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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.
| 101. |
If a, b and c are in G.P and x and y, respectively , be arithmetic means between a,b and b,c thenA. `a/x+c/y=2`B. `a/x+c/y=c/a`C. `1/x +1/y=2/b`D. `1/x+1/y=2/ac` |
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Answer» Correct Answer - A::C a,b,c are in G.P. Hence, `b^(2)=ac` (1) x is A.M. of a and b. Hence, 2x=a+b (2) y is A.M. of b and c. Hence, 2y=b+c (3) `thereforea/x+c/y=axx2/(a+b)+cxx2/(b+c)` [Using (2) and (3)] `=2[(ab+ac+ac+bc)/(ab+ac+b^(2)+bc)]` =2 [Using (1)] Again, `1/x+1/y=2/(a+b)+2/(b+c)` `=(2(a+c+2b))/(ab+ac+b^(2)+bc)` `=(2(a+c+2b))/(ab+2b^(2)+bc) (becauseb^(2)=ac)` `=(2(a+c+2b))/(b(a+c+2b))=2/b` |
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| 102. |
Consider a sequence `{a_n}` with a_1=2 & `a_n =(a_(n-1)^2)/(a_(n-2))` for all ` n ge 3` terms of the sequence being distinct .Given that `a_2 " and " a_5` are positive integers and `a_5 le 162`, then the possible values (s) of `a_5` can beA. 136B. 64C. 32D. 2 |
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Answer» Correct Answer - A::C Given `a_(1)=2,(a_(n))/(a_(n-1))=(a_(n-1))/(a_(n-2))` Hence,`a_(1),a_(2),a_(3),a_(4),a_(5),..` in G.P. Let `a_(2)=x`. Then for n=3, we have `(a_(3))/(a_(2))=(a_(2))/(a_(1))` `rArra_(2)^(2)=a_(1)a_(3)` `rArra_(3)=x^(2)/2` i.e., `2,x,x^(2)/2,x^(3)/4,x^(4)/8`,... with common ratio `r=x/2` Given `x^(4)/8` are integers. So if x is even, then only `x^(4)/8` will be an integer. Hence, possible value of x is 4 and 6. (`xne2` as terms are distinct) Hence, possible value of `a_(5)=x^(4)/8` is `4^(4)/8,6^(4)/8` |
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| 103. |
If `p(x)=(1+x^2+x^4++x^(2n-2))//(1+x+x^2++x^(n-1))`is a polomial in `x`, then find possible value of `ndot` |
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Answer» Correct Answer - n is odd `p(x)=((1-x^(2n))/(1-x^(2)))((1-x)/(1-x^(n))=(1+x^(n))/(1+x))` As p(x) is a polynomial , x=-1 must be a zero of `1+x^(n)`, i.e., `1+(-1)^(n)=0`. Hence, n is odd. |
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| 104. |
If first and `(2n-1)^(th)` terms of A.P., G.P. and H.P. are equal and their nth terms are a,b,c respectively, thenA. a=b=cB. `a ge be ge c`C. `a+b =c`D. `ac- b^2 =0 ` |
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Answer» Correct Answer - B::D Let x be the first term and y be the (2n-1)th term of A.P., G.P. and H.P. whose nth terms are a,b,c, respectively. According to the property of A.P., G.P. and H.P., xa,y are in A.P. x,b,y are in G.P. x,c,y are in H.P. Using this information, we get `a=(x+y)/2=A.M.` `b=sqrt(xy)=G.M.` and `c=(2xy)/(x+y)`=H.M. We know that A.M.,G.M. and H.M. are in G.P.. `thereforeb^(2)=ac` Also, A.M.`ge`G.M.`ge`H.M. `rArragebgec` |
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| 105. |
If `a ,b ,a n dc`are in A.P. `p ,q ,a n dr`are in H.P., and `a p ,b q ,a n dc r`are in G.P., then `p/r+r/p`is equal to`a/c-c/a`b. `a/c+c/a`c. `b/q+q/b`d. `b/q-q/b`A. A.PB. G.PC. G.PD. none of these |
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Answer» Correct Answer - D a,b, and c are in A.P. Hence, 2b=a+c `a/(bc)+2/b=(a+2c)/(bc)ne2/c` `rArra/(bc),1/c,2/b` are not in A.P. `(bc)/a+b/2=(2bc+ab)/(2a)nec` Hence, the given numbers are not in H.P. Again, `a/(bc)2/b=(2a)/(b^(2)c)ne1/c^(2)` Therefore, the given numbers are not in G.P. |
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| 106. |
If `a ,b ,a n dc`are in A.P. `p ,q ,a n dr`are in H.P., and `a p ,b q ,a n dc r`are in G.P., then `p/r+r/p`is equal to`a/c-c/a`b. `a/c+c/a`c. `b/q+q/b`d. `b/q-q/b`A. `a/c-c/a`B. `a/c+c/a`C. `b/q+q/b`D. `b/q-q/b` |
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Answer» Correct Answer - B `p/r+r/p=(p^(2)+r^(2))/(pr)=((p+r)^(2)-2pr)/(pr)` `=((4p^(2)r^(2))/q^(2)-2pr)/(pr)[{:(because" p,q,r are in H.P."),(" "thereforeq=(2pr)/(p+r)):}]` `=(4pr)/q^(2)-2=(4b^(2))/(ac)-2` `[because` ap,bq,cr are in A.P. `rArrb^(2)q^(2)=acpr]` `=((a+c)^(2))/(ac)-2[a,b,c` are in A.P`rArr2b=a+c]` `=a/c+c/a` |
