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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.
| 101. |
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,-1,0,1`B. 0,1,2,3,C. `-1,0,1,2,`D. none of these |
| Answer» Correct Answer - C | |
| 102. |
The sum of squares of three distinct real numbers which form an increasing GP is `S^2` (common ratio is r). If sum of numbers is `alpha S`, then if `r=3` then `alpha^2` cannot lie inA. `1ltalpha^(2)lt3`B. `(1)/(3)ltalpha^(2)lt3`C. `1ltalphalt3`D. `(1)/(3)ltalphalt3` |
| Answer» Correct Answer - B | |
| 103. |
Let `S_k ,k=1,2, ,100 ,`denotes thesum of the infinite geometric series whose first term s `(k-1)/(k !)`and the common ratio is `1/k`, then the value of `(100^2)/(100 !)+sum_(k=2)^(100)(k^2-3k+1)S_k`is _______. |
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Answer» `S_(k)=(a)/(1-r)=((k-1)/(k!))/(1-(1)/(k))=(k)/(k!)=(1)/((k-1)!)` Now, `sum_(k=2)^(100)|(k^(2)-3k+1)S_(k)|=sum_(k=2)^(100)|(k^(2)-3k+1)*(1)/((k-1)!)|` `=sum_(k=2)^(100)|((k-1))/((k-2))=(k)/((k-1)!)|` `=|(1)/(0!)-(2)/(1!)|+|(2)/(1!)-(3)/(2!)|+|(3)/(2!)-(4)/(3!)|+"........."+|(99)/(98!)-(100)/(99!)|` `=((2)/(1!)-(1)/(0!))+((2)/(1!)-(3)/(2!))+((3)/(2!)-(4)/(3!))+"........."+((99)/(98!)-(100)/(99!))` `=3-(100)/(99)=3-((100)^(2))/(100!)` `:.((100)^(2))/(100!)+sum_(k=2)^(100)|(k^(2)-3k+1)S_(k)|=3`. |
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| 104. |
Let `S_(k),k=1,2, . . . ,100`, denote the sum of the infinite geometric series whose first term is `(k-1)/(k!)` and the common ratio `(1)/(k)`. Then the value of `(100^(2))/(100!)+sum_(k=1)^(100) |(k^(2)-3k+1)S_(k)|`, isA. 3B. 6C. 8D. 9 |
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Answer» Correct Answer - A We have, `S_(k)=((k-1)/(k!))/(1-(1)/(k))=(1)/((k-1)!)` `:." "underset(k=2)overset(100)sum|(k^(2)-3k+1)S_(k)|` `=underset(k=2)overset(100)sum|{:({(k-1)^(2)-k})/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:(k-1)/((k-2)!)-(k)/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:((k-2)+1)/((k-2)!)-((k-1)+1)/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:(1)/((k-3)!)-(1)/((k-2)!)+(1)/((k-2)!)-(1)/((k-2)!):}|` `=underset(k=2)overset(100)sum{(1)/((k-3)!)-(1)/((k-2)!)}underset(k=2)overset(100)sum{(1)/((k-2)!)-(1)/((k-1)!)}` `={(1-(1)/(1!))+((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+ . . . +((1)/(97!)-(1)/(98!))}` `+{(1-(1)/(1!))+((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+ . . . +((1)/(98!)-(1)/(99!))}` `=(1-(1)/(98!))+(1-(1)/(99!))=2(1)/(98!)-(1)/(99!)=2-(100)/(99!)` Hence, `((100)^(2))/(100!)+underset(k=1)overset(100)sum|{:(k^(2)-3k+1)S_(k):}|` `(100)/(99!)+S_(1)+underset(k=2)overset(100)sum|{:(k^(2)-3k+1)S_(k):}|` `=(100)/(99!)+1+2-(100)/(99!)=3` |
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| 105. |
Let `S_(n),n=1,2,3,"…"` be the sum of infinite geometric series, whose first term is n and the common ratio is `(1)/(n+1)`. Evaluate `lim_(n to oo)(S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1))/(S_(1)^(2)+S_(2)^(2)+"......"+S_(n)^(2))`. |
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Answer» `:. S_(n)=(n)/(1-(1)/(n+1)) impliesS_(n)=n+1` `S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1)` `=sum_(r=1)^(n)S_(r)S_(n-r-1)=sum_(r=1)^(n)(r+1)(n-r+2)` `=sum_(r=1)^(n)[(n+1)r-r^(2)+(n+2)]` `=(n+1)sumr-sum_(r=1)^(n)r^(2)+(n+2)sum_(r=1)^(n)1` `=(n+1)sumn-sumn^(2)+(n+2)*n` `=((n+1)n(n+1))/(2)-(n(n+1)(2n+1))/(6)+(n+2)n` `(n)/(6)(n^(2)+9n+14)" " "....(i)"` and `S_(1)^(2)+S_(1)^(2)+"...."+S_(n)^(2)=sum_(r=1)^(n)S_(r)^(2)=sum_(r=1)^(n)(r+1)^(2)=sum_(r=0)^(n)(r+1)^(2)-1^(2)` `=((n+1)(n+2)(2n+3))/(6)-1` `(n)/(6)(2n^(2)+9+13)" " ".....(ii)"` From Eqs. (i) and (ii), we get `lim_(n to oo)(S_(1)S_(n)+S_(2)S_(n-1)+S_(3)S_(n-2)+"..."+S_(n)S_(1))/(S_(1)^(2)+S_(2)^(2)+"......"+S_(n)^(2))` `=lim_(n to oo)((n)/(6)(n^(2)+9n+14))/((n)/(6)(2n^(2)+9n+13))=lim_(n to oo)((1+(9)/(n)+(14)/(n^(2))))/((2+(9)/(n)+(13)/(n^(2))))` `(1+0+0)/(2+0+0)=(1)/(2)`. |
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| 106. |
