(n+1))/2}` `=1/6[(n^(2)(n+1)^(2))/2+(n(n+1)(2n+1))/6+(n(n+1))/2]` `=1/6xx1/2n(n+1)[n(n+1)[n(n+1)+(2n+1)+1]` `=(n(n+1))/12[n^(2)+3n+2]` For n= 22 `S_(22)=(22xx23)/12[22^(2)+66+2]` `=(22xx23)12[552]` `=22xx23xx46` =23276 |
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| 89. |
Find the sum `3/2-5/6+7/(18)-9/(54)+oodot` |
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Answer» Correct Answer - `15/16` `S=3/2-5/6+7/18-9/54+..oo` `rArrS=3(1/2)+5(-1/6)+7(1/18)+9(-1/54)+..` (1) `rArr-1/3S=3(-1/6)+5(1/18)+9(-1/54)+..` (2) Subtracting (2) from (1), we have `4/3S=3(1/2)+[2(-1/6)+2(1/18)+2(-1/54)+…oo]` `=3/2+(2(-1/6))/(1-(-1/3))` `=3/2-(1/3)/(4/3)` `=3/2-1/4` `=5/4` `rArrS=15/16` |
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| 90. |
The A.M. of two given positive numbers is 2. If the larger number isincreased by 1, the G.M. of the numbers becomes equal to the A.M. of thegiven numbers. Then find the H.M. |
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Answer» Correct Answer - `3/2` Let a,b`(agtb)` be the two numbers. Given, `(a+b)/2=2` or a+b=4 (1) and `sqrt((a+1)b)=2` or (a+1)b=4 or `b^(2)-5b+4=0` [Using (1)] or b=1,4 But `bne4`, therefore, b=1, and hence a=3. Hence, harmonic mean, `H=(2xx3xx1)/(3+1)=3/2` |
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| 91. |
The product of the three numbers in G.P. is 125 and sum of theirproduct taken in pairs is `(175)/2`. Find them. |
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Answer» Correct Answer - 5/2, 5,11 Let three terms in G.P. be a/r,a,ar. Given that the product is 125. `therefore` a=5 Also, `a^(2)(rxx1+1xx1/r+rxx1/r)=175/2` or `25(r^(2)+r+1)=175/2r` or `2(r^(2)+r+1)=7r` or `2r^(2)-5r+2=0` or `(r-2)(2r-1)=0` or `r=2,1/2` Hence, the numbers are 10,5,5/2 or 5/2,5,10. |
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| 92. |
The sum of the first n terms of the series `1^2+ 2.2^2+3^2 +2.4^2+....` is `(n(n+1)^2)/2` when n is even. Then the sum if n is odd , is |
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Answer» Correct Answer - `(n^2(n+1))/(2)` When n is even, last term of the series is `n^(2)`. So, we have `1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+5^(2)+2xx6^(2)+..+(n-1)^(2)+2xxn^(2)` `=(n(n+1)^(2))/2` ..(1) When n is odd, the series is `1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+5^(2)+2xx6^(2)+....+2(n-1)^(2)+n^(2)` `therefore` Sum of series=`[1^(2)+2xx2^(2)+3^(2)+2xx4^(2)+...+2xx(n-1)^(2)]+n^(2)` `=((n-1)n^(2))/2+n^(2)` [In (1) replacing n by (n-1)] `=(n^(2)(n+1))/2` |
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| 93. |
If the sum of the first `100` terms of an `AP` is `-1` and the sum of even terms lying in first `100` terms is `1`, then which of the following is not true ?A. Common difference of the sequence is `(3)/(50)`B. First term of the sequence is `(-149)/(50)`C. `100^(th)` term `=(74)/(25)`D. None of these |
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Answer» Correct Answer - D `(d)` `x_(1)+x_(2)+x_(3)+……+x_(100)=(100)/(2)(x_(1)+x_(100))=-1` `impliesx_(1)+x_(100)=-(1)/(50)` `x_(2)+x_(4)+…+x_(100)=(50)/(2)(x_(1)+d+x_(100))=1` `impliesx_(1)+x_(100)+d=(1)/(25)` `impliesd=(3)/(50)` ltbr `x_(1)+x_(1)+99d=(-1)/(50)` `impliesx_(1)=(-149)/(50)` `x_(100)=x_(1)+99d` `=(-149)/(50)+99xx(3)/(50)` `=(74)/(25)` |
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| 94. |
Find two numbers whose arithmetic mean is 34 and the geometric mean is16. |
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Answer» Correct Answer - 64 and 4 Let the two numbers be a and b. Then, A.M.=34 `rArr(a+b)/2=34` or a+b=68 G.M=16 `rArrsqrt(ab)=16` or ab=256 `therefore(a-b)^(2)=(a+b)^(2)-4ab` or `(a-b)^(2)=(68)^(2)-4xx256=3600` or a-b=60 (2) On solving (1) and (2), we get a= 64 and b=4. Hence, the required numbers are 64 and 4. |