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| 107. |
यदि `(a^(n)+b^(n))/(a^(n-1)+b^(n-1)),` a तथा b के मध्य समांतर माध्य हो तो n का मान ज्ञात कीजिए | |
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Answer» Correct Answer - n=1 According to the given condition, we have `(a+b)/2=(a^(n)+b^(n))/(a^(n-1)+b^(n-1))` or `(a+b)(a^(n-1)+b^(n-1))=2(a^(n)+b^(n))` or `a^(n)+ab^(n-1)+ba^(n-1)+b^(n)=2a^(n)+2b^(n)` or `ab^(n-1)+a^(n-1)b=a^(n)+b^(n)` or `ab^(n-1)-b^(n)=a^(n)-a^(n-1)b` or `b^(n-1)(a-b)=a^(n-1)(a-b)` or `b^(n-1)=a^(n-1)` or `(a/b)^(n-1)=1=(a/b)^(0)` `rArrn-1=0` or n=1 |
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| 108. |
If a,b,c are in A.P and `a^2,b^2,c^2` are in H.P then which is of the following is /are possible ?A. `ax^2 +bx+c =0`B. `ax^2bx+c=0`C. `a,b-c/2` form a G.PD. `a-b, c/2` from a G.P |
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Answer» Correct Answer - A::C Given that a,b,c are in A.P. `rArr2b=a+c` …(1) and `a^(2),b^(2),c^(2)` are in H.P. `rArr1/b^(2)-1/a^(2)=1/c^(2)-1/b^(2)` `rArr((a-b)(a+b))/(b^(2)a^(2))=((b-c)(b+c))/(b^(2)c^(2))` `rArr ac^(2)+bc^(2)=a^(2)b+a^(2)c` [`because a-b=b-c`] `rArrac(c-a)+b(c-a)(c+a)=0` `rArr(c-a)(ab+bc+ca)=0` `rArrc-a=0` or ab+bc+ca=0 For c=a, from (1),a=b=c For (a+c)b+ca=0, from (1), `2b^(2)+ca=0` `rArrb^(2)=a((-c)/2)` This implies that a,b,-c/2 are in G.P. |
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| 109. |
If `1/(b-a)+1/(b-c)=1/a+1/c ,`then`a ,b ,a n dc`are in H.P.`a ,b ,a n dc`are in A.P.`b=a+c``3a=b+c`A. a,b, and c are in H.PB. a,b, and c are in A.PC. b=a+cD. 3a= b+c |
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Answer» Correct Answer - A::B `1/(b-a)+1/(b-c)=1/a+1/c` or `1/(b-a)-1/c=1/a-1/(b-c)` or `(c-b+a)/(c(b-a))=(b-c-a)/(a(b-c))` `rArrc-b+a=0` or `1/(c(b-a))=1/(a(c-b))` `rArrb=a+c` or bc-ac=ac-ab `rArrb=a+c`or `b=(2ac)/(a+c)` |
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| 110. |
If a, b, c, d are distinct integers in A. P. Such that `d=a^2+b^2+c^2`, then a + b + c + d is |
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Answer» Correct Answer - 2 `a+3k=a^(2)+(a+k)^(2)+(a+2k)^(2)` (1) (where k is the common difference of A.P) `rArr5k^(2)+3(2a-1)k+3a^2)-a=0` (2) `rArr9(2a-1)^(2)-20(3a^(2)-a)ge0(becausek is "real")` `rArr24a^(2)+16a-9le0` `rArr-1/3-(sqrt(70))/12ltalt-1/3+(sqrt(70))/12` `rArrk=0`,3/5 [Not possible] When a=-1 `5k^(2)-9k+4=0` `rArrk=1,4/5rArrk=1` (since k is an integer) `thereforea=-1,b=0,c=1,d=2` `rArra+b+c+d=2` |
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| 111. |
If `x ,y ,a n dz`are distinct prime numbers, then`x , y , a n dz`may be in A.P. but not in G.P.`x , y , a n dz`may be in G.P. but not in A.P.`x , y , a n dz`can neither be in A.P. nor inG.P.none of theseA. x,y and z may be in A.P but not in G.PB. x,y and z may be in G.P but not in A.PC. x,y and z can neither be inD. none of these |
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Answer» Correct Answer - A x,y,z are in G.P. `hArr y^(2)=xz` `hArr` x is a factor of y (not possible) Taking x=3,y=5,z=7, we have x,y,z are in A.P. Thus x,y,z may be in A.P. but not in G.P. |
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| 112. |
If `x^2+9y^2+25 z^2=x y z((15)/2+5/y+3/z),t h e n``x ,y ,a n dz`are in H.P. b. `1/x ,1/y ,1/z`are in A.P.c. `x ,y ,z`are in G.P.d. `1/a+1/d=1/b=1/c`A. x,y and z are in H.PB. `1/x, 1/y,1/z ` are in G.PC. x,y,z are in G.PD. `1/x,1/y,1/z` are in G.P |
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Answer» Correct Answer - A::C `x^(2)+9y^(2)+25z^(2)=15yz+5zx+3xy` `rArr(x)^(2)+(3y)^(2)+(5z)^(2)-(x)(3y)-(3y)(5z)-(x)(5z)=0` `rArr1/2[(x-3y)^(2)+(3y-5z)^(2)+(x-5z)^(2)]=0` `rArrx-3y=0,3y-5z=0,x-5z=0` x=3y=5z `rArrx:y:z=1/1:1/3:1/5` Therefore,1/x,1/y and 1/z are in A.P. and x,y, and z are in H.P. |