If `S_(1), S_(2), S_(3),...,S_(n)` are the sums of infinite geometric series, whose first terms are 1, 2, 3,.., n and whose common rations are `(1)/(2), (1)/(3), (1)/(4),..., (1)/(n+1)` respectively, then find the values of `S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ...+ S_(2n-1)^(2)`. |
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Answer» Correct Answer - `(1)/(6) (2n) (2n +1) (4n +1) -1` Consider an infinite GP with first term 1,2,3,.., n and common ratios `(1)/(2), (1)/(3), (1)/(4),..., (1)/(n +1)` `:. S_(1) = (1)/(1 -1//2) = 2` `{:(" "S_(2) = (2)/(1 -1//3) = 3),(" "vdots " "vdots " "vdots),(S_(2n -1) = (2n -1)/(1-1//2n) = 2n):}` `:. S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ...+ S_(2n-1)^(2)` `= 2^(2) + 3^(2) + 4^(2) + ...+ (2n)^(2)` `= (1)/(6) (2n) (2n+1) (4n+1) -1` |
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| 107. |
The minimum value of `((a^2 +3a+1)(b^2+3b + 1)(c^2 3c+ 1))/(abc)`The minimum value of , where `a, b, c in R` isA. `(11^(3))/(2^(3))`B. 125C. 25D. 27 |
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Answer» Correct Answer - B Let `A=((a^(2)+3a+1)(b^(2)+3b+1)(c^(2)+3c+1))/(abc)` `=((a^(2)+3a+1)/(a))((b^(2)+3b+1)/(b))((c^(2)+3c+1)/(c))` `=(a+3+(1)/(a))(b+3+(1)/(b))(c+3+(1)/(c))` where `a,b,c in R^(+)`. Applying `Am ge GM` on a and `(1)/(a)` `a+(1)/(a)ge 2 " " implies a+(1)/(b)+3ge 5` Similarly, `b+(1)/(b)ge 2 " " implies b+(1)/(b)+3ge 5` and `c+(1)/(c)ge 2 " " implies c+(1)/(c)+3ge 5` `:.(a+(1)/(a)+3)(b+(1)/(b)+3)(c+(1)/(c)+3)ge 125` So, `Age 5*5*5 " " implies Age 125` minimum value of A is 125. |
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| 108. |
If `cos (x-y),, cosx and cos(x+y)` are in H.P., are in H.P., then `cosx*sec(y/2)`=A. `pmsqrt(2)`B. `(1)/(sqrt(2))`C. `-(1)/(sqrt(2))`D. None of these |
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Answer» Correct Answer - A `:.cos(x-y),cosx,cos(x+y)` are in HP. `:.cosx=(2cos(x-y)cos(x+y))/(cos(x-y)+cos(x+y))` `implies cosx=(2cos^(2)x-sin^(2)y)/(2cosx cosy)` `implies cos^(2)x cosy =cos^(2)x-sin^(2)y` `implies cos^(2)x(1-cosy)=sin^(2)y` `=(1+cos y)(1-cosy )` `implies cos^(2)x=(1+cosy)" "[:.1-cosyne0]` `implies cos^(2)x= 2 cos^(2)""(y)/(2)` `implies cos^(2)xsec^(2)((y)/(2))=2` `:. cos x sec ((y)/(2))= pm sqrt(2)` |
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| 109. |
Let `S_(k)`, where `k = 1,2`,....,100, denotes the sum of the infinite geometric series whose first term is `(k -1)/(k!)` and the common ratio is `(1)/(k)`. Then, the value of `(100^(2))/(100!) +sum_(k=2)^(100) | (k^(2) - 3k +1) S_(k)|` is.... |
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Answer» Correct Answer - 4 We have, `S_(k) = ((k-1)/(k!))/(1 - (1)/(k)) = (1)/((k -1)!)` Now, `(k^(2) -3k +1)S_(k)= {(k -2) (k-1) -1} xx (1)/((k -1)!)` `= (1)/((k -3)!) - (1)/((k -1)!)` `rArr underset(k=1)overset(100)sum ! (k^(2) - 3k + 1) S_(k) |=1 + 1 +2 - ((1)/(99!) + (1)/(98!)) = 4 - (100^(2))/(100!)` `rArr (100^(2))/(100!) + underset(k=1)overset(100)sum | (k^(2) -3k +1) S_(k)| = 4` |
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| 110. |
Three non-zero real numbers from an A.P. and the squares of these numbers taken in same order from a G.P. Then, the number of all possible value of common ratio of the G.P. isA. 1B. 2C. 3D. none of these |
| Answer» Correct Answer - C | |
| 111. |
If `x >1,y >1,z >1`are in GP, then `1/(1+1nx),1/(1+1ny),1/(1+1nz)`are in(1998, 2M)AP (b) HP(c) GP (d)none of theseA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - C `:.x,y,z`are in GP `" " " "[x,y,zgt1]` `:.1nx,1ny,1nz` are in AP. and `2x,4x,6x` are also in AP. `" " " "[xgt1]` By property, `2x+1nx,4x+1ny,6x+1nz` are also in AP. `(1)/(2x+1nx),(1)/(4x+1ny),(1)/(6x+1nz)` are in HP. |
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| 112. |
If a,b,c,d,e be 5 numbers such that a,b,c are in A.P; b,c,d are in GP & c,d,e are in HP then prove that a,c,e are in GPA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 113. |
