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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.
| 151. |
If the middle term amongst any odd number (n) consecutive terms of an A.P, is m, then their sum is(a) `2m^2n` (b)`(mn)/2` (c)`mn` (d)`mn^2`A. 2mmnB. mn/2C. mnD. `mn^(2)` |
| Answer» Correct Answer - C | |
| 152. |
In the sequence 1, 2, 2, 3, 3, 3, 4, 4,4,4,....., where n consecutive terms have the value n, the 150 term isA. 17B. 16C. 18D. none of these |
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Answer» Correct Answer - A Let `150^(th)` term be equal to n. Then, `1+2+3+ . . . .+(n-1)lt150lt1+2+3+ . . . .+n` `rArr" "(n(n-1))/(2)lt150lt(n(n+1))/(2)` `rArr" "n(n-1)lt300ltn(n+1)` `rArr" "n^(2)-n-300lt0andn^(2)+n-300gt0` `rArr" "((1-sqrt(1201))/(2)ltnlt(1+sqrt(1201))/(2))` `and,(nlt(-1-sqrt(1201))/(2)orgt(-1+sqrt(1201))/(2))` `rArr" "(-16.8ltnlt17.82)and(nlt-17.82or,ngt16.8)` `rArr" "n=17`. Hence, `150^(th)` term is equal to 17. |
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| 153. |
`sum_(r=1)^(n) r^(2)-sum_(r=1)^(n) sum_(r=1)^(n) ` is equal to |
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Answer» Correct Answer - C We have, `underset(r=1)overset(n)sumr^(2)-underset(r=1)overset(n)sumunderset(r=1)overset(m)sumr` `=underset(r=1)overset(n)sumr^(2)-underset(m=1)overset(n)sum(m(m+1))/(2)` `=underset(r=1)overset(n)sumr^(2)-(1)/(2)underset(m=1)overset(n)summ^(2)-(1)/(2)underset(m=1)overset(n)summ` `(1)/(2)underset(r=1)overset(n)sumr^(2)-(1)/(2)underset(r=1)overset(n)sumr=(1)/(2){underset(r=1)overset(n)sumr^(2)-underset(r=1)overset(n)sumr}` |
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| 154. |
In the sequence 1, 2, 2, 3, 3, 3, 4, 4,4,4,....., where n consecutive terms have the value n, the 150 term is |
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Answer» Let the 150th term = n Then, `1+2+3+"...."+(n-1)lt150lt1+2+3+"....."+n` `implies ((n-1)n)/(2)lt150lt(n(n+1))/(2)` `impliesn(n-1)lt300 lt n(n+1)` Taking first two members `impliesn(n-1)lt300 implies n^(2)-n- 300 lt0` `implies (n-(1)/(2))lt300+(1)/(4)` `implies 0 lt n lt (1)/(2)+(sqrt(1201))/(2)` `implies 0 lt n lt 17.8 " " "......"(i)` and taking last two members, `n(n=1)gt300` `implies (n+(1)/(2))^(2)gt300+(1)/(4)` `therefore " " ngt -(1)/(2)+(sqrt1201)/(2)` `implies " "ngt 16.8" " "....."(ii)` From Eqs. (i)and (ii), we get 16.8ltnlt17.8 `implies " "n=17 ` |
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| 155. |
If the sequence `1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ...`where `n` consecutive terms has value n then `1025^th` term isA. `2^(9)`B. `2^(10)`C. `2^(11)`D. `2^(8)` |