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| 113. |
If a,b,c are three distinct numbers in G.P., b,c,a are in A.P and a,bc, abc, in H.P then the possible value of b isA. `3+4sqrt(2)`B. `3-4 sqrt(2)`C. `4+ 3 sqrt(2)`D. `4-3sqrt(2)` |
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Answer» Correct Answer - C::D Let `a=A/r,b=A and C=Ar` Now b,c,a are in A.P.. `therefore2c=a+b` `rArr2Ar=A/r+A` `rArr2r^(2)-r-1=0` `rArrr=1 or r=-1/2` Sine a,b, are distinctt, `r=-1/2` `thereforea=-2A,b=A,c=-A/2` a,bc,abc are in H.P. `therefore 1/a,1/(bc),1/(abc)` are in A.P. `rArr 2/(bc)=1/(abc)+1/a` `rArr-4/A^(2=1/A^(3)-1/(2A)` `rArr-4/(A^(2))=(2-A^(2))/(2A^(3))` `rArr-8A=2-A^(2)` `rArrA^(2)-8A-2=0` `rArrA=pm3sqrt2` |
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| 114. |
If `a`, `b`, `c` are real numbers forming an `A.P.` and `3+a`, `2+b`, `3+c` are in `G.P.` , then minimum value of `ac` isA. `-4`B. `-6`C. `3`D. None of these |
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Answer» Correct Answer - B `(b)` `:. (2+b)^(2)=(3+a)(3+c)` ,brgt `:. 4+b^(2)+4b=9+3(a+c)+ac` `:.4+b^(2)+4b=9+6b+ac` ,brgt `:.ac=b^(2)-2b-5=(b-1)^(2)-6` `:.ac ge -6` |
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| 115. |
If p,q and r are in A.P then which of the following is / are true ?A. pth,qth and rth terms of A.P are in A.PB. pth,qth,and rht terms of G.P are in G.PC. pth , qth , and rht terms of H.P are in H.PD. none of these |
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Answer» Correct Answer - A::B::C If p,q,r are in A.P., then pth,qth and rth, terms are equidistant terms which are always in the same series of which they are terms. |
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| 116. |
If a, b and c are in H.P., then the value of `((ac+ab-bc)(ab+bc-ac))/(abc)^2` isA. `((a+c)(3a -c))/(4a^2 c^2)`B. `2/(bc)-1/b^2`C. `2/(bc)-1/b^2`D. `((a-c)(3a+c))/(4a^2c^2)` |
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Answer» Correct Answer - A::B `(1/b+1/c-1/a)(1/c+1/a-1b)` `=(1/b+1/c-2/b+1/c)(1/c+1/b-1/c)` Also by eliminating b, we get the given expression `((a+c)(3a-c))/(4a^(2)c^(2))` |
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| 117. |
Given that `x+y+z =15` when a,x,y,z,b are in A.P and `1/x+1/y+1/z=5/3 when a,x,y,z,b are in H.P .ThenA. G.M of a and b is 3B. one possible value of a + 2b is 11C. A.M of a and b is 6D. greatest value of a-b is 8 |
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Answer» Correct Answer - A::B::D `x+y+z=3((a+b)/2)` `rArr15-3((a+b))/2` `rArra+b=10` `1/x+1/y+1/z=(3(1/a+1/b))/2` `rArr5/3=(3(a+b))/(2ab)=(3xx10)/(2ab)` `rArrab=9` (2) From (1) and (2), a=9,b=1 or a=1 and b=9. Hence, G.M. `=sqrt(ab)=3,a+2b=11` or 19. |
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| 118. |
`1/(sqrt(2)+sqrt(5))+1/(sqrt(5)+sqrt(8))+1/(sqrt(8)+sqrt(11))+ n`terms is equal to`(sqrt(3n+2)-sqrt(2))/3`b. `n/(sqrt(2+3n)+sqrt(2))`c. less than`n`d. less than`sqrt(n/3)`A. `((sqrt(3n +2))-sqrt(2))/(3)`B. `(n)/(sqrt(2+3n)+sqrt(2))`C. less than nD. less than `sqrt(n/3)` |
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Answer» Correct Answer - A::B::C `1/(sqrt2+sqrt5)+1/(sqrt5+sqrt8)+1/(sqrt8+sqrt(11))+…` n terms `=(sqrt5-sqrt2)/3+(sqrt8-sqrt5)/3+….+(sqrt(5+(n-1)3)-sqrt(2+(n-1)3))/3` `=(sqrt(3n+2)-sqrt2)/3` `=(3n+2-2)/(3(sqrt(3n+2)+sqrt2))` `=n/(sqrt(3n+2)+sqrt2)` `=n/(sqrt(2+3n)+sqrt2)ltn/(sqrt(3n))ltn` |
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| 119. |
If `S_n=1^2-2^2+3^2-4^2+5^2-6^2+ ,t h e n``S_(40)=-820`b. `S_(2n)> S_(2n+2)`c. `S_(51)=1326`d. `S_(2n+1)> S_(2n-1)`A. `S_(40)=-820`B. `S_(2n) gt S_(2n+2)`C. `S_(51)=1326`D. `S_(2n +1) gt S_(2n-1)` |