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| 95. |
Find the sum of `n`terms of the series `1+4/5+7/(5^2)+10+5^3+dot` |
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Answer» Correct Answer - `5/4+15/16 (1-(1)/(5^(n-1)))-((3n-2))/(4(5^(n-1))` Clearly, the given series is an arthimatic-geometrico-geometric series whose corresponding A.P. and G.P. are, respectively, 1,4,7,10,1,1/5,`1//5^(2),1//5^(3)`,… The nth term of A.P is `[1+(n-1)xx3]=3n-2`. The nth term of G.P. is `[1xx(1//5)^(n-1)]=(1//5)^(n-1)` So, the nth erm of the given series is `(3n-2)xx(1//5^(n-1))=(3n-2)//5^(n-1)` Let `S_(n)=1+4/5+7/5^(2)+10/5^(3)+..+(3n-5)/(5^(n-2))+(3n-2)/(5^(n-1))` `1/5S_(n)=1/5+4/5^(2)+7/5^(3)+...+((3n-5))/(5^(n-1))+(3n-2)/5^(n)` Subracting (2) from (1), we get `S_(n)-1/5S_(n)=1+[3/5+3/5^(2)+3/5^(2)+...+3/(5^(n-1))]-((3n-2))/5^(n)` or `4/5S_(n)=1+3/5((1-(1/5)^(n-1)))/((1-1/5))-((3n-2))/5^(n)` `=1+3/5([1-1/(5^(n-1))])/((4/5))-((3n-2))/5^(n)` `=1+3/4(1-1/(5^(n-1)))-((3n-2))/5^(n)` or `S_(n)=5/4+15/16(1-1/(5^(n-1)))-((3n-2))/(4(5^(n-1))` |
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| 96. |
The number of terms of an A.P. is even, the sum of odd terms is 24, ofthe even terms is 3, and the last term exceeds the first by 10 1/2 find thenumber of terms and the series.A. 8B. 4C. 6D. 10 |
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Answer» Correct Answer - D Given, `a_(1),a_(2),a_(3),..` are terms of A.P `therefore(a_(1)+a_(2)+…+a_(p))/(a_(1)+a_(2)+…+a_(q))=(p^(2))/(q^(2))` `rArr(p/2[2a_(1)+(p-1)d])/(q/2[2a_(1)+(q-1)d])=(p^(2))/(q^(2))` or `(2a_(1)+(p-1)d)/(2a_(1)+(q-1)d)=p/q` or `[2a_(1)+(p-1)d]q=[2a_(1)+(q-1)d]p` or `2a_(1)(q-p)=d[(q-1)p-(p-1)q]` or `2a_(1)(q-p)=d(q-p)` or `2a_(1)=d` `therefore(a_(6))/(a_(21))=(a_(1)+5d)/(a_(1)+20d)=(a_(1)+10a_(1))/(a_(1)+40a_(1))=11/41` |
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| 97. |
If the arithmetic means of two positive number a and b `(a gt b )` is twice their geometric mean, then find the ratio a: b |
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Answer» Correct Answer - `(2+ sqrt(3)):(2-sqrt(3))` Let A be the A.M. and G be the G.M. of a and b. Then the numbers a and b are roots of the equation `x^(2)-4Gx+G^(2)=0`b `[because a=2G]` `rArrx=(4Gpmsqrt(16G^(2)-4G^(2)))/2` `=(4Gpm2sqrt3G)/2=(2pmsqrt3)G` Hence, a:b=`(2+sqrt3):(2-sqrt3)` |
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| 98. |
Let `a=111 1(55d igi t s),b=1+10+1=^2++10^4,c=1+10^5+10^(10)+10^(15)++10^(50),`then`a=b+c`b. `a=b c`c. `b=a c`d. `c=a b`A. a+b+cB. a=bcC. b=acD. c=ab |
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Answer» Correct Answer - B `a=1+10+10^(2)+….+10^(54)` `=(10^(55)-1)/(10-1)=(10^(55)-1)/(10^(5)-1)xx(10^(5)-1)/(10-1)=bc` |
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| 99. |
Consider the ten numbers `a r ,a r^2, a r^3, ,a r^(10)dot`If their sum is 18 and the sum of their reciprocals is 6, then theproduct of these ten numbers is`81`b. `243`c. `343`d. 324A. 81B. 243C. 343D. 324 |
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Answer» Correct Answer - B Given `(ar(r^(10)-1))/(r-1)=18` (1) Also `(1/(ar)(1-1/r^(10)))/(1-1/r)=6` or `1/(ar^(11))cdot((r^(10)-1)r)/(r-1)=6` or `1/(a^(2)r^(11))cdot(ar(r^(10)-1))/(r-1)=6` From (1) and (2), `1/(a^(2)r^(11))xx18=6` or `a^(2)r^(11)=3` Now P`=a^(10)^(55)=(a^(2)r^(11))^(5)=3^(5)=243` |
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| 100. |
Let `E=1/(1^2)+1/(2^2)+1/(3^2)+`Then,`E3//2`c. `E >2`d. `E |
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Answer» Correct Answer - A::B::D `Elt1+1/((1)(2))+1/((2)(3))+…` `=1+(1-1/2)+(1/2-1/3)+….=2` `Egt1+1/((2)(3))+1/((3)(4))+..` `=1+(1/2-1/3)+(1/3-1/4)+….=3/2` |
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