If first and `(2n-1)^th` terms of an AP, GP. and HP. are equal and their nth terms are a, b, c respectively, then(a) a=b=c (b)a+c=b (c) a>b>c and `ac-b^2=0` (d) none of theseA. `a = b = c`B. `a ge b ge c`C. `a + c = b`D. `ac - b^(2) = 0` |
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Answer» Correct Answer - A::B::D Since, first and `(2n -1)th` terms are equal. Let first term be x and `(2n -1)`th term be y shoe middle term is `l_(n)`. Thus, in arithmetic progression, `t_(n) = (x + y)/(2) = a` In geometric progression, `t_(n) = sqrt(xy) = b` In harmonic progression, `t_(n) = (2xy)/(x + y) = c` `rArr b^(2) = ac and a gt b gt c` [using `AM gt GM gt HM`] Here, equality holds (i.e., `a = b =c`) only if all terms are same. Hence, options (a), (b) and (d) are correct |
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| 114. |
if `(m+1)th, (n+1)th and (r+1)th` term of an AP are in GP.and m, n and r in HP. . find the ratio of first term of A.P to its common differenceA. `-(2)/(n)`B. `(2)/(n)`C. `-(n)/(2)`D. `(n)/(2)` |
| Answer» Correct Answer - A | |
| 115. |
if `(m+1)th, (n+1)th and (r+1)th` term of an AP are in GP.and m, n and r in HP. . find the ratio of first term of A.P to its common differenceA. `n//2`B. `-n//2`C. `n//3`D. `-n//3` |
| Answer» Correct Answer - B | |
| 116. |
If a,b,c are in GP,`a-b,c-a,b-c` are in HP, then `a+4b+c` is equal toA. `0`B. 1C. `-1`D. None of these |
| Answer» Correct Answer - A | |
| 117. |
The sixth term of an `A.P., a_1,a_2,a_3,.............,a_n` is `2`. If the quantity `a_1a_4a_5`, is minimum then then the common difference of the `A.P.`A. `x=8//5`B. `x=5//4`C. `x=2//3`D. `x=4//5` |
| Answer» Correct Answer - C | |
| 118. |
If `a_1,a_2,a_3,....a_n` are positive real numbers whose product is a fixed number c, then the minimum value of `a_1+a_2+....+a_(n-1)+2a_n` is |
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Answer» `therefore AM ge GM` ` therefore (a_(1)+a_(2)+"......."+a_(n-1)+3a_(n))/(n)ge (a_(1)a_(2)"...."a_(n-1)3a_(n))^((1)/(n))=(3c)^((1)/(n))` ` implies a_(1)+a_(2)+"......."+a_(n-1)+3a_(n)ge n (3c)^((1)/(n))` Hence, the minimum value of ` a_(1)+a_(2)+"...."+a_(n-1)+3a_(n) " is " n(3c)^((1)/(n))`. |
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| 119. |
Statement 1 `4,8,16,` are in GP and 12,16,24 are in HP. Statement 2 If middle term is added in three consecutive terms of a GP, resultant will be in HP.A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1C. Statement 1 is true, Statement 2 is falseD. Statement 1 is false, Statement 2 is true |
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Answer» Correct Answer - A If a,b,c are in GP. Then, `b^(2)=ac` If middle term is added, then `a+b,2b " and " c+b` are in GP. `(I-II)/(II-III)=(a+b-2b)/(2b-(c+b)) " " [" here ",I=a+b,II=2b,III=c+b]` `=(a-b)/(b-c)=(ab-b^(2))/(b^(2)-bc)=(ab-ac)/(ac-bc)" " [:.b^(2)=ac]` `=(a(b-c)(a+b)(b+c))/(c(a-b)(a+b)(b+c))` `=(a(b^(2)-c^(2))(a+b))/(c(a^(2)-b^(2))(b+c))` `=(a(ac-c^(2))(a+b))/(c(a^(2)-ac)(b+c)),(a+b)/(b+c)=(I)/(II)` Hence, `a+b,2b,b+c` are in HP. Hence, both statements are true and Statement 2 is correct explanation for Statement 1. |
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| 120. |
If `a_1,a_2,a_3` are 3 positive consecutive terms of a GP with common ratio K . then all the values of K for which the inequality `a_3>4a_2-3a_1`, is satisfiedA. `1ltrlt3`B. `-3ltrlt-1`C. `rgt3orrlt1`D. none of these |
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Answer» Correct Answer - C Let `a_(1),a_(2),a_(3)` be first three consecutive terms of G.P. with common ratio r. Then, `a_(2)=a_(1)randa_(3)=a_(1)r^(2)`. Now, `a_(3)gt4a_(2)-3a_(1)` `rArr" "a_(1)r^(2)gt4a_(1)r-3a_(1)` `rArr" "r^(2)gt4r-3` `rArr" "r^(2)-4r+3gt 0rArr(r-1)(r-3)gt0rArrrlt1 or rgt3` |
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| 121. |
Find a three digit numberwhose consecutive digits form a GP. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order. Now, if we increase the second digit of the required number by 2, then the resulting digits will form an AP. |