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Answer» Correct Answer - A::B::C::D Clearly, nth term of the given series is negative or positive accordingly as n is even or odd, respectively. Case I: When n is even: In this case, the given series is `S_(n)=1^(2)-2^(2)+3^(2)-4^(2)+…+(n-1)^(2)-n^(2)` `=(1^(2)-2^(2))+(3^(2)-4^(2))+..+((n-1)^(2)-n^(2))` `=(1-2)(1+2)+(3-4)(3+4)+...+((n-1)-(n))(n-1+n)` `=-(1+2+3+4+...+(n-1)+n)` `=-(n(n+1))/2` (1) Case II, When n is odd: In this case, the given series is `S_(n)=(1^(2)-2^(2))+(3^(2)-4^(2))+..+{(n-2)^(2)-(n-1)^(2)}+n^(2)` =`(1-2)(1+2)+(3-4)(3+4)+...+((n-2)-(n-1))xx((n-2)+(n-1))+n^(2)` `=-((n-1)(n-1+1))/2+n^(2)=(n(n+1))/2` (2) `rArrS_(40)=-820` [Using (1)] `S_(51)=1326` [Using (2)] Also, `S_(2n)gtS_(2n+2)` [From (1)] `S_(2n+1)gtS_(2n+1)` [From (2)] |
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| 120. |
For the series, `S=1 +(1)/(1+3)(1+2)^2+(1)/((1+3+7))(1+2+3)^2+(1)/((1+3+5+7))(1+2+3+4)^2+...`A. `7^(th) " term is " 16`B. `7^(th) " term " is 18`C. Sum of first 10 terms is `(505)/(4)`D. Sum of first 10 terms is `(405)/(4)` |
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Answer» Correct Answer - A::C `S=1+1/((1+3))(1+2)^(2)+1/((1+3+5))(1+2+3)^(2)+1/((1+3+5+7))(1+2+3+4)^(2)+…` The `r^(th)` term is given by `T_(r)=1/r^(2)(1+2+..+r)^(2)=1/r^(2){(r(r+1))/2}^(2)=(r^(2)+2r+1)/4` `thereforeT_(7)=16` and `S_(10)=sum_(r=1)^(10)T_(r)` =`1/4{((10)(10+1)(20+1))/6+(10)(10+1)+10}` `=505/4` |
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| 121. |
If `Sigma_(r=1)^(n) r(r+1)(2r +3)=an^4+bn^3+cn^2+dn +e ` thenA. a-b=d-cB. e=0C. `a,b-2//3,c-1 " are in " in A.P`D. `(b+d)//a` is an integer |
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Answer» Correct Answer - A::B::C::D `an^(4)+bn^(3)+cn^(3)+dn+e` `=2sum_(r=1)^(n)r(r+1)(r+2)-sum_(r=1)^(n)r(r+1)` `=2/4n(n+1)(n+2)(n+3)-1/3n(n+1)(n+2)` `=1/6(3n^(4)+16n^(3)+27n^(2)+14n)` |
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| 122. |
If sum of an indinite `G.P p,1,1//p,1//p^2`…=9/2.. Is then value of p isA. 2B. `3//2`C. 3D. `9//2` |
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Answer» Correct Answer - B::C We have `p/(1-1//p)=9/2` or `2p^(2)-9p+9=0` or p=3/2,3 |
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| 123. |
If `a ,b, c ,d`are in G.P, then `(b-c)^2+(c-a)^2+(d-b)^2`is equal to`A. `(a-d)^(2)`B. `(ad)^(2)`C. `(a+d)^(2)`D. `(a//d)^(2)` |
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Answer» Correct Answer - A Let r be the common ratio of the G.P., a,b,c,d. Then, `b=ar,c=ar^(2) and d=ar^(3)` ltbrRgt `therefore(b-c)^(2)+(c-a)^(2)+(d-b)^(2)` `=(ar-ar^(2))^(2)+(ar^(2)-a)^(2)+(ar^(3)-ar)^(2)` `=a^(r )r^(2)(1-r)^(2)+a^(2)(r^(2)-1)^(2)+a^(2)r^(2)(r^(2)-1)^(2)` `=a^(2)(r^(6)-2r^(3)+1)` `=a^(2)(1-r^(3))^(2)` `=(a-ar^(3))^(2)` `=(a-d)^(2)` |
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| 124. |
If `H_1.,H_2,…,H_20` are 20 harmonic means between 2 and 3, then `(H_1+2)/(H_1-2)+(H_20+3)/(H_20-3)=`A. 20B. 21C. 40D. 38 |
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Answer» Correct Answer - C `(H_(1)+2)/(H_(1)-2)+(H_(20)+3)/(H_(20)-3)=(1/2+1/H_(1))/(1/2-1/H_(1))+(1/3+1/H_(20))/(1/3-1/H_(20))` `=(1/2+1/2+d)/(1/2-d-1/2)+(1/3+1/3-d)/(1/3+d-1/3)` `=(1+d)/(-d)+(2/3-d)/d` `=(2/3-1)/d-2` `=2xx21-2` [as also,`1/3=1/2+21d`] =40 |
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| 125. |
Find the value of `{:(" "SigmaSigma),(1 le i le j):} " "i xx (1/2)^j` |
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Answer» Correct Answer - 2 `S=sumsum_(ileiltj)ixx(1/2)^(j)` `=1xx[(1/2)^(2)+(1/2)^(3)+(1/2)^(4)+…]` `+2xx[(1/2)^(3)+(1/2)^(4)+(1/2)^(5)+..]` `+3xx[(1/2)^(4)+(1/2)^(5)+(1/2)^(6)+…]` ….. `=1xx((1/2)^(2))/(1-(1/2))+2xx((1/2)^(3))/(1-(1/2))+3xx((1/2)^(4))/(1-(1/2))+..` `thereforeS=1xx(1/2)+2xx(1/2)^(2)+3xx(1/2)^(3)+...` `therefore (1/2)S=1xx(1/2)^(2)+2xx(1/2)^(3)+3xx(1/2)^(4)...` Subtracting, we get `(1/2)S=(1/2)+(1/2)^(2)+(1/2)^(3)+(1/2)^(4)+...=(1/2)/(1-1/2)=1` `therefore` S=2 |
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| 126. |
