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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Let `S_(n)` denote the sum of n terms of the series `1^(2)+3xx2^(2)+3^(2)+3xx4^(2)+5^(2)+3xx6^(2)+7^(2)+ . . . .` Statement -1: If n is odd, then `S_(n)=(n(n+1)(4n-1))/(6)` Statement -2: If n is even, then `S_(n)=(n(n+1)(4n+5))/(6)`A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.C. Statement -1 is true, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
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Answer» Correct Answer - A Let n be even and let n=2m. Then, `S_(n)=1^(2)+3xx2^(2)+3^(2)+3xx4^(2)+5^(2)+3xx6^(2)+ . . . +(2m-1)^(2)+3xx(2m)^(2)` `rArr" "S_(n)={1^(2)+3^(2)+5^(2)+ . . . .+(2m-1)^(2)}+3{2^(2)+4^(2)+6^(2)+ . . . +(2m)^(2)}` `rArr" "s_(n)=underset(r=1)overset(2m)sumr^(2)+2underset(r=1)overset(m)sum(2r)^(2)` `rArr" "S_(n)=(2m(2m+1)(4m+1))/(6)+(8m(m+1)(2m+1))/(6)` `rArr" "S_(n)(2m(2m+1))/(6){(4m+1)+4(m+1)}` `rArr" "S_(n)=(1)/(6)(2m)(2m+1)(4(2m)+5)=(n(n+1)(4n+5))/(6)` So, statement -2 is true. If n is odd, then `S_(n)=1^(2)+3xx2^(2)+3^(2)+3xx4^(2)+ . . . +3xx(n-1)^(2)+n^(2)` `rArr" "S_(n)=((n-1)(n-1+1)(4(n-1)+5))/(6)+n^(2)` Replacing n by n-1 in statement-2. `rArr" "S_(n)(n(n-1)(4n+1))/(6)+n^(2)` `rArr" "S_(n)(n)/(6){(n-1)(4n+1)+6n}` `rArr" "S_(n)=(n)/(6)(4n^(2)+3n-1)=(n(n+1)(4n-1))/(6)` So, statement -1 is true and statement -2 is a correct explanation for statement -1. |
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| 2. |
The odd value of n for which `704 + 1/2 (704) + 1/4 (704) + ... upto n terms = 1984 - 1/2 (1984) + 1/4 ( 1984) - .... upto n terms is :A. 5B. 3C. 4D. 10 |
| Answer» Correct Answer - A | |
| 3. |
The positive integer `n`for which `2xx2^2xx+3xx2^3+4xx2^4++nxx2^n=2^(n+10)`is`510`b. `511`c. `512`d. 513A. 510B. 512C. 513D. 508 |
| Answer» Correct Answer - C | |
| 4. |
If the first two terms of a H.P. are `2//5and12//23` respectively. Then, largest term isA. 5th termB. 6th termC. 4th termD. 6th term |
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Answer» Correct Answer - A Let the H.P. be `(1)/(a),(1)/(a+d),(1)/(a+2d),(1)/(a+3d), . . .` Then , ,`(1)/(a)=(2)/(5)and(1)/(a+b)=(12)/(23)rArra=(5)/(2)andd=-(7)/(12)` Now, `n^(th)` term of the H.P. `=(1)/(a+(n-1)d)=(12)/(37-7n)` So,, the nth term is largest when 37-7n has the least value. Clearly, `(12)/(37-7n)` is least for n=5. Hence, 5th term is the largest term. |
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| 5. |
consider an infinite geometric series with first term `a` and comman ratio `r` if the sum is `4` and the second term is `3/4` then find `a&r`A. `a = 4//7, r= 3//7`B. `a = 2, r = 3//8`C. `a = 3//2, r = 1//2`D. `a = 3, r =1//4` |
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Answer» Correct Answer - D Since, sum = 4 and second term `= (3)/(4)` It is given first term a and common ratio r `rArr (alpha)/(1-r) = 4, alphar = (3)/(4)` `rArr r = (3)/(4alpha)` `rArr (alpha)/(1 - (3)/(4 alpha)) = 4` `rArr (4 alpha^(2))/(4alpha -3) = 4` `rArr (alpha -1) (alpha -3) = 0` `rArr alpha = 1 or 3` When `alpha = 1, r = 3//4` and when `alpha = 3, r = 1//4` |
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| 6. |
If `1^2+2^2+3^2++2003^2=(2003)(4007)(334)a n d(1)(2003)+(2)(2002)+(3)(2001)++(2003)(1)=(2003)(334)(x),t h e nx`equals`2005`b. `2004`c. `2003`d. 2001A. 2005B. 2004C. 2003D. 2001 |
| Answer» Correct Answer - A | |
| 7. |
Find the sum of the series `1+2^(2)x+3^(2)x^(2)+4^(2)x^(3)+"...."" upto "infty|x|lt1`. |