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Answer» Correct Answer - B Let the `1025^(th)` term be equal to `2^(n)`. Then, `1+2+4+8+ . . . . +2^(n-1)lt1025le1+2+4+8+ . . . . . +2^(n)` `rArr" "2^(n)-1lt1025le2^(n+1)-1` `rArr" "n=10` Hence, `1025^(th)` term is equal to `2^(10)`. |
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| 156. |
If the coefficient of x in the expansion of `prod_(r=1)^(110)(1+rx)` is `lambda(1+110)(1+10+10^(2))`, then the value of ` lambda` is |
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Answer» `:.sum_(r=1)^(110)(1+rx)=(1+x)(1+2x)(1+3x)"...."(1+110x)` `=1^(100)+(z+2x+3x+"...."+110x)1^(109)+"....."` So, coefficient of x in `sum_(r=1)^(110)(1+rx)=(1+2+3+"......"+110)=(110xx111)/(2)=55xx111=6105` Now, `lambda(1+10)(1+10+10^(2))=lambda(11)(111)` `implieslambda(11)(111)=6105 implies lambda=5`. |
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| 157. |
If `(10)^9+""2(11)^1(10)^8+""3(11)^2(10)^7+""ddot""+""10(11)^9=k(10)^9`, then k is equal to(1) `(121)/(10)`(2) `(441)/(100)`(3) 100(4) 110 |
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Answer» `(10)^9[1 + 2(11/10)^1 + 3 (11/10)^2 + ....+ (11/10)^9]` `k = 1 + 2*11/10 + 3*(11/10)^2 + ... + 100(11/10)^9` also,`11/10k = 11/10 + 2(11/10)^2 + .......... + 9(11/10)^9 + 10 (11/10)^2 ` subtracting eqn`(1)-(2):` `-k/10 = 1 + 11/10 + (11/10)^2 + ...... + (11/10))^9 - 10(11/10)^10` `S_n = (a(1-r^n))/(1-r)` `= 1 xx((11/10)^10 -1)/(11/10 - 1)` `-k/10 = 10 [(11/10)^10 - 1] - 10(11/10)^10` `= 10 xx(11/10)^10 - 10 -10(11/10)^10` `-k/10 = -10` `k=100` option 3 is correct |
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| 158. |
`a, b, c` are sides of a triangle and `a, b, c` are in GP If `log a- log 2b, log 2b-log 3c and log 3c-log a` are in AP then |
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Answer» a,b,c are sides of triangles `b^2=ac-(1)` A,B,C are in AP `2B=A+C` `2(log2b-log3C)=loga-log2b-logc-loga` `2log((2b)/(3c))=loga/(2b)+log(3c)/a` `log((2b)/(3c))^2=log(a/(2b)*(3c)/a)` `((2b)/(3c))^2=(a)/(2b)*(3c)/(a)` `8b^3=27c^3` `2b=3c-(2)` `b^2=ac` `b=3/2c` `9/4c^2=9c` `9c=4a` `c=4/9a` `2b=3c` `b=2/3a` |
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| 159. |
Let ?, b, c be in an AP and `a^2, b^2, c^2` be in GP. If a < b < c and `a + b+ c=3/2` then the value of a isA. `(1)/(2sqrt2)`B. `(1)/(2sqrt3)`C. `(1)/(2) - (1)/(sqrt3)`D. `(1)/(2) - (1)/(sqrt2)` |
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Answer» Correct Answer - D Since, a, b and c are in an AP Let `a = A -D, b = A, c = A + D` Given, `a + b + c = (3)/(2)` `rArr (A -D) + A + (A +D) = (3)/(2)` `rArr 3A = (3)/(2) rArr A = (1)/(2)` `:.` The number are `(1)/(2) -D, (1)/(2), (1)/(2) + D` Also, `((1)/(2) -D)^(2), (1)/(4), ((1)/(2) + D)^(2)` are in GP. `:. ((1)/(4))^(2) = ((1)/(2) - D)^(2) ((1)/(2) + D)^(2) rArr (1)/(16) = ((1)/(4) - D^(2))^(2)` `rArr (1)/(4) - D^(2) = +- (1)/(sqrt2)` `:. a = (1)/(2) +- (1)/(sqrt2)` so, out of the given values, `a = (1)/(2) - (1)/(sqrt2)` is the right choice |