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Answer» Let the three digits be `a,ar,ar^(2)`, then according to hypothesis `100a+10ar+ar^(2)-792=100ar^(2)+10ar+a` ` implies 99a(1-r^(2))=792` ` implies a(1+r)(1-r)=8 " " "......(i)"` and `a,ar+2,ar^(2)` are in AP. Then, `2(ar+2)=a+ar^(2)` ` implies a(r^(2)-2r+1)=4 implies a(r-1)^(2)=4" " "......(ii)"` On dividing Eq.(i) by Eq. (ii), we get `(r+1)/(r-1)=-2 impliesr=(1)/(3)` From Eq. (ii),`a=9` Thus, digits are `9,3,1` and so the required number is 931. |
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| 122. |
Suppose four distinct positive numbers `a_1, a_2, a_3, a_4,` are in G.P. Let `b_1=a_1,b_2=b_1+a_2.b_3=b_2+a_3 and b_4=b_3+a_1.`A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement IB. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement IC. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true |
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Answer» Correct Answer - C Let `a_(1) = 1, a_(2) = 2, rArr a_(3) = 4, a_(4) = 8` `:. B_(1) = 1, b_(2) =3, b_(3) = 7, b_(4) = 15` Clearly, `b_(1), b_(2), b_(3), b_(4)` are not in HP. Hence, Statement II is false Statement I is already true |
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| 123. |
The product of three consecutive terms of a GP is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then the sum of the original three terms of the given GP is: (a) 36 (b) 32 (c) 24 (d) 28A. 36B. 28C. 32D. 24 |
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Answer» Correct Answer - B Let the three consecutive terms of a GP are `(a)/(r)`, a and ar. Now, according to the question, we have `(a)/(r).a.ar = 512` `rArr a^(3) = 512` `rArr a = 8`...(i) Also, after adding 4 to first two terms, we get `(8)/(r) + 4, 8 + 4, 8r` are in AP `rArr 2(12) = (8)/(r) + 4 + 8r` `rArr 24 = (8)/(r) + 8r + 4 " " 20 = 4 ((2)/(r) + 2r)` `rArr 5 = (2)/(r) + 2r " " 2r^(2) - 5r + 2 = 0` `rArr 2r^(2) - 4r - r + 2 = 0` `rArr 2r (r -2) - 1 (r -2) = 0` `rArr (r - 2) (2r -1) = 0` `rArr (r -2) (2 -1) = 0` `rArr r = 2, (1)/(2)` Thus, the terms are either 16, 8 4 or 4, 8, 16. Hence, required sum = 28 |
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| 124. |
If `a_1, a_2, a_3,.....a_n` are in H.P. and `a_1 a_2+a_2 a_3+a_3 a_4+.......a_(n-1) a_n=ka_1 a_n`, then k is equal toA. `n(a_(1)-a_(n))`B. `(n-1)(a_(1)-a_(n))`C. `na_(1)a_(n)`D. `(n-1)a_(1)a_(n)` |
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Answer» Correct Answer - D `(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a_(2))="....."=(1)/(a_(n))-(1)/(a_(n-1))=d" "[" say "]` Then `a_(1)a_(2)=(a_(1)-a_(2))/(d),a_(2)a_(3)=(a_(2)-a_(3))/(d)".....",a_(n-1)a_(n)=(a_(n-1)-a_(n))/(d)` `:.a_(1)a_(2)+a_(2)a_(3)+"....."+a_(n-1)a_(n)=(a_(1)-a_(n))/(d)` Also, `(1)/(a_(n))=(1)/(a_(1))+(n-1)d implies (a_(1)-a_(n))/(d)=(n-1)a_(1)a_(n)` `:.a_(1)a_(2)+a_(2)a_(3)+"......"+a_(n-1)a_(n)=(n-1)a_(1)a_(n)`. |
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| 125. |
The first and second terms of a `GP` are `x^-4` and `x^n` respectively. If `x^52` is the eighth terms of the same progression, then `n` is equal toA. 13B. 4C. 5D. 3 |
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Answer» Correct Answer - B The common ratio of the G.P. is `x^(n+4)` `:." "x^(52)="Eighth term"` `rArr" "x^(52)-x^(-4)(x^(n+4))^(7)rArr56=7(n+4)rArr7n=28rArrn=4` |
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| 126. |
If `a , b , c ,d`and `p`are distinct real numbers such that(1987, 2M)`(a^2+b^2+c^2)p^2-2(a b+b c+c d)P+(b^2+c^2+d^2)geq0,t h e na , b , c , d`are in AP(b) are in GPare in HP(d) satisfy `a b=c d`A. A.PB. G.PC. H.PD. ab=cd |
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Answer» Correct Answer - B We have, `(a^(2)+b^(2)+c^(2))p^(2)-2p(ab+bc+cd)+(b^(2)+c^(2)+d^(2))le0` `rArr" "(ap-b)^(2)+(bp-c)^(2)+(cp-d)^(2)le0` `rArr" "ap-b=0=bp-c=cp-d` `rArr" "(b)/(a)=(c)/(b)=(d)/(c)=prArr` a,b,c,d are in G.P. |