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Answer» Here, the numbers `1^(2),2^(2),3^(2),4^(2),"...."i.e.1,4,9,16"...."` are not in AP but ` 1,4-1=3,9-4=5,16-9=7,"...."` are in AP. Let `S_(infty)=1+2^(2)x+3^(2)x^(2)+4^(2)x^(3)+"...."" upto "infty` `=1+4x+9x^(3)+16x^(3)+"...."" upto "infty "......(i)"` Multiplying both sides of Eq. (i) by x, we get `S_(infty)=x+4x^(2)+9x^(3)+16x^(4)+"...."" upto "infty "......(ii)"` Subtracting Eq.(ii) from Eq.(i), we get `(1-x)S_(infty)=1+3x+5x^(2)+7x^(3)+"...."" upto "infty "......(iii)"` Again, multiplying both sides of Eq. (iii) by x, we get `x(1-x)S_(infty)=x+3x^(2)+5x^(3)+7x^(4)+"...."" upto "infty "......(iv)"` Subtracting Eq. (iv) from Eq. (iii),we get `(1-x)(1-x)S_(infty)=1+2x+2x^(2)+2x^(3)+"...."" upto "infty` `=1+2(x+x^(2)+x^(3)+"...."" upto "infty)` `=1+2((x)/(1-x))=((1+x)/(1-x))` ` therefore S_(infty)=((1+x))/((1-x)^(3))` |
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| 8. |
If the two terms of a H.P. are `2//5and12//23` respectively, then the largest term isA. 6B. 12C. 5D. 7 |
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Answer» Correct Answer - A Proceeding as in Illustration 1, we have Largest term `=(12)/(37-7xx5)=6` |
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| 9. |
Let a,b,c be in A.P. and `|a|lt1,|b|lt1|c|lt1.ifx=1+a+a^(2)+ . . . ."to "oo,y=1+b+b^(2)+ . . . ."to "ooand,z=1+c+c^(2)+ . . . "to "oo`, then x,y,z are inA. APB. GPC. HPD. none of these |
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Answer» Correct Answer - C We have, `x=(1)/(1-a),y=(1)/(1-b),z=(1)/(1-c)` Now, a,b,c are in A.P. `:." "1-a,1-b,1-c` are in A.P. `:." "(1)/(1-a),(1)/(1-b),(1)/(1-c)` are in H.P. `rArrx,y,z` are in H.P. |
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| 10. |
An infinite GP has first term `x`and sum 5, then `x`belongs to(2004, 1M)`x |
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Answer» Correct Answer - C We know that, the sum of infinite terms of GP is `S_(oo) = {((alpha)/(1 -r)","|r| lt1),(oo"," |r| ge 1):}` `:. S_(oo) = (x)/(1 -r) = 5 " " [|r| lt 1]` or `1 - r = (x)/(5)` `rArr r = (5 -x)/(5)` exists only when `|r| lt1` i.e. `-1 lt (5 -x)/(5) lt 1` or `10 lt - x lt 0` `rArr 0 lt x lt 10` |
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| 11. |
If f is a function satisfying `f(x+y)=f(x)f(y)`for all `x ,y in X`such that `f(1)=3`and `sum_(x=1)^nf(x)=120`, find the value of n. |
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Answer» We know, `a^(m+n) = a^ma^n` It means, `f(x)` can be `a^x` to satisfy the given condition. Then, `f(x+y) = f(x)f(y)`. `:. f(x) = a^x` We are given, `f(1) = 3 => a^1 = 3=> a = 3` `f(2) = 3^2 = 9` `f(3) = 3^3 = 27` So, the series will be, `3,9,27...` It will form a GP with `a = 3 and r = 3` Now, Sum of first n terms in a GP, `S_n = (a(r^n -1)/(r-1))` We are given, `S_n = 120` `:. (3(3^n -1))/(3-1) = 120=>3^n-1 = 80=> 3^n = 81` So, value of `n` is `4`.`=>3^n = 3^4=> n = 4` |
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| 12. |
If `|alpha|lt1,|beta|lt1` `1-alpha+alpha^(2)-alpha^(3)+ . . . . ."to "oo=s_(1)` `1-beta+beta^(2)-beta^(3)+ . . . ."to "oo=s_(2)`, then `1-alphabeta+alpha^(2)beta^(2)+ . . . . ."to "oo` equalsA. `s_(1)s_(2)`B. `(s_(1)s_(2))/(1+s_(1)s_(2))`C. `(s_(1)s_(2))/(1-s_(1)-s_(2)+2s_(1)s_(2))`D. `(1)/(1+s_(1)s_(2))` |