Find the sum of infinite series `(1)/(1xx3xx5)+(1)/(3xx5xx7)+(1)/(5xx7xx9)+….` |
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Answer» Correct Answer - `1/(12)` `T_(r)=1/((2r-1)(2r+1)(2r+3))` `=1/4cdot((2r+3)-(2r-1))/((2r-1)(2r+1)(2r+3))` `=1/4[1/((2r-1)(2r+1))-1/((2r+1)(2r+3))]` `=1/4[V(r-1)-V(r )],` where V( r)=`1/((2r+1)(2r+3))` `thereforesum_(r=1)^(n)T_(r)=sum_(r=1)^(n)1/4[V(r-1)-V(r )]` `=1/4[V(0)-V(n)]` `=1/4[1/3-1/((2n+1)(2n+3))]` `therefore` Sum of infinite terms=`1/4[1/3-0]=1/12` |
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| 127. |
Statement 1 : The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+….+(361 +380 +400) is 8000 Statement 1: `Sigma_(k=1)^(n) (k^3-(k-1)^3)=n^3`, for any natural number n.A. Statement 1 is fasle ,statement 2 is trueB. Statement 1 is true ,statement 2 is true , statement 2 is a correct explanation for statement 1.C. Statement 1 is true, statements 2 is true statement 2 is not a correct explanation for statement 1D. Statement 1 is true, statement 2 is false |
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Answer» Correct Answer - B `T_(n)=(n-1)^(2)+(n-1)n+n^(2)` `=((n-1)^(3)-n^(3)))/((n-1)-n)=n^(3)-(n-1)^(3)` `T_(1)=1^(3)-0^(3)` `T^(2)=2^(3)-1^(3)` . . . . . . . . . . . . `T_(20)=20^(3)-19^(3)` Adding, `S_(20)=20^(3)-0^(3)=8000` |
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| 128. |
Let `S=(sqrt(1))/(1+sqrt1+sqrt(2))+sqrt(2)/(1+sqrt(2)+sqrt(3))+(sqrt(3))/(1+sqrt(3)+sqrt(4))+...+(sqrt(n))/(1+sqrt(n)+(sqrtn+1))=10` Then find the value of n. |
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Answer» Correct Answer - n=24 `T_(r)=(sqrtr)/(1+sqrtr+sqrt(r+1))=(sqrtr{1+sqrtr-sqrt(r+1)})/(1+r+2sqrtr-(r+1))` `=1/2{1+sqrtr-sqrt(r+1)}` `thereforeS_(n)=1/2(n+1)-sqrt(n+1)=10` Let `sqrt(n+1)`=x `thereforex^(2)-x=20` `rArrx^(2)-x-20=0` `rArrx=sqrt(n+1)=5` `thereforen=24` |
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| 129. |
If ` a^2+2bc ,b^2+2ca, c^2+2ab` are in A.P. then :-A. `(a-b)(c-a),(a-b)(b-c),(b-c)(c-a)` are in A.PB. b-c,c-a,a-b are in H.PC. a+b,b+c,c+a are in H.PD. `a^2,b^2,c^2` are in H.P |
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Answer» Correct Answer - A::B `a^(2)+2bc,b^(2)+2ca,c^(2)+2b` `thereforeab+bc+ca-a^(2)-2bc,ab+bc+ca-b^(2)-2ca`, `ab+bc+ca-c^(2)-2ab` are in A.P. `rArr(c-a)(a-b),(a-b)(b-c),(b-c)(c-a)` are in A.P. `((c-a)(a-b))/((a-b)(b-c)(c-a)),((a-b)(b-c))/((a-b)(b-c)(c-a)),` `((b-c)(c-a))/((a-b)(b-c)(c-a))` are in A.P. `rArr b-c,c-a,a-b` are in H.P. |
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| 130. |
If `a_(n+1)=1/(1-a_n)` for `n>=1` and `a_3=a_1`. then find the value of `(a_2001)^2001`. |
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Answer» Correct Answer - `-1` We have, `a_(n+1)=1/(1-a_(n))` `thereforea_(2)=1/(1-a_(1))` and `a_(3)=1/(1-a_(2))=1/(1-1/(1-a_(1))=(1-a_(1))/(-a_(1))` Since `a_(3)=a_(1),` we have `1-a_(1)/(-a_(1))=a_(1)` `rArra_(1)^(2)-a_(1)+1=0` `rArra_(1)=-omega` or `-omega^(2)`, where `omega` is cube root of unity. Now, `a_(5)=1/(1-a_(4))=1/(1-1/(1-a_(3)))` `=(1-a_(3))/(-a_(3))` `=(1-a_(1))/(-a_(1))=a_(1)=a_(3)` and so on `thereforea_(1)=a_(3)=a_(5)....a_(2001)` Thus, `(a_(2001))^(2001)=(-omega)^(2001)` or `(-omega^(2))^(2001=-1` or `(-1)^(2001)(omega^(3))^(1334)=-1` (`becauseomega^(3)=1`) |
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| 131. |
Two consecutive numbers from 1, 2, 3, ..., n are removed, then arithmetic mean of the remaining numbers is `105/4` then `n/10` must be equal toA. [45,55]B. [52,60]C. [41,49]D. none of these |
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Answer» Correct Answer - A Let m and (m+1) be the removed numbers from 1,2,…,n. Then, the sum of the remaining numbers is n(n+1)/2-(2m+1). From given condition, `105/4=((n(n+1))/2-(2m+1))/((n-2))` or `2n^(2)-103n-8m+206`=0 Since n and m are integers, so n must be even. Let n=2k. Then, `m=(4k^(2)+103(1-k))/4` Since m is an integer, then 1-k must be divisble by 4. Let k=1+4t. Then we get n=8t+2 and `m=16t^(2)-95t+1`. Now, `1lemltn` `rArr1le16t^(2)-95t+1lt8t+2` Solving, we get t=6. Hence, n=50 and m=7 Hence, the removed numbers are 7 and 8. Also, sm of all numbers is 50(50+1)/2=1275. |