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| 127. |
Let `{a_n}` bc a G.P. such that `a_4/a_6=1/4` and `a_2+a_5=216`. Then `a_1=`A. `12or,(108)/(7)`B. 10C. `7or,(54)/(7)`D. none of these |
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Answer» Correct Answer - A Let r be the common ratio of the G.P. Then, `(a_(4))/(a_(6))=(1)/(4)rArr(1)/(r^(2))=(1)/(4)rArrr=pm2` Now, `a_(2)+a_(5)=216` `rArr" "a_(1)(r+r^(4))=216` `rArr" "18a_(1)=216or,14a_(1)=216rArra_(1)=12,(108)/(7)` |
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| 128. |
Shamshad Ali buys a scooter for Rs. 2200. He pays Rs. 4000 cash andagrees to pay the balance in annual instalments of Rs. 1000 plus 10% intereston the unpaid amount. How much the scooter will cost him? |
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Answer» `rs 2200 -> ` cost `1yr -> 1000 + 10/100 xx 18000 = 1000 + 1800` `2 yr-> 1000 + 10/100 xx 17000 = 1000 + 1700` `n yr -> 1000 + 10/100 (18000 - (n-1)1000)` `= 1000+ 1800 + 100 - n100` `= 2700 - n(100)` `S_n = [ sum 2900 - n(100)] + 4000` `= 2900n - 100(n(n+1))/2 + 4000` `= 52200 - 500 xx 9 xx 19 + 4000` `= 56200 - 17100` `= 39200 rs` Answer |
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| 129. |
If the number x,y,z are in H.P. , then `sqrt(yz)/(sqrt(y)+sqrt(z)),sqrt(xz)/(sqrt(x)+sqrt(z)),sqrt(xy)/(sqrt(x)+sqrt(y))` are inA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - A `:.x,y,z` are in HP. `:.(1)/(x),(1)/(y),(1)/(z)` are in AP. `:.(1)/(x)-(1)/(y)=(1)/(y)-(1)/(z)" " ".......(i)"` `(sqrtyz)/(sqrt(y)+sqrt(z))=(1)/((1)/(sqrt(y))+(1)/(sqrt(z)))=a" " [" say "]` `(sqrtzx)/(sqrt(x)+sqrt(z))=(1)/((1)/(sqrt(z))+(1)/(sqrt(x)))=b" " [" say "]` and `(sqrtxy)/(sqrt(x)+sqrt(y))=(1)/((1)/(sqrt(x))+(1)/(sqrt(y)))=c" " [" say "]` `:.(a-b)/(b-c)=((1)/((1)/(sqrt(y))+(1)/(sqrt(z)))-(1)/((1)/(sqrt(z))+(1)/(sqrt(x))))/((1)/((1)/(sqrt(z))+(1)/(sqrt(x)))-(1)/((1)/(sqrt(x))+(1)/(sqrt(y))))=((1)/(x)-(1)/(y))/((1)/(y)-(1)/(z))=(a)/(a)` `" " " " [" from Eq. (i) "]` Hence, `(sqrtyz)/(sqrt(y)+sqrt(z)),(sqrt(zx))/(sqrt(z)+sqrt(x)),(sqrt(xy))/(sqrt(x)+sqrt(z))` are in AP. |
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| 130. |
If x,y,z are in A.P. then : yz,zx,xy are in (a) A.P (b) G.P (c) H.P (d) no definite sequenceA. A.P.B. G.P.C. H.P.D. no definite sequence |
| Answer» Correct Answer - C | |
| 131. |
If `sum_(i=1)^(n)a_(i)^(2)=lambda, AAa_(i)ge0` and if greatest and least values of `(sum_(i=1)^(n)a_(i))^(2)` are `lambda_(1)` and `lambda_(2)` respectively, then `(lambda_(1)-lambda_(2))` isA. `nlambda`B. `(n-1)lambda`C. `(n+2)lambda`D. `(n+1)lambda` |
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Answer» Correct Answer - B `:.`AM of 2nd powes `ge` 2nd power of AM `:.(a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2))/(n) ge((a_(1)+a_(2)+a_(3)+"..."a_(n))/(n))^(2)` `implies (lambda)/(n)ge((sum_(i=1)^(n)a_(i))/(n))^(2) :.ge(sum_(i=1)^(n)a_(i))^(2)le n lambda " " ".......(i)"` Also,`(a_(1)+a_(2)+a_(3)+"..."a_(n))^(2)=a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2)+2suma_(1)a_(2)=lambda+2suma_(1)a_(2)gelambda` `:.(sum_(i=1)^(n)a_(i))^(2)le lambda" " ".......(ii)"` From Eqs. (i) and (ii), we get `lambdale(sum_(i=1)^(n)a_(i))^(2)le n lambda` `:.lambda_(1)=nlambda" and "lambda_(2)=lambda` Then, `lambda_(1)-lambda_(2)=(n-1)lambda` |
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| 132. |
If `x > 0, y > 0, z>0 and x + y + z = 1` then the minimum value of `x/(2-x)+y/(2-y)+z/(2-z)` isA. 0.2B. 0.4C. 0.6D. 0.8 |
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Answer» Correct Answer - C Since, AM of `(-1)th` powers `ge(-1)th` powers of AM `:.((2-x)^(-1)+(2-y)^(-1)+(2-z)^(-1))/(3)ge(((2-x)+(2-y)+(2-z))/(3))^(-1)` `=[(6-(x+y+z))/(3)]^(-1)=((6-1)/(3))^(-1)=(3)/(5)" " [:.x+y+z=1]` `((2-x)^(-1)+(2-y)^(-1)+(2-z)^(-1))/(3)ge (3)/(5)` or `(1)/(3)[(1)/(2-x)+(1)/(2-y)+(1)/(2-z)]ge (3)/(5)` `implies (1)/(2-x)+(1)/(2-y)+(1)/(2-z)ge (9)/(5)` or `(2)/(2-x)+(2)/(2-y)+(2)/(2-z)ge (18)/(5)` or `1+(x)/(2-x)+1+(y)/(2-y)+1+(z)/(2-z)ge (18)/(5)` or `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge (18)/(5)-3` Hence, `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge (3)/(5)=0.6` `implies (x)/(2-x)+(y)/(2-y)+(z)/(2-z)ge 0.6` Thus, minimum value of `(x)/(2-x)+(y)/(2-y)+(z)/(2-z)` is 0.6. |