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Answer» Correct Answer - C We have, `1-alpha+alpha^(2)-alpha^(3)+ . . . . ."to "oo=s_(1)` `and,1-beta+beta^(2)-beta^(3)+ . . . ."to "oo=s_(2)` `rArr" "(1)/(a+alpha)=s_(1)and(1)/(1+beta)=s_(2)` `rArr" "alpha=(1)/(s_(1))-1andbeta=(1)/(s_(2))-1` `:." "1-alphabeta+alpha^(2)beta^(2)-a^(3)beta^(3)+ . . ."to "oo` `=(1)/(1+alphabeta)+(1)/(1+((1)/s_(1)-1)((1)/(s_(2))-1))=(s_(1)s_(2))/(1-(s_(1)+s_(2))+2s_(1)s_(2))` |
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| 13. |
The 10th common term between the series 3+7+11+ . . . And 1+6+11+ . . . ., isA. 191B. 193C. 211D. none of these |
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Answer» Correct Answer - A Given series are 3+7+11+15+19+23+27+31+ . . . . . . and, 1+6+11+16+21+26+31+ . . . . . . As illustration 8. The common differences of the two series are 4 and 5 respectively. Therefore, the series of common terms will form an A.P. with common difference 20 and first term 11. Hence, 10th common term `=11+(10-1)xx20=191`. |
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| 14. |
An infinite G.P has first term x and sum 5 then x belongs to ?A. `xlt-10`B. `-10ltxlt0`C. `0ltxlt10`D. `xgt0` |
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Answer» Correct Answer - C Let r be the common ratio of the G.P. Then, `-1ltrlt1`. Now, Sum=5 `rArr" "(x)/(1-r)=5` `rArr" "1-r=(x)/(5)` `rArr" "r=1-(x)/(5)` `rArr" "-1lt1-(x)/(5)lt" "[because-1ltrlt1]` `rArr" "-2lt-(x)/(5)lt0rArr2gt(x)/(5)gt0rArr10gtxgt0rArrltxlt10` |
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| 15. |
If f(x) is a function satisfying f(x+y)=f(x)f(y) for all x,y `in` N such that f(1)=3 and `sum_(x=1)^(n) f(x)=120`. Then, the value of n isA. 4B. 5C. 6D. none of these |
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Answer» Correct Answer - A We have, `f(x+y)=f(x)f(y)" for all "x,yinN`. `:." "f(x)={f(1)}^(x)=3^(x)" "[becausef(1)=3]` Now, `underset(x=1)overset(n)sumf(x)=120` `rArr" "underset(x=1)overset(n)sum3^(x)=120` `rArr" "(3(3^(n)-1))/((3-1))=120rArr3^(n)-1=80rArr3^(n)=3^(4)rArrn=4` |
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| 16. |
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of swrvice will be Rs. 11040 afterA. 19 monthsB. 20 monthsC. 21 monthsD. 18 months |
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Answer» Correct Answer - C Let the time taken to save Rs. 11040 be `(n+3)` months. For first 3 months, he saves Rs. 200 each month. `:. " In " (n+3)" month "` `3xx200+(n)/(2)[2(240)+(n-1)x40]=11040` `implies 600+(n)/(2)[40(12+n-1)]=11040` `implies 600+20n(n+11)=11040` `implies n^(2)+11n-522=0` `implies (n-18)(n+29)=0` `:. n=18, " neglecting " n=-29` `:. " Total time " =n+3=21 " months "`. |
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| 17. |
If `a_(1),a_(2),a_(3),"......"` be in harmonic progression with `a_(1)=5` and `a_(20)=25`. The least positive integer n for which `a_(n)lt0` isA. 22B. 23C. 24D. 25 |
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Answer» Correct Answer - D `:.a_(1),a_(2),a_(3),"......."` are in HP. `:.(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)),"......."` Let D be the common difference of this AP, then `(1)/(a_(20))=(1)/(a_(1))+(20-1)D` `implies D((1)/(25)-(1)/(5))/(19)=-(4)/(25xx19)` and `(1)/(a_(n))=(1)/(a_(1))+(n-1)D=(1)/(5)-(4(n-1))/(25xx19)=((95-4n+4)/(25xx19))` `=((99-4n)/(25xx19))lt0" " [:.a_(n)lt0]` `implies 99-4nlt0 implies ngt24.75`. |
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| 18. |