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| 132. |
Two consecutive numbers from 1,2,3 …., n are removed.The arithmetic mean of the remaining numbers is 105/4 . The removed numbersA. lie between 10 and 20B. are less than 1500C. are less than 1500D. none of these |
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Answer» Correct Answer - C Let m and (m+1) be the removed numbers from 1,2,…,n. Then, the sum of the remaining numbers is n(n+1)/2-(2m+1). From given condition, `105/4=((n(n+1))/2-(2m+1))/((n-2))` or `2n^(2)-103n-8m+206`=0 Since n and m are integers, so n must be even. Let n=2k. Then, `m=(4k^(2)+103(1-k))/4` Since m is an integer, then 1-k must be divisble by 4. Let k=1+4t. Then we get n=8t+2 and `m=16t^(2)-95t+1`. Now, `1lemltn` `rArr1le16t^(2)-95t+1lt8t+2` Solving, we get t=6. Hence, n=50 and m=7 Hence, the removed numbers are 7 and 8. Also, sm of all numbers is 50(50+1)/2=1275. |
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| 133. |
If the sum of the first ten terms of an `A.P` is four times the sum of its first five terms, the ratio of the first term to the common difference is: |
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Answer» Correct Answer - `(1)/(2)` Let a be the first term and d be the common difference of A.P. According to the question. `S_(10)=4S_(5)` `rArr5[2a+9d]=4xx5/2[2a+4d]` `rArr10a+45d=20a+40d` `10a-5d=0` `rArra/d=1/2` |
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| 134. |
If the surm of the first ten terms of the series,`(1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+........`, is `16/5m` ,then m is equal toA. 101B. 100C. 99D. 102 |
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Answer» Correct Answer - A `S_n=(8/5)^5+(12/5)^2+(16/5)^2+(20/5)^2+(24/5)^4+...` `=8^2/5^2+12^2/5^2+16^2/5^2+20^2/5^2+24^2/5+.....` `therefore T_n=(4n+4)^2/5^2` `therefore S_n =1/5^2 underset(n=1)overset(10)Sigma16 (n+1)^2` `=16/25[(underset(n=1)overset(11)Sigman^2)-1]` `=16/25[(11xx12xx23)/(6)-1]` `=16/25xx 505 =16/5m` `rArr m= 101` |
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| 135. |
The sum of the first 9 terms of the series `1^3/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 +3^3)/(1 + 3 +5)` ..... is :A. 71B. 96C. 142D. 192 |
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Answer» Correct Answer - B `T_(n)=(1^3)+2^(3)+3^(3)+….+n^(3))/(1+3+5+…+(2n-1))=((n^(2)(n+1)^(2))/4)/(n/2(1+2n-1))` `rArrT_(n)=((n^(2)(n+1)^(2))/4)/(n^(2))` `thereforeT_(n)=1/4(n+1)^(2)` `T_(n)=1/4[n^(2)+2n+1]` `S_(n)=sum_(n=1)^(n)T_(n)` `S_(n)=1/4[(n(n+1)(2n+1))/6+n(n+1)+n]` Putting n=9, we get `S_(9)=1/4[(9xx10xx19)/6+9xx10+9]` `=1/4[285+90+9]=384/4=96` |
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| 136. |
If the 2nd , 5th and 9thterms of anon-constant A.P. are in G.P., then thecommon ratio of this G.P. is :(1) `8/5`(2)`4/3`(3)1 (4) `7/4`A. `4/3`B. 1C. `7/4`D. `8/5` |
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Answer» Correct Answer - A Given a+d,a+4d,a+8d are in G.P. `therefore (a+4d)^(2)=(a+d)(a+8d)=a^(2)+9ad+8d^(2)` `rArr8d^(2)=adrArra=8d` `therefore9d,12d,16d` are G.P. Common ratio `r=12/9=4/3` |
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| 137. |
If in a progression `a_1, a_2, a_3, e t cdot,(a_r-a_(r+1))`bears a constant atio with `a_rxxa_(r+1)`, then the terms of the progression are ina. A.P b. G.P. c. H.P. d. none of theseA. A.PB. G.PC. H.PD. none of these |
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Answer» Correct Answer - C `(a_(r)-a_(r+1))/(a_(r)a_(r+1))=k` (constant) `rArr1/(a_(r+1))-1/a_(r)=k` `1/a_(1),1/a_(2),….,1/a_(n)` are in A.P. `rArra_(1),a_(2),a_(3),…` are in H.P. |