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| 133. |
If `x_1 , x_2, ..., x_n` are any real numbers and n is anypositive integer, thenA. `n underset(i=1)overset(n)sum x_(i)^(2) lt (underset(i=1)overset(n)sum x_(i))^(2)`B. `n underset(i=1)overset(n)sum x_(i)^(2) ge (underset(i=1)overset(n)sum x_(i))^(2)`C. `n underset(i=1)overset(n)sum x_(1)^(2) ge n (underset(i=1)overset(n)sum x_(i))^(2)`D. None of these |
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Answer» Correct Answer - B Since, `x_(1) , x_(2),..,x_(n)` are positive real numbers. `:.` Using nth power mean inequality `(x_(1)^(2) + x_(2)^(2) + ...+ x_(n)^(2))/(n) ge ((x_(1) + x_(2) + ...+ x_(n))/(n))^(2)` `rArr (n^(2))/(n) (underset(i=1)overset(n)sum x_(1)^(2)) ge (underset(i=1)overset(n)sum x_(i))^(2) rArr n (underset(i=1)overset(n)sumx_(i)^(2)) ge (underset(i=1)overset(n) sum x_(i))^(2)` |
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| 134. |
The largest term common to the sequences `1, 11 , 21 , 31 , to100`terms and `31 , 36 , 41 , 46 , to100`terms is`381`b. `471`c. `281`d. none of theseA. 281B. 381C. 471D. 521 |
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Answer» Correct Answer - D Sequence `1,11,21,31,"…"` has common difference =10 and sequence `31,36,41,46,"…"` has common difference =5. Hence, the sequence with common terms has common dofference LCM of 10 and 5 which is 10. The first common term os 31. Hence, sequence is `31,41,51,61,71,".."" " "....(i)"` Now, `t_(100)` of forst sequence `=1+(100-1)10=991` and `t_(100)` of second sequence `=31+(100-1)5=526` Value of largest common term lt 526 ` :. t_(n) " of Eq. " (i) is 31+(n-1)10=10n+21` `t_(50)=10xx50+21=521` is the value of largest common term. Aliter Let mth term of the jfirst sequence be equal to the nth term of the second sequence, then `1+(m-1)10=31+(n-1)5` `implies 10m-9=5n+26 implies 10m-35=5n` `implies 2m-7=n le 100 implies 2m le 107` `implies m le 53(1)/(2)` `:.` Largest value of m=53 `:.` Value of largest term `=1+(53-1)10=521` |
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| 135. |
Let `A_1 , G_1, H_1`denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For `n >2,`let `A_(n-1),G_(n-1)` and `H_(n-1)` has arithmetic, geometric and harmonic means as `A_n, G_N, H_N,` respectively.A. `G_(1) gt G_(2) gt G_(3) gt ...`B. `G_(1) lt G_(2) lt G_(3) lt...`C. `G_(1) = G_(2) = G_(3) = ...`D. `G_(1) lt G_(3) lt G_(5) lt .... And G_(2) gt G_(4) gt G_(6) gt ...` |
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Answer» Correct Answer - C Let a and b are two numbers. Then, `A_(1) = (a +b)/(2) , G_(1) = sqrt(ab), H_(1) = (2ab)/(a +b)` `A_(n) = (A_(n-1) + H_(n-1))/(2)` `G_(n) = sqrt(A_(n-1) H_(n-1))`, `H_(n) = (2A_(n-1) H_(n-1))/(A_(n-1) + H_(n-1))` Clearly, `G_(1) = G_(2) = G_(3) = ...= sqrt(ab)` |
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| 136. |
Let `A_1 , G_1, H_1`denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For `n >2,`let `A_(n-1),G_(n-1)` and `H_(n-1)` has arithmetic, geometric and harmonic means as `A_n, G_N, H_N,` respectively.A. `H_(1) gt H_(2) gt H_(3) gt ...`B. `H_(1) lt H_(2) lt H_(3) lt ...`C. `H_(1) gt H_(3) gt H_(5) gt ...and H_(2) lt H_(4) lt H_(6) lt ...`D. `H_(1) lt H_(3) lt H_(5) lt...and H_(2) gt H_(4) gt H_(6) gt ....` |
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Answer» Correct Answer - B As above `A_(1) gt H_(2) gt H_(1), A_(2) gt H_(3) gt H_(2)` `:. H_(1) lt H_(2) lt H_(3) lt`... |
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| 137. |
Let `A_1 , G_1, H_1`denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For `n >2,`let `A_(n-1),G_(n-1)` and `H_(n-1)` has arithmetic, geometric and harmonic means as `A_n, G_N, H_N,` respectively.A. `G_(1)gtG_(2)gtG_(3)gt"......."`B. `G_(1)ltG_(2)ltG_(3)lt"......."`C. `G_(1)=G_(2)=G_(3)="......."`D. `G_(1)ltG_(3)ltG_(5)lt"......." and "G_(2)gtG_(4)gtG_(6)gt"......."` |