Let `a_(n)` be the nth term of an AP, if `sum_(r=1)^(100)a_(2r)=alpha " and "sum_(r=1)^(100)a_(2r-1)=beta`, then the common difference of the AP isA. `(alpha-beta)/(200)`B. `alpha-beta`C. `(alpha-beta)/(100)`D. `beta-alpha` |
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Answer» Correct Answer - C Given, `a_(2)+a_(4)+a_(6)+"......."+a_(200)=alpha" " "....(i)"` and `a_(1)+a_(3)+a_(5)+"......."+a_(199)=beta" " "....(ii)"` On subtracting Eq. (ii) from Eq. (i), we get `(a_(2)-a_(1))+(a_(4)-a_(3))+(a_(6)-a_(5))+"......."+(a_(200)-a_(199))=alpha-beta` `implies d+d+d+"....."+d=alpha-beta implies 100d=alpha-beta` `:. d=((alpha-beta))/(100)`. |
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| 19. |
Let `a_(n)` be the `n^(th)` term of the G.P. of positive numbers. Let `sum_(n=1)^(100) a_(2n)=alpha and sum_(n=1)^(100) a_(2n-1)=beta`, such that `a!=beta`, then the common ratio isA. `alpha//beta`B. `beta//alpha`C. `sqrt(alpha//beta)`D. `sqrt(beta//alpha)` |
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Answer» Correct Answer - A Let a be the first term and r be the common ratio of the given G.P. Then, `alpha=underset(n=1)overset(100)suma_(2n)=a_(2)+a_(4)+ . . . +a_(200)=ar^(3)+ . . . .ar^(199)` `rArr" "alpha=ar(1+r^(2)+r^(4)+ . . .+r^(198))` `and,beta=underset(n=1)overset(100)suma_(2n-1)=a_(1)+a_(3)+ . . .+a_(199)=a+ar^(2)+ . . .ar^(198)` `rArr" "beta=a(1+r^(2)+ . . .+r^(198))` Clearly, `alpha//beta=r`. |
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| 20. |
Let `a_(n)` be the nth term an A.P. if `sum_(r=1)^(100) a_(2r)=alpha and sum_(r=1)^(100)a_(2r-1)=beta`, them the common difference of the A.P., isA. `(alpha-beta)/(100)`B. `beta-alpha`C. `(alpha-beta)/(200)`D. `alpha-beta` |
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Answer» Correct Answer - A We have, `alpha=underset(r=1)overset(100)suma_(2r)andbetaunderset(r=1)overset(100)suma_(2r-1)` `:." "alpha-beta=underset(r=1)sum100(a_(2r)-a_(2r-1))` `rArr" "alpha-beta=100d`, where d is the common difference `rArr" "d=(alpha-beta)/(100)` |
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| 21. |
Let `a_(n)` be the nth term of an AP, if `sum_(r=1)^(100)a_(2r)=alpha " and "sum_(r=1)^(100)a_(2r-1)=beta`, then the common difference of the AP isA. `alpha-beta`B. `beta-alpha`C. `(alpha-beta)/(2)`D. None of these |
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Answer» Correct Answer - D Given that, `sum_(r=1)^(100)a_(2r)=alpha` ` implies a_(2)+a_(4)+"....."+a_(200)=alpha" " "....(i)"` and `sum_(r=1)^(100)a_(2r-1)=beta` `implies a_(1)+a_(3)+"......"+a_(199)=beta " " "………(ii)"` On subtracting Eq.(ii) from Eq.(i), we get ` (a_(2)-a_(1))+(a_(4)-a_(3))+"......."+(a_(200)-a_(199))=alpha -beta` `d+d+"......"` upto 100 terms `=alpha -beta` [beacause `a_(n)` be the nth term of AP with common difference d] `100d=alpha-beta` `d=(alpha-beta)/(100)`. |
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| 22. |
Consider the pattern shown below: `{:(" Row ",1,1,,,),(" Row ",2,3,5,,),(" Row ",3,7,9,11,),(" Row ",4,13,15,17,19):}etc.` The number at the end of row 60 isA. 3659B. 3519C. 3681D. 3731 |
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Answer» Correct Answer - A If last term of nth row is `T_(n)`, then Let `S=1+5+11+19+"...."T_(n)` `(S=ul1+ul5+ul11+ul19+"....+"ulT_(n-1)+ulT_(n))/(0=1+4+6+8"....."n" terms "-T_(n))` `T_(n)=1+2(2+3+4+"....."(n-1)" terms" )` ,brgt `=1+2((n-1))/(2)[2*2+(n-2)*1]` `=1+(n+1)(n+2)` `=1+n^(2)+n-2` `implies T_(n)= n^(2)+n-1` `:. T_(60)=(60)^(2)+60-1=3600+59=3659`. |
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| 23. |