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| 138. |
Let `a_1,a_2,a_3….., a_(101)` are in G.P with `a_(101) =25 ` and `Sigma_(i=1)^(201) a_i=625` Then the value of `Sigma_(i=1)^(201) 1/a_i` eaquals _______. |
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Answer» Correct Answer - 1 Let a be the first fterm and r be the common ratio of G.P. Then `a(1-r^(201))/(1-r)=625` (1) Now `sum_(r=1)^(201)1/(a_(i))=1/(a_(1))+1/(a_(2))+..+1/(a_(201))` `=1/a+1/(ar)+..+1/(ar^(200))` `=(1/a((1/r)^(201)-1))/((1/r-1))` `=1/a((1-r^(201))/(1-r))1/r^(200)` `=1/axx625/axx1/r^(200)` [from (1)] `=625/((ar^(100))^(2))` `=625/(a_(101))^(2)` `=625/625=1` |
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| 139. |
Let `a_1,a_2,a_3,……,a100` be an arithmetic progression with `a_1=3 and S_p=Sigma_(i=1)^(p) a_i, 1 le p le 100`. For any integer n with `1 le n le 20 , let m=5 n`. If `(S_m)/(S_n)` does not depend on .n then `a_2` is _________. |
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Answer» Correct Answer - 6 `a_1,a_2,a_3,.....,a_100` is an A.P `a_1+3,S_p=underset(i=1)overset(p)Sigma a_a,1 le p le 100` `(S_m)/(S_(n))=(S_(5n))/(S_n)=((5n)/(2)(6+(5n-1)d))/(n/2(6+(n-1)d))` `S_m/S_n` is independent of n of 6-d =0 `rArr` d=6 |
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| 140. |
If `a_(1),a_(2)a_(3),….,a_(15)` are in `A.P` and `a_(1)+a_(8)+a_(15)=15`, then `a_(2)+a_(3)+a_(8)+a_(13)+a_(14)` is equal toA. `25`B. `35`C. `10`D. `15` |
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Answer» Correct Answer - A `(a)` Given `A.P.` is `a_(1),a_(2),a_(3),…….,a_(15)` `a_(1)+a_(15)=a_(2)+a_(14)=….=2a_(8)` `a_(1)+a_(15)+a_(8)=(3)/(2)(a_(1)+a_(15))=15` `impliesa_(1)+a_(15)=10` `a_(2)+a_(3)+a_(8)+a_(13)+a_(14)=2(a_(1)+a_(15))+a_(8)` `=2(10)+5=25` |
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| 141. |
Between the numbers `2` and `20`, `8` means are inserted. Then their sum isA. `88`B. `44`C. `176`D. None of these |
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Answer» Correct Answer - A `(a)` `a,A_(1),A_(2),………A_(n)`, `b` are in `A.P.` where `a=2`, `b=20`, `n=8` `:.` Sum of the means `=(n)/(2)(a+b)=(8)/(2)(2+20)=88` |
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| 142. |
If `a_1,a_2,a_3,...` are in A.P. and `a_i>0` for each i, then `sum_(i=1)^n n/(a_(i+1)^(2/3)+a_(i+1)^(1/3)a_i^(1/3)+a_i^(2/3))` is equal toA. `(n)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))`B. `(n+1)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))`C. `(n-1)/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))`D. None of these |
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Answer» Correct Answer - C `(c )` Let `d` be common difference of `A.P.impliesd=a_(i)-a_(i-1)` Now `(1)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(a_(i+1)^(1//3)-a_(i)^(1//3))/(a_(i+1)-a_(i))` `=(1)/(d)[a_(i+1)^(1//3)-a_(i)^(1//3)]` Thus `sum_(i=1)^(n)(n)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(1)/(d)sum_(i=1)^(n-1)(a_(i+1)^(1//3)-a_(i)^(1//3))` `=(1)/(d)(a_(n)^(1//3)-a_(1)^(1//3))` `=(1)/(d)((a_(n)-a_(1)))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))` `=((n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))` |
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| 143. |
If `S_n,` denotes the sum of `n` terms of an `A.P.`, then `S_(n+3)-3S_(n+2)+3S_(n+1)-S_n=`A. `2s_n`B. `S_(n+1)`C. `3S_n`D. 0 |
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Answer» Correct Answer - D `S_(n+3)-3S_(n+2)+3S_(n+1)-S_(n)` `=(S_(n+3)-S_(n+2))-2(S_(n+2)-S_(n-1))+(S_(n+1)-S_(n))` =`T_(n+3)-2T_(n+2)+T_(n+1)` `=(T_(n+3)-T_(n+2))-(T_(n+2)-T_(n+1))` =d-d=0 |
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| 144. |
Find the sum `Sigma_(r=1)^(oo) (r-2)/((r+2)(r+3)(r+4))` |
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Answer» Correct Answer - `1/12` `T_(r)=(r-2)/((r+2)(r+3)(r+4))=(r(r+4)-(r+1)(r+2))/((r+2)(r+3)(r+4))` `=r/((r+2)(r+3))-((r+1))/((r+3)(r+4))` `therefore S_(n)=sum_(r=1)^(n)T_(r)=1/12-((n+1))/((n+3)(n+4))` `thereforeS_(oo)=1/12` |