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Answer» Correct Answer - C `A_(1)=(a+b)/(2),G_(1)=sqrt(ab),H_(1)=(2ab)/(a+b)` `A_(n)=(A_(n-1)+H_(n-1))/(2),G_(n)=sqrt(A_(n-1)H_(n-1))" and " H_(n)=(2A_(n-1)H_(n-1))/(A_(n-1)+H_(n-1))` Clearly, `G_(1)=G_(2)=G_(3)="......"=sqrt(ab)`. |
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| 138. |
Let `A_1 , G_1, H_1`denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For `n >2,`let `A_(n-1),G_(n-1)` and `H_(n-1)` has arithmetic, geometric and harmonic means as `A_n, G_N, H_N,` respectively.A. `A_(1) gt A_(2) gt A_(3) gt ...`B. `A_(1) lt A_(2) lt A_(3) lt ...`C. `A_(1) gt A_(3) gt A_(5) gt .... and A_(2) lt A_(4) lt A_(6) lt...`D. `A_(1) lt A_(3) lt A_(5)lt ....and A_(2) gt A_(4) gt A_(6) gt ...` |
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Answer» Correct Answer - A `A_(2)` is AM of `A_(1) and H_(1) and A_(1) gt H_(1)` `rArr A_(1) gt A_(2) gt H_(1)` `A_(3)` is AM of `A_(2) and H_(2) and A_(2) gt H_(2)` `{:(rArr,A_(2),gt,A_(3),gt,H_(2),),(,vdots,,vdots,,vdots,),( :.,A_(1),gt,A_(2),gt,A_(3),gt...):}` |
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| 139. |
If a, b, c are distinct positive real numbers in G.P and `log_ca, log_bc, log_ab` are in A.P, then find the common difference of this A.PA. 3B. `3//2`C. `1//2`D. `2//3` |
| Answer» Correct Answer - B | |
| 140. |
If m is the AM of two distinct real numbers l and n `(l,ngt1)` and `G_(1),G_(2)" and "G_(3)` are three geometric means between l and n, then `G_(1)^(4)+2G_(2)^(4)+G_(3)^(4)` equalsA. `4l^(2)m^(2)n^(2)`B. `4l^(2)mn`C. `4lm^(2)n`D. `4lmn^(2)` |
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Answer» Correct Answer - C Given, `m=(l+n)/2rArr l+n =2n" ...(i)"` and l, `G_1, G_2, G_3,` n are in GP `:." "G_1/l =G_2/G_1=G_2/G_2=n/G_3` `rArr" "G_1G_3 = "In," G_1^2 = lG_2, G_2^2 =G_3 G_1, G_3^2 =nG_21" (ii)"` Now, `G_1^4 +G_2^4 +G_3^4 =l^2 G_2^2+2G_2^4+n^2G_2^2` `=G_2^2(l^2 +2G_2^2 +n^2)" "["from eq.(ii)"]` `=G_3 G_1 (l^2 +2G_2^2 +n^2) " " [from Eq.(ii) ]` `=G_3 G_1 (l^2 +2G_3 G_1 + n^2)` = In `(l^2+"2 In"+n^2)" " [from Eq.(ii)]` = In `(l+n)^2` = In `(2m)^2" "[from Eq. (i)]` `=4//m^2n` |
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| 141. |
For `0 |
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Answer» Correct Answer - False Since, `(log_(a) x + (1)/(log_(a) x))/(2) gt 1, " using " AM ge GM` Here, equality holds only when `x =a` which is not possible. So, `log_(a) x + log_(x) a` is greater than 2 Hence, it is a false statement |
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| 142. |
If `a ,b ,c ,da n dp`are distinct real numbers such that `(a^2+b^2+c^2)p^2-2(a b+b c+c d)p+(b^2+c^2+d^2)lt=0,`then prove that `a ,b ,c , d`are in G.P.A. are in APB. are in GPC. are in HPD. satisfy `ab = cd` |
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Answer» Correct Answer - B Here, `(a^(2) + b^(2) + c^(2)) p^(2) -2 (ab + bc + cd) p + (b^(2) + c^(2) + d^(2)) le 0` `rArr (a^(2) p^(2) - 2abp + b^(2)) + (b^(2) p^(2) -2bcp + c^(2)) + (c^(2) p^(2) -2cdp + d^(2)) le 0` `rArr (ap - b)^(2) + (bp -c)^(2) + (cp -d)^(2) le 0` [since, sum of squares is never less than zero] Since, each of the squares is zero. `:. (ap = b)^(2) = (bp -c)^(2) + (cp -d)^(2) = 0` `rArr p = (b)/(a) = (c)/(b) = (d)/(c)` `:.` a, b, c, d are in GP. |
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| 143. |
The next term of the G.P. `x ,x^2+2,a n dx^3+10`is`(729)/(16)`b. `6`c. `0`d. `54` |
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Answer» Correct Answer - C::D According to the question, `x,x^(2)+2" and "x^(3)+10` are in GP. So, `(x^(2)+2)^(2)=x(x^(3)+10)=0` `implies x^(4)+4+4x^(2)-x^(4)-10x=0` `implies 4x^(2)-10x+4=0` `implies 2x^(2)-5x+2=0` `implies 2x^(2)-4x-x+2=0` `implies 2x(x-2)-1(x-2)=0` `implies(x-2)(2x-1)=0` `implies x=2` or `x=(1)/(2)` For`x=2`,first 3 terms are2,6,18. So, 4th term of GP `=2*(3)^(3)=54` For `x=(1)/(2)`, first 3 terms are `(1)/(2), (9)/(4),(81)/(8)`. So, `T_(4)=(1)/(2)((9)(2))^(3)=(1)/(2)xx(729)/(16)` |