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| 160. |
- If `log(5c/a),log((3b)/(5c))`and `log(a/(3b))`are in AP, where a, b, c are in GP, then a, b, c are the lengths ofsides of(A) an isosceles triangle(B) an equilateral triangle(D) none of these(C) a scalene triangleA. an isosceles triangleB. an equilateral triangleC. a scalene triangleD. none of these |
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Answer» Correct Answer - D It is given that `log((5c)/(a)),log((3b)/(5c))andlog((a)/(3b))` are in A.P., `rArr" "2log((3b)/(5c))=log((5c)/(a))+log((a)/(3b))` `rArr" "((3b)/(5c))^(2)=(5c)/(a)xx(a)/(3b)rArr3b=5c` Also, a,b,c are in G.P. `:." "b^(2)=acrArr((5c)/(3))^(2)=acrArr25c^(2)=9acrArr25c=9a` `:." "(9a)/(5)=5c=3b` `rArr" "(1)/(5//9)=(b)/(1//3)=(c)/(1//5)` `rArr" "(a)/(5)=(b)/(3)=(c)/(9//5)=lamda("say")rArra=5lamda,b=3lamdaandc=(9)/(5)lamda` We observe that `b+clta`. Thus, a,b,c cannot form the sides of a triangle. |
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| 161. |
If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle isA. `3:4:5`B. `4:5:6`C. `5:6:7`D. `7:8:9` |
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Answer» Correct Answer - B Let the sides of the triangle be a-d,a and a+d, with `agtdgt0`. Clearly, a-d is the smallest and a+d is the largest side. So, A is the smallest angle and C is the largest angle. It is given that C=2A. Thus, the angles of the triangle are A,2A and `pi-3A`. Applying the law of sines, we obtain `(a-d)/(sinA)=(a)/(sin(pi-3A))=(a+d)/(sin2A)` `rArr" "(a-d)/(sinA)=(a)/(sin3A)=(a+d)/(sin2A)` `rArr" "(a-d)/(sinA)=(a)/(3sinA-4sin^(3)A)=(a+d)/(2sinAcosA)` `rArr" "(a-d)/(1)=(a)/(3-4sin^(2)A)=(a+d)/(2cosA)` `rArr" "3-4sin^(2)A=(a)/(a-d)and2cosA=(a+d)/(a-d)` `rArr" "4cos^(2)A-1=(a)/(a-d)and2cosA=(a+d)/(a-d)` `rArr" "((a+d)/(a-d))^(2)-1=(a)/(a-d)rArra=5d`. Thus, the sides of the triangle are a-d,a,a+d i.e. 4d,5d,6d. Hence, the ratio of the sides of the triangle is `4d:5d:6d" i.e. "4:5:6`. |
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| 162. |
If n is a root of `x^(2)(1-ac)-x(a^(2)+c^(2))-(1+ac)=0` and if n harmonic means are inserted between a and c, find the difference between the first and the last means. |
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Answer» Let `H_(1),H_(2),H_(3),"....",H_(n),` are n harmonic means, then `a,H_(1),H_(2),H_(3),"....",H_(n),b` are in HP. `:.(1)/(a),(1)/(H_(1)),(1)/(H_(2)),(1)/(H_(3)),"....",(1)/(H_(n)),(1)/(b)` are in AP. If d be the common difference, then `(1)/(c )=(1)/(a)+(n+2-1)d` `:. d=((a-c))/(ac(n+1))" " "......