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| 145. |
Find the sum `Sigma_(r=1)^(oo)(3n^2+1)/((n^2-1)^3)` |
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Answer» Correct Answer - `9/16` `T_(n)=(3n^(2)+1)/((n^(2)-1)^(3))=1/2(6n^(2)+2)/((n^(2)-1)^(3))` `=1/2(((n+1)^(3)-(n-1)^(3))/((n+1)^(3)(n-1)^(3)))` `=1/2(1/((n-1)^(3))-1/(n+1)^(3))` `thereforeS=1/2[(1/1^(3)-1/3^(3))+(1/2^(3)-1/3^(3))+(1/2^(3)-1/4^(3))+(1/3^(3)-1/5^(3))` `+(1/4^(3)-1/6^(3))+...]` `=1/2(1+1/8)=9/16` |
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| 146. |
Find the sum `Sigma_(r=1)^(oo) (r)/(r^4+1/4)` |
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Answer» Correct Answer - `840/841` `T_(r)=r/(r^(4)+1/4)=r/(r^(4)+r^(2)+1/4-r^(2))=r/((r^(2)+1/2)^(2)-r^(2))` `=1/2[(2r)/((r^(2)+r+1/2)(r^(2)-r+1/2))]` `=1/2[((r^(2)+r+1/2)-(r^(2)-r+1/2))/((r^(2)+r+1/2)(r^(2)-r+1/2))]` `=1/2[1/(r^(2)-r+1/2)-1/(r^(2)+r+1/2)]` `=[1/(2r^(2)-2r+1)-1/(2r^(2)+2r+1)]` `thereforesum_(r=1)^(20)T_(r)=sum_(r=1)^(20)[1/(2r^(2)-2r+1)-1/(2r^(2)+2r+1)]` `=1/(2(1)^(2)-2(1)+1)-1/(2(20)^(2)+2(20)+1)` `=1-1/841-840/841` |
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| 147. |
The value of x that satisfies the relation `x=1 -x+x^2-x^3+x^4-x^5+….oo is `A. `2 cos 36^@`B. `2 cos 144^@`C. `2 sin 18^@`D. `2 cos 18^@` |
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Answer» Correct Answer - C `x=1-x+x^(2)-x^(3)+x^(4)-x^(5)+..oo` `thereforex=1/(1+x)` `rArrx^(2)+x-1=0` `rArrx=(-1pmsqrt5)/2` But for infinite G.P., x=`(-1-sqrt5)/(2)` is rejected. `thereforex=(-1+sqrt5)/2=2sin18^(@)` |
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| 148. |
If |`|a| lt 1 "and " |b| lt 1` then the sum of the series `1+(1+a)b+(1+a+a^2)b^2+(1+a+a^2+a^3)b^3+…..is `A. `(1)/((1-a)(1-b))`B. `(1)/((1-a)(1-ab))`C. `(1)/((1-b)(1-ab))`D. `(1)/((1-a)(1-b)(1-ab))` |
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Answer» Correct Answer - C We have `1+(1+a)b+(1+a+a^(2))b^(2)+(1+a+a^(2)+a^(3))b^(3)+…oo` `=sum_(n=1)^(oo)(1+a+a^(2)+…+a^(n-1))b^(n-1)` `=sum_(n=1)^(oo)((1-a^(n))/(1-a))b^(n-1)` `=sum_(n=1)^(oo)(b^(n-1))-a/(1-a)sum_(n=1)^(oo)(ab)^(n-1)` `=1/(1-a)[1+b+b^(2)+..oo]-a/(1-a)[1+ab+(ab)^(2)+...oo]` `=1/(1-a)xx1/(1-b)-a/((1-a)(1-ab))` `=1/((1-ab)(1-b))` |
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| 149. |
If `x=9^(1//3)xx9^(1//9)xx 9^(1//27)xx….,y=4^(1//3)xx-4^(1//9)xx 4^(1//27)x….,` and `z=Sigma_(r=1)^(oo) (1+i)^(r)` then arg (x+yz) is equal to |
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Answer» Correct Answer - C `x=9^(1/3+1/9+1/27+..)`=`9^((1/3)/(1-1/3))=9^(1/2)=3` `y=4^(1/3-1/9+1/27+..)=4^((1/3)/(1+1/3))=4^(1/4)=sqrt2` `z=sum_(r=1)^(oo)(1+i)^(-r)=1/(1+i)+1/((1+i)^(2))+1/((1+i)^(3))+..` `=(1/(1+i))/(1-1/(1+i))=1/i=-i` Let `alpha=x+yz=3-isqrt2` `thereforeargalpha=-tan^(-1)(sqrt2/3)` |
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| 150. |
Which of the following can be terms (not necessarily consecutive) ofany A.P.?a. 1,6,19b. `sqrt(2),sqrt(50),sqrt(98)`c. `log2,log16 ,log128`d. `sqrt(2),sqrt(3),sqrt(7)`A. 1,6,19B. `sqrt(2).sqrt(50),sqrt(98)`C. log 2,log 16 , log128D. `sqrt(2),sqrt(3),sqrt(7)` |
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Answer» Correct Answer - A::B::C Let a,b,c be pth,qth and rth terms of A.P. Then a=A+(p-1)D,b=A+(q-1)D,c=A+(r-1)D `rArr(r-q)/(q-p)=(c-b)/(b-a)` is rational number. now for 1,6,19,`(r-q)/(q-p)=(19-6)/(6-1)` is a rational number. For `sqrt2,sqrt(50),sqrt(98),(r-q)/(q-p)=(sqrt(98)-sqrt(50))/(sqrt(50)-sqrt2)=(7sqrt2-5sqrt2)/(5sqrt2-sqrt2)` `=1/2` is a rational number. For log2, log 16, log 128 `(r-q)/(q-p)=(log128-log16)/(log16-log2)=(7log2-4log2)/(4log2-log2)=1` is a rational number. But for `sqrt2,sqrt3,sqrt7,(r-q)/(q-p)` is not a rational number. |
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