If `S_(1),S_(2)andS_(3)` denote the sum of first `n_(1)n_(2)andn_(3)` terms respectively of an A.P., then `(S_(1))/(n_(1))(n_(2)-n_(3))+(S_(2))/(n_(2))+(n_(3)-n_(1))+(S_(3))/(n_(3))(n_(1)-n_(2))`= |
| Answer» Correct Answer - A | |
| 24. |
If `sum_(r=1)^(oo)(1)/((2r-1)^(2))=(pi^(2))/(8)`, then `sum_(r=1)^(oo) (1)/(r^(2))` is equal toA. `(pi^(2))/(24)`B. `(pi^(2))/(3)`C. `(pi^(2))/(6)`D. none of these |
| Answer» Correct Answer - C | |
| 25. |
If `|a| < 1 and |b| < 1,` then the sum of the series `a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+.....oo` isA. `(a)/(1-a)+(ab)/(1-ab)`B. `(a^(2))/(1-a^(2))+(ab)/(1-ab)`C. `(b)/(1-b)+(a)/(1-a)`D. `(b^(2))/(1-b^(2))+(ab)/(1-ab)` |
| Answer» Correct Answer - B | |
| 26. |
If` 1/1^2+1/2^2+1/3^2....oo=pi^2/6` then `1/1^2+1/3^2+1/5^2....=`A. `pi^(2)//8`B. `pi^(2)//12`C. `pi^(2)//3`D. `pi^(2)//2` |
| Answer» Correct Answer - A | |
| 27. |
`a ,b ,c`are positive real numbers forming a G.P. ILf `a x 62+2b x+c=0a n ddx^2+2e x+f=0`have a common root, then prove that `d//a ,e//b ,f//c`are in A.P.A. A.P.B. G.PC. H.P.D. none of these |
| Answer» Correct Answer - A | |
| 28. |
the value of `[(0.16)^(log_0.25 (1/3+1/3^2+1/3^3+....................+oo))]^(1/2)` isA. 2B. 3C. 4D. 1 |
| Answer» Correct Answer - C | |
| 29. |
If `x,|x+1|,|x-1|` are first three terms of an A.P., then the sum of its first 20 terms isA. 360, 180B. 350, 180C. 150, 100D. 180, 150 |
| Answer» Correct Answer - B | |
| 30. |
(i) 1,3,5,7,… (ii) `pi,pi+e^(pi),pi+2e^(pi),"...."` (iii) `a, a-b, a-2b,a-3b,"…."` |
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Answer» (i) Here, 2nd term - 1st term = 3rd term - 2nd term `="…."` `implies 3-1=5-3="..."=2,` which is a common difference. (ii) Here, 2nd term - 1st term = 3rd term -2nd term` ="…."` `implies (pi+e^(pi))-pi=(pi+2e^(pi))-(pi+e^(pi))="...."` `" "=e^(pi),` which is a common difference. (iii) Here, 2n term -1st term = 3rd term -2nd term ="...." `implies (a-b)-a=(a-2b)-(a-b)="...."` `" "=-b,` which is a common difference. |
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| 31. |
If the sum of the first n terms of series be `5n^(2)+2n`, then its second term isA. `(56)/(15)`B. `(27)/(14)`C. 17D. 16 |
| Answer» Correct Answer - C | |
| 32. |
For the series, `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`+...7th term is 167th term is 18Sum of first 10 terms is `(505)/4`Sum of first 10 terms is `(45)/4`A. 7th term is 16B. 7th term is 18C. 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)+"......."` `T_(n)=(1)/(1+3+5+7+"........"" n terms ")*(1+2+3+"......."" n terms " )^(2)` `=(1)/([(n)/(2)[2*1+(n-1)*2]])*((n(n+1))/(2))^(2)=((n+1)^(2))/(4)` (a) `T_(7)=((7+1)^(2))/(4)=(64)/(4)=16` (b) `S_(10)sum_(n=1)^(10)((n+1)/(2))^(2)=(1)/(4)sum_(n=1)^(10)(n^(2)+2n+1)` `=(1)/(4)(sum_(n=1)^(10)n^(2)+2sum_(n=1)^(10)n+sum_(n=1)^(10)1)` `=(1)/(4)((10xx11xx21)/(6)+(2xx10xx11)/(2)+10)` `=(1)/(4)(385+110+10)=(505)/(4)` |
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| 33. |
(i) Write `sum_(r=1)^(n)(r^(2)+2)` in expanded form. (ii) Write the series `(1)/(3)+(2)/(4)+(3)/(5)+(4)/(6)+"…"+(n)/(n+2)` in sigma form. |
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Answer» (i) On putting `r=1,2,3,4,"....","n in"(r^(2)+2),` we get `3,6,11,18,"…",` `(n^(2)+2)` Hence,`sum_(r=1)^(n)(r^(2)+2)=3+6+11+18+"...."+(n^(2)+2)` (ii) The rth terms of series `=(r)/(r+2).` Hence, the give series can be written as `(1)/(3)+(2)/(4)+(3)/(5)+(4)/(6)+"...."+(n)/(n+2)=sum_(r=1)^(n)((r)/(r+2))` |