(i)"` `implies (1)/(h_(1))=(1)/(a)+d` and `(1)/(h_(n))=(1)/(c)-d` `:.h_(1)-h_(n)=(a)/(1+ad)-(c)/(1-cd)=(a)/(1+(a-c)/(a(n+1)))-(c)/(1-(a-c)/(a(n+1)))` `=(ac(n+1))/(an+a)-(ac(n+1))/(an+c)=ac(n+1)((1)/(cn+a)-(1)/(an+c))` `=ac(n+1)((an+c-cn-a)/(acn^(2)+(a^(2)+c^(2))n+ac))` `=(ac(a-c)(n^(2)-1))/(acn^(2)+(a^(2)+c^(2))n+ac)" " "....(ii)"` But given n is a root of `x^(2)(1-ac)-x(a^(2)+c^(2))-(1+ac)=0` Then, `n^(2)(1-ac)-n(a^(2)+c^(2))-(1+ac)=0` `acn^(2)+(a^(2)+c^(2))n+ac=n^(2)-1` then from Eq. (ii), `h_(1)-h_(1)=(ac+(a-c)(n^(2)-1))/((n^(2)-1))=ac(a-c)` |
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| 163. |
The natural number `a` for which `sum_(k=1,n) f(a+k)=16(2^n-1)` where the function f satisfies the relation `f(x+y)=f(x).f(y)` for all natural numbers x,y and further `f(1)=2` is:-A) 2B) 3C) 1D) none of these |
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Answer» Given, `f(x+y)=f(x)f(y)" " "....(i)"` and `f(1)=2" " "……(ii)"` On putting `x=y=1` in Eq. (i), we get `f(1+1)=f(1)f(1)=2*2` `:.f(2)=2^(2) " " "…..(iii)"` Now, on putting `x=1,y=2` in Eq. (i), we get `f(1+2)=f(1)f(2)=2*2^(2)" " [" from Eqs. (ii) and (iii) "]` `:. f(3)=2^(3)` On, putting `x=1,y=2` in Eq. (i), we get `f(2+2)=f(2)f(2)=2^(2)*2^(2) " " [" from Eq. (iii) "]` `:. f(4)=2^(4)` `vdots" " vdots " " vdots " "` Similarly, `f(lambda)=2^(lambda),lambda in N` `:.f(a+k)=2^(a+k),a+k inN` `:.sum_(k=1)^(n)f(a+k)=16(2^(n)-1) implies sum_(k=1)^(n)2^(a+k)=16(2^(n)-1)` `implies 2^(n)sum_(k=1)^(n)2^(k)=16(2^(n)-1)` `implies 2^(n)(2^(1)+2^(2)+2^(3)+"...."2^(n))=16(2^(n)-1)` `implies 2^(n)*(2(2^(n)-1))/((2-1))=16(2^(n)-1)` `implies 2^(a+1)=16=2^(4)` `impliesa+1=4` `:. a=3`. |
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| 164. |
If the sides of a triangle are in G.P., and its largest angle is twicethe smallest, then the common ratio `r`satisfies the inequality`0A. `0ltrltsqrt(2)`B. `1ltrltsqrt(2)`C. `1ltrlt2`D. none of these |
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Answer» Correct Answer - B Let the sides of the triangle be `a//r`, a and ar, with `agt0gtandrgt1`. Let `alpha` be the smallest angle, so that the largest angle is `2alpha`. Then, `alpha` is opposite to the side `a//r and2alpha` is opposite to the side ar. Applying sine rule, we get `(a//r)/(sinalpha)=(ar)/(sin2alpha)` `rArr" "(sin2alpha)/(sinalpha)=r^(2)` `rArr2cosalpha=r^(2)` `rArr" "r^(2)lt2" "[becausealpha!=:.2cosalphalt2]` `rArr" "rltsqrt(2)` Hence, `1ltrltsqrt(2)` |
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| 165. |