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| 144. |
The G.M. of two positive numbers is 6. Their arithmetic mean A and harmonic mean H satisfy the equation `90A+5H=918`, then A may be equal to (A) `5/2` (B) 10 (C) 5 (D) `1/5`A. `(1)/(5)`B. 5C. `(5)/(2)`D. 10 |
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Answer» Correct Answer - A::D `:.G=6` and `G^(2)=AH` `implies H=(36)/(A)` Given, `90A+5H=918` `implies 90A+5xx(36)/(A)=918" " implies 5A+(10)/(A)=51` `implies 5A^(2)-51A+10=0" " implies (A-10)(5A-1)=0` `:.A=10,(1)/(5)` |
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| 145. |
whether the statement is true or false:` n_(1), n_(2),...,n_(p)` are p positive integers, whose sum is an even number, then the number of odd integers among them is odd. |
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Answer» Correct Answer - False Since, `n_(1), n_(2),..., n_(p)` are p positive integers, whose sum is even and we known that, sum of any two odd integers is even. `:.` Number of odd integers must be even. Hence, it is a false statement |
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| 146. |
If n is an odd integer greater than or equal to 1, then the value of `n^3 - (n-1)^3 + (n-1)^3 - (n-1)^3 + .... + (-1)^(n-1) 1^3`A. `((n+1)^(2)(2n-1))/(4)`B. `((n-1)^(2)(2n-1))/(4)`C. `((n+1)^(2)(2n+1))/(4)`D. none of these |
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Answer» Correct Answer - A We have, `n^(3)-(n-1)^(3)+(n-2)^(3)+ . . . . . . +(-1)^(n-1)1^(3)` `=1^(3)-2^(3)+3^(3)-4^(3)+ . . .. +n^(3)" "[because" n is odd"]` `=[1^(3)+2^(3)+3^(3)+4^(3)+ . . . +n^(3)]-2[2^(3)+4^(4)+ . . . .+(n-1)^(3)]` `={(n(n+1))/(2)}^(2)-2^(4){1^(3)+2^(3)+ . . . . +((n-1)/(2))^(3)}` `={(n(n+1))/(2)}^(2)-16{(1)/(2)((n-1)/(2))((n-1)/(2))((n-1)/(2)+1)}^(2)` `={(n(n+1))/(2)}^(2)-(1)/(4){(n+1)(n-1)}^(2)` `=((n+1)^(2))/(4){n^(2)-(n-1)^(2)}=((2n-1)(n+1)^(2))/(4)` |
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| 147. |
If a, b, c, d, e, f are in А.Р., then e-c is equal toA. 2(c-a)B. 2(d-b)C. 2(f-d)D. 2(d-c) |
| Answer» Correct Answer - D | |
| 148. |
If `x and y` are positive real numbers and `m, n` are any positive integers, then `(x^n y^m)/((1+x^(2n))(1+y^(2m))) lt 1/4` |
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Answer» Correct Answer - False Using `AM ge GM`, `(1 + x^(2n))/(2) ge sqrt(1.x^(2n))` `rArr (1 + x^(2n))/(2) ge x^(n)` `rArr (x^(n))/(1 + x^(2n)) le (1)/(2)` `:. (x^(n).y^(m))/((1 + x^(2n)) (1 + y^(2m))) le (1)/(4)` Hence, it is false statement |
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| 149. |
Consecutive odd integers whose sum is `25^2-11^2` areA. `n=14`B. `n=16`C. first odd number is 23D. last odd number is 49 |
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Answer» Correct Answer - A::C::D Let n consecutive odd numbers be `2k+1,2k+3,2k+5,"....."2k+2n-1` According to question, sum of these n numbers `=(n)/(2)[2k+1+2k+2n-1]=n(2k+n)` `=n^(2)+2kn=(n+k)^(2)-k^(2)` Given that, `(n+k)^(2)-k^(2)=25^(2)-11^(2)` `implies n+k=25` and `k=11 " "impliesmn=14` and `k=11` So, first term `=2k+1=23` Last term `=2k+2n-1=22+28-1=22+27=49` |
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| 150. |
The sum of the products of 2n numbers `pm1,pm2,pm3, . . . . ,n` taking two at time isA. `-underset(r=1)overset(n)sumr`B. `underset(r=1)overset(n)sumr^(2)`C. `-underset(r=1)overset(n)sumr^(2)`D. none of these |
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Answer» Correct Answer - C Let S be the required sum. We know that `2underset(i)sumunderset(iltj" ")underset(j)sumx_(i)x_(j)=(underset(i=1)overset(n)sumx_(i))^(2)-underset(i=1)overset(n)sum x_(i)^(2)` `:." "2S=(1-1+2-2+3-3+ . . . .+n-n)^(2)-2underset(r=1)overset(n)sumr^(2)` `rArrS=-underset(r=1)overset(n)sumr^(2)` |
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