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| 34. |
Let `a_1, a_2, a_3, ...a_(n)` be an AP. then: `1 / (a_1 a_n) + 1 / (a_2 a_(n-1)) + 1 /(a_3a_(n-2))+......+ 1 /(a_(n) a_1) = `A. 2B. `a_(1)_+a_(n)`C. `2(a_(1)+a_(n))`D. `(2)/(a_(1)+a_(n))` |
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Answer» Correct Answer - D We know that the sum of the terms equidistant from the beginning and the end an A.P. is always same i.e. `a_(1)+a_(n)=a_(2)+a_(n-1)=a_(3)+a_(n-2)= . . .=a_(k)+a_(n-(k+1))= . . .` `:." "(a_(1)+a_(n)){(1)/(a_(1)a_(n))+(1)/(a_(2)a_(n-1))+(1)/(a_(3)a_(n-1))+ . . . +(1)/(a_(n)a_(1))}` `=(a_(1)+a_(n))/(a_(1)a_(n))+(a_(1)+a_(n))/(a_(2)a_(n-1))+(a_(1)+a_(n))/(a_(3)a_(n-1))+ . . .+(a_(1)+a_(n))/(a_(n)a_(1))` `=(a_(1)+a_(n))/(a_(1)a_(n))+(a_(2)+a_(n-1))/(a_(2)a_(n-1))+(a_(3)+a_(n-2))/(a_(3)a_(n-2))+ . . . (a_(1)+a_(n))/(a_(1)a_(n))` `=((1)/(a_(n))+(1)/(a_(1)))+((1)/(a_(n-1))+(1)/(a_(2)))+((1)/(a_(n-2))+(1)/(a_(3)))+. . . +((1)/(a_(n-1))+(1)/(a_(2)))+((1)/a_(n)+(1)/(a_(1)))` `=2{(1)/(a_(1))+(1)/(a_(2))+(1)/(a_(3))+ . . .(1)/(a_(n))}` `:." "(1)/(a_(1)a_(n))+(1)/(a_(2)a_(n-1))+(1)/(a_(3)a_(n-2))+ . . . +(1)/(a_(n)a_(1))` `=(2)/(a_(1)+a_(n)){(1)/(a_(1))+(1)/(a_(2))+ . . .+(1)/(a_(n))}` `:." "lamda=(2)/(a_(1)+a_(n))` |
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| 35. |
If `f:NtoR,` where `f(n)=a_(n)=(n)/((2n+1)^(2))` write the sequence in ordered pair from. |
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Answer» Here, `a_(n)=(n)/((2n+1)^(2))` On putting `n=1,2,3,4,"….."` successively, we get `a_(1)=(1)/((2*1+1)^(2))=(1)/(9), a_(2)=(2)/(2*2+1)^(2)=(2)/(25)` `a_(3)=(3)/(2*3+1)^(2)=(3)/(49),a_(4)=(4)/(2*4+1)^2=(4)/(81)` `" "vdots " "vdots" "vdots` Hence, we obtain the sequence `(1)/(9),(2)/(25),(3)/(49),(4)/(81),"...."` Now, the sequence in ordered pair form is `{(1"",(1)/(9)),(2"",(20)/(25)),(3"",(3)/(49)),(4"",(4)/(81)),"..."}` |
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| 36. |
If the sum of n terms of a series is`2n^(2)+5n` for all values of n, find its 7th term. |
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Answer» Given, `S_(n)=2n^(2)+5n` `implies S_(n-1)=2(n-1)^(2)+5(n-1)=2n^(2)+n-3` `therefore" "T_(n)=S_(n)-S_(n-1)=(2n^(2)+5n)-(2n^(2)+n-3)=4n+3` Hence, `T_(7)=4xx7+3=31` |
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| 37. |
The Fibonacci sequence is defined by `1=a_1=a_2( and a)_n=a_(n-1)+a_(n-2),n >2`. Find `(a_(n+1))/(a_n),`for n = 1, 2, 3, 4, 5. |
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Answer» `because " "a_(1)=1=a_(2)` `therefore " "a_(3)=a_(2)+a_(1)=1+1=2,` `a_(4)=a_(3)+a_(2)=2+1=3` `a_(5)=a_(4)+a_(3)=3+2=5` and `a_(6)=a_(5)+a_(4)=5+3=8` `therefore" "(a_(2))/(a_(1))=1,(a_3)/(a_2)=(2)/(1)=2,(a_4)/(a_3)=(3)/(2),(a_5)/(a_4)=(5)/(3)`and `(a_6)/(a_5)=(8)/(5)` |
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| 38. |
If a,b,c are non-zero real numbers, then the minimum value of the expression `((a^(8)+4a^(4)+1)(b^(4)+3b^(2)+1)(c^(2)+2c+2))/(a^(4)b^(2))` equalsA. 12B. 24C. 30D. 60 |
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Answer» Correct Answer - C Let `P=((a^(8)+4a^(4)+1)(b^(4)+3b^(2)+1)(c^(2)+2c+2))/(a^(4)b^(2))` `(a^(4)+4+(1)/(a^(4)))(b^(2)+3+(1)/(b^(2))){(c+1)^(2)+1}` `:.a^(4)+4+(1)/(a^(4))ge6,b^(2)+3+(1)/(b^(2))ge5 " and "(c+1)^(2)+1ge 1` `[:. x+(1)/(x)ge2 " for "xgt0]` `:.P ge 6*5*1=30 implies Pge30` Hence, the required minimum value is 30. |
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| 39. |