Consider a series `1/2+1/(2^2)+2/(2^3)+3/(2^4)+5/(2^5)+.............+(lambdan)/(2^n).` If `S_n` denotes its sum to `n` tems, then `S_n` cannot beA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A::B::C::D `:. S_(n)=(1)/(2)+(1)/(2^(2))+(2)/(2^(3))+(3)/(2^(4))+(5)/(2^(5))+"...."+(lambdan)/(2^(n))` ` =(3)/(4)+(1)/(4)((1)/(2)+(1)/(2^(2))+(2)/(2^(3))+(3)/(2^(4))+(5)/(2^(5))+"...."+(lambdan)/(2^(n))) +(1)/(2)((1)/(2)+(1)/(2^(2))+(2)/(2^(3))+"...."+(lambdan)/(2^(n))) -(1)/(4)-(lambda_(n))/(2^(n+2))-(lambda_(n))/(2^(n+1))` `implies S_(n)=(3)/(4)+(1)/(4)S_(n)+(1)/(2)S_(n)-(1)/(4)-(lambda_(n))/(2^(n+2))-(lambda_(n))/(2^(n+1))` `implies =(1)/(4)S_(n)=(1)/(2)-(lambda_(n))/(2^(n+2))-(lambda_(n))/(2^(n+1))implies S_(n)=2-(lambda_(n))/(2^(n+2))-(lambda_(n))/(2^(n-1))lt2` |
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| 166. |
Three successive terms of a G.P. will form the sides of a triangle if the common ratio r satisfies the inequalityA. `(sqrt(3)-1)/(2)ltrlt(sqrt(3)+1)/(2)`B. `(sqrt(5)-1)/(2)ltrlt(sqrt(5)+1)/(2)`C. `(sqrt(2)-1)/(2)ltrlt(sqrt(2)+1)/(2)`D. none of these |
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Answer» Correct Answer - B Let the lengths of the sides of the triangle be `a,ar,ar^(2)`. We have the following three cases : CASE I When r=1 In this case, the lengths of sides of the triangle are a,a,a i.e. the triangle is equilateral. CASE II When `rgt1` In this case, the length of the largest side is `ar^(2)`. Therefore, the triangle will be formed, if `a+argtar^(2)` `rArr" "r^(2)-r-1lt0` `rArr" "(1-sqrt(5))/(2)ltrlt(1+sqrt(5))/(2)` `rArr" "rlt(1+sqrt(5))/(2)" "[becausergt1]` . . ..(i) CASE III When `rlt1` In this case, the length of the largest side is a. So, the triangle will be formed, if `ar+ar^(2)gta` `rArr" "r^(2)+r-1gt0` `rArr" "rlt(-1-sqrt(5))/(2)or,rgt(-1+sqrt(5))/(2)` `rArr" "(sqrt(5)-1)/(2)ltrlt1" "[becauserlt1]` . . . (ii) From (i) and (ii), we obtain : `(sqrt(5)-1)/(2)ltrlt(sqrt(5)+1)/(2)` |
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| 167. |
If three successive terms of as G.P. with commonratio `rgt1` form the sides of a triangle and [r] denotes the integral part of x the `[r]+[-r]=` (A) 0 (B) 1 (C) -1 (D) none of these |
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Answer» Correct Answer - C Let the sides of a triangle be `a,ar,ar^(2)`. Since, `rgt1`. Therefore, `ar^(2)` is the largest side. `:." "a+argtar^(2)` `rArr" "ar^(2)-a-arlt0` `rArr" "r^(2)-r=1lt0` `rArr" "(1-sqrt(5))/(2)ltrlt(1+sqrt(5))/(2)` `rArr" "1ltrlt(1+sqrt(5))/(2)rArr[r]=1" "[becausergt1]` Also, `-(1+sqrt(5))/(2)lt-rlt-1rArr[-r]=-2` Hence, [r]+[-r]=1-2=-1. |
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| 168. |
If `a_(1),a_(2),a_(3)(a_(1)gt0)` are three successive terms of a GP with common ratio r, the value of r for which `a_(3)gt4a_(2)-3a_(1)` holds is given byA. `1ltrlt3`B. `-3ltrlt-1`C. `rlt1` or `rgt3`D. None of these |
| Answer» Correct Answer - B | |
| 169. |
in a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals-A. `(1)/(2)(1-sqrt(5))`B. `(1)/(2)sqrt(5)`C. `sqrt(5)`D. `(1)/(2)(sqrt(5)-1)` |
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Answer» Correct Answer - D Let geometric progression is `a,ar,ar^(2),"…."," " [a,rgt0]` `:.a=ar+ar^(2)` `implies r^(2)+r+1=0 implies r=(-1pmsqrt(5))/(2)` `:.r=(sqrt(5)-1)/(2)`. |
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| 170. |
If pth,qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a(q-r)+b(r-p)+c(p-q)=0A. 1B. `-1`C. 0D. `(1)/(2)` |
| Answer» Correct Answer - C | |
| 171. |
If `S`denotes the sum to infinity and `S_n`the sum of `n`terms of the series `1+1/2+1/4+1/8+ ,`such that `S-S_n |
| Answer» Correct Answer - D | |
| 172. |
If `a, b, c` are in `GP`, then the equations `ax^2 +2bx+c = 0` and `dx^2 +2ex+f =0` have a common root if `d/a , e/b , f/c `are inA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - A Given that a,b,c are GP. Then, `b^(2)=ac " " "…..(i)"` and equations `ax^(2)+2ex+c=0` and `dx^(2)+2ex+f=0 " have a common root. ""………..(A)"` Now, `ax^(2)+2bx+c=0` `implies ax^(2)+2sqrt(ac)" " x+c=0 " " [" by Eq. (i) "]` `implies (sqrt(ax)+sqrt(c ))^(2)=0 " " implies sqrt(ax)+sqrt(c )=0` `implies x=-(sqrt(c ))/(sqrt(a)) " " [" repeated "]` By the condition `(A), (-(sqrt(c ))/(sqrt(a)))` be the root of `dx^(2)-2ex+f=0` So, it saftisfy the equation `d(-sqrt(c/(a)))^(2)+2e(-sqrt(c/(a)))+f=0` ` implies (dc)/(a)-2esqrt(c/(a))+f=0 " " implies (d)/(a)-(2e)/(sqrt(ac))+(f)/(c )=0` ` implies (d)/(a)-(2e)/(b)+(f)/(c )=0 " " implies (d)/(a)+(f)/(c )=2((e)/(b))` So, `(d)/(a),(e)/(b),(f)/(c )` are in AP. |
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| 173. |
The sum of first two terms of an infinite G.P. is 1 and every terms is twice the sum of the successive terms. Its first terms isA. `1//3`B. `2//3`C. `3//4`D. `1//4` |
| Answer» Correct Answer - C | |
| 174. |
The natural numbers arearranged innthe form given below The rth group containing `2^(r-1)` numbers. Prove that sum of the numbers in the nth group is `2^(n-2)[2^(n)+2^(n+1)-1]`. |
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Answer» Let 1st term of the r th goup be `T_(r)` and the 1st terms of successive rows are `1,2,4,8,"…..",` respectively. `T_(r)=1*2^(r-1)=2^(r-1)` Hence, the sum of the numbers in te rth group is `=(2^(r-1))/(2){2*2^(r-1)+(2^(r-1)-1)*1}` `[:. " number of terms in rth group is " 2^(r-1)]` `=2^(r-2){2^(r)+2^(r-1)-1}` Hence, sim of the numbers in the nth group is `2^(r-2)[2^(n)+2^(n-1)-1]`. |
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| 175. |
Let `S_n` denotes the sum of the terms of n series `(1lt=nlt=9)` `1+22+333+.....999999999`, isA. `S_(n)-S_(n-1)=(1)/(9)(10^(n)-n^(2)+n)`B. `S_(n)=(1)/(9)(10^(n)-n^(2)+2n-2)`C. `9(S_(n)-S_(n-1))=n(10^(n)-1)`D. None of these |