Let a,b,c be positive real numbers in H.P. Statement -1: `(a+b)/(2a-b)+(c+b)/(2c-b)ge4` Statement-2: `(a)/(b)+(b)/(c)+(c)/(a)ge3`A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.C. Statement -1 is true, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
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Answer» Correct Answer - B Since a,b,c are in H.P. Therefore, `b=(2ac)/(a+c)`. Now, `(a+b)/(2a-b)+(c+b)/(2c-b)=(a+(2ac)/(a+c))/(2a-(2ac)/(a+c))+(c+(2ac)/(a+c))/(2c-(2ac)/(a+c))` `=(1+3c)/(2a)+(c+3a)/(2c)` `=(1)/(2)+(3c)/(2a)+(1)/(2)+(3a)/(2c)` `=1+(3)/(2)((1)/(c)+(c)/(a))` `ge1+3=4" "[AMgtGM,(a)/(c)+(c)/(a)ge2]` So, statement-1 is true. Using `AMgeGm`, we obtain `(a^(2)c+b^(2)a+c^(2)b)/(3)ge(a^(2)cxxb^(2)axxc^(2)b)^(1//3)` `rArr" "a^(2)c+b^(2)a+c^(2)bge3abcrArr(a)/(b)+(b)/(c)+(c)/(a)ge3` So, statement -2 is true. But, statement -2 is not a correct explanation for statement -1. |
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| 40. |
In the quadratic equation `ax^2 + bx + c = 0`, if `Delta = b^2-4ac and alpha + beta, alpha^2 + beta^2, alpha^3 + beta^3` are in GP. where `alpha, beta` are the roots of `ax^2 + bx + c =0`, then |
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Answer» `ax^2 + bx+c = 0` `alpha and beta` are roots `alpha + beta = -b/a` `alpha*beta= c/a` given, `/_ = b^2- 4ac` & `alpha+ beta , alpha^2+beta^2, alpha^3+ beta^3` are in GP. :. the condition will be `(alpha^2 + beta^2)^2 = (alpha+ beta)*(alpha^3+ beta^3)` solving further: `[(alpha+beta)^2 - 2alpha beta]^2 = (alpha+beta)*[(alpha+beta)^3 - 3alpha beta (alpha+ beta)]` `[b^2/a^2 - 2c/a]^2 = (-b/a)[-b^3/a^3 +3*c/a*b/a]` `[(b^2-2ac)/a^2]^2 = (-b/a)[(-b^3+3abc)/a^3]` `[(b^2-2ac)/a^2] = b^2/a^4[b^2 - 3ac]` `(b^2 - 2ac)^2/cancel(a^4) = b^2/cancel(a^4)[b^2 -3ac]` `cancel(b^4) + 4a^2c^2 - 4b^2ac= cancel(b^4) - 3b^2ac` `4a^2c^2 - b^2ac = 0` `-ac(b^2 - 4ac) = 0` `-ac. /_ = 0` so` /_ = 0` `c/_ = 0` option (c) is correct |
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| 41. |
If for a G.P., a=8 and `t_(4)`=64 then : r=A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - B | |
| 42. |
If three positive real numbers a,b,c are in AP such that abc=4, then the minimum value of b is |
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Answer» `therefore a,b,c " are in AP."` Let `a=A-D, b=A, c=A+D` Then, `a=b-D, c=b+D` Now, `abc=4` `(b-D)b(b+D)=4` `implies b(b^(2)-D^(2))=4` `implies b^(2)-D^(2)ltb^(2)` `implies b(b^(2)-D^(2)) lt b^(3) implies 4ltb^(3)` `therefore " " bgt (4)^((1)/(3))" or " bgt (2)^((2)/(3))` Hence, the minimum value of b is `(2)^((2)/(3))` |
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| 43. |
If `a, b, c `are positive real numbers such that `ab^2c^3= 64` then minimum value of `1/a+2/b+3/c` is equal to |
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Answer» Given `ab^2c^3=64` now we have to find minimum value of `1/a+2/b+3/c`. this can be done by the concept fo` AM>=GM`, where AM and GM are the arithmetic and geometric mean. Let there be 6 terms that are`1/a,1/b,1/b,1/c,1/c,1/c`, applying `AM>=GM` on them. we get, `(1/a+2/b+3/c)/6>=(1/(ab^2c^3))^(1/6)` `(1/a+2/b+3/c)/6>=3` hence minimum value is 3. |
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| 44. |
If `y z+z x+x y=12 ,w h e r ex ,y ,z`are positive values, find the greatest value of `x y zdot` |
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Answer» Given, `yz+zx+xy=12` (constant), the value of `(yz)(zx)(xy)` is greatest when `yz=zx=xy` Hence, `n=3` and `k=12` Hence, greatest value of `(yz)(zx)(xy)` is `((12)/(3))^(3)` i.e.64. `:.` Greatest value of `x^(2)y^(2)z^(2)` is 64. Thus, greatest value of xyz is 8. Aliter Given `yz+zx+xy=12`, the greatest value of (yz)(zx)(xy) is greatest when `yz=zx=xy=c " " [" say "]` Since, `yz+zx+xy=12` `:. c+c+c=12` `implies 3c=12` r `c=4` `:. yz=zx=xy=4` Hence, greatest value of (yz)(zx)(xy) is `4*4*4` i.e. greatest value of `x^(2)y^(2)z^(2)` is 64. Hence, greatest value of xyz is 8. |
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| 45. |
For a real number `x, [x]` denotes greatest integer function, then find value of `[1/2]+[1/2+1/100]+[1/2+2/100]+....+[1/2+99/100]`A. `-153`B. `-133`C. `-131`D. `-135` |
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Answer» Correct Answer - B Given series is `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ... ... +[-(1)/(3) - (99)/(100)]` [where, [x] denotes the greatest integer `le x`] Now, `[-(1)/(3)], [-(1)/(3) - (1)/(100)] , [-(1)/(3)-(2)/(100)],... + [-(1)/(3) - (66)/(100)]` all the term have value `-1` and `[-(1)/(3) - (67)/(100)], [-(1)/(3) -(68)/(100)], ..., [-(1)/(3) - (99)/(100)]` all the term have value `-2` So, `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ... + [-(1)/(3) - (66)/(100)] ` `= -1 -1 -1-1...67` times. `= (-1) xx 67 = -67` and `[-(1)/(3) - (67)/(100)] + [-(1)/(3) - (68)/(100)]+..+ [-(1)/(3) - (99)/(100)]` `= -2 -2-2 -2`...33 times `= (-2) xx 33 = -66` `:. [-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + ....+ [-(1)/(3) - (99)/(100)]` `= (-67) + (-66) = - 133` Alternate Solution `:. [-x] = -[x] -1, " if " x !in` Integer, and `[x] + [x + (1)/(n)] + [x + (2)/(n)] + ...+ [x + (n-1)/(n)] = [nx], n in N` So given series `[-(1)/(3)] + [-(1)/(3) - (1)/(100)] + [-(1)/(3) - (2)/(100)] + [-(1)/(3) - (2)/(100)] + ....+ [(-1)/(3) - (99)/(100)]` `= (-[(1)/(3)] -1) + (-[(1)/(3) + (1)/(100)] -1) + (-[(1)/(3) + (2)/(100)] -1) + ...+ (-[(1)/(3) + (99)/(100)] -1)` `= (-1) xx 100 - [(1)/(3) xx 100] = - 100 - 33 = - 133` |
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| 46. |
Statement 1 `a+b+c=18(a,b,cgt0)`, then the maximum value of abc is 216. Statement 2 Maximum value occurs when `a=b=c`.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 Statement 1 `a+b+c=18,a,b,cgt0` Applying `AM ge GM` for a,b,c `(a+b+c)/(3)ge root(3)sqrt(abc) implies root(3)sqrt(abc)le 6 implies abc le 216` Maximum value of abc is 216 which occurs at `a=b=c`. Statement 2 is the correct explanation for Statement 1. |
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| 47. |
If `(1-p)(1+3x+9x^2+27 x^3+81 x^4+243 x^5)=1-p^6p!=1`, then the value of `p/xi s``1/3`b. `3`c. `1/2`d. `2`A. `1//2`B. 2C. `1//4`D. 4 |
| Answer» Correct Answer - B | |
| 48. |
If `a+b+c=3` and `agt0,bgt0,cgt0` then the greatest value of `a^(2)b^(2)c^(2)` isA. `(3^(4)*2^(10))/(7^(7))`B. `(3^(10)*2^(4))/(7^(7))`C. `(3^(2)*2^(12))/(7^(7))`D. `(3^(12)*2^(2))/(7^(7))` |
| Answer» Correct Answer - C | |
| 49. |
If a, b, c are three positive real numbers such that `abc^(2)` has the greatest value `(1)/(64)`, thenA. `a=b=(1)/(2),c=(1)/(4)`B. `a=b=c=(1)/(3)`C. `a=b=(1)/(4),c=(1)/(2)`D. `a=b=c=(1)/(4)` |
| Answer» Correct Answer - A | |
| 50. |
The sum of the series `2/3+8/9+(26)/(27)+(80)/(81)+`to `n`terms is`n-1/2(3^(-n)-1)`(b) `n-1/2(1-3^(-n))`(c) `n+1/2(3^n-1)`(d) `n-1/2(3^n-1)`A. `n-(1)/(2)(3^(-n)-1)`B. `n-(1)/(2)(1-3^(-n))`C. `n+(1)/(2)(3^(n)-1)`D. `n-(1)/(2)(3^(n)-1)` |
| Answer» Correct Answer - A | |