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Answer» Correct Answer - C `:. S_(n)=1+22+333+"..."+ubrace(" nnn""...."n)_(" n terms ")` `:. S_(n)-S_(n-1)=ubrace(" nnn""...."n)_(" n times ")=ubrace(111"...."1)_(" n times ")` `=n(10^(n-1)+10^(n-2)+"….."+10+1)=(n(10^(n)-1))/(10-1)` `:. 9(S_(n)-S_(n-1))=n(10^(n)-1)` |
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| 176. |
If x,y,z are in AP and `tan^(-1)x,tan^(-1)y,tan^(-1)z` are also in AP, thenA. `2x=3y=6z`B. `6x=3y=2z`C. `6x=4y=3z`D. `x=y=z` |
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Answer» Correct Answer - D `:.` x,y,z are in AP. Let `x=y-d,z=y+d " " "…….(i)"` Also, given `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP. `:.2tan^(-1)y=tan^(-1)x+tan^(-1)z` `implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))` `implies (2y)/(1-y^(2))=(x+z)/(1-xz) implies (2y)/(1-y^(2))=(2y)/(1-(y^(2)-d^(2)))` `implies y^(2)=y^(2)-d^(2)" " [" fromEq.(i) "]` `:. d=0` From Eq. (i), x=y and z=y `:. x=y=z` Aliter `:. x,y,z" are in AP. "" " "…….(i)"` `:. 2y=x+z" " "........(ii)"` Also, `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP. `:.2tan^(-1)y=tan^(-1)x+tan^(-1)z` `implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))` `implies (2y)/(1-y^(2))=(x+z)/(1-xz)=(2y)/(1-xz)" "[" from Eq.(ii) "]` `implies y^(2)=zx` `:. x,y,z " are in GP. "" "".........(iii)"` From Eqs. (i) and (ii) x,y,z are in AP and also in GP, then x=y=z. |
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| 177. |
Find the sum of the product of every pair of the first `n` of natural number |
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Answer» We find that `S=1*2+1*3+1*4+"...."+2*3+2*4+"......."+3*4+3*5+"....."(n-1)*n"......(i)"` `:.[1+2+3+"...."+(n-1)+n]^(2)=1^(2)+2^(2)+3^(2)+"...."+(n-1)^(2)+n^(2) +2[1*2+1*3+1*4+"...."+2*3+2*4+"...."+3*4+3*5+"......."+(n-1)*n]` `(sumn)^(2)=sumn^(2)+25" " [" from Eq. (i) "]` ` implies S=((sumn)^(2)=sumn^(2))/(2)` `({(n(n+1))/(2)}^(2)-(n(n+1)(2n+1))/(6))/(2)` `((n^(2)(n+1)^(2))/(4)-(n(n+1)(2n+1))/(6))/(2)` ,`(n(n+1))/(24)[3n(n+1)-2(2n+1)]` `(n(n+1)(3n^(2)-n+2))/(24)` Hence, `S=((n-1)n(n+1)(3n+2))/(24)` |
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| 178. |
Solve the following equaions for x and y `log_(10)x+(1)/(2)log_(10)x+(1)/(4)log_(10)x+"...."=y " and " (1+3+5+"...."+(2y-1))/(4+7+10+"...."+(3y-1))=(20)/(7log_(10)x)`. |
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Answer» From the first equation ` log_(10)x{1+(1)/(2)+(1)/(4)+"…."+oo}=y` `implies log_(10)x{(1)/(1-(1)/(2))}=y` `implies 2log_(10)x=y" " "….(i)"` From the second equation `(1+3+5+"...."+(2y-1))/(4+7+10+"...."+(3y-1))=(20)/(7log_(10)x)` `implies ((y)/(2)(1+2y-1))/((y)/(2)(4+3y+1))=(20)/(7log_(10)x)` `implies (2y)/(3y+5)=(20)/(7log_(10)x)` `implies 7y(2log_(10)x)=60y+100` `implies 7y(y)=60y+100 " " [" from Eq. (i) "]` `implies 7y^(2)-60y-100=0` `:. (y-10)(7y+10)=0` `:. y=10,yne (-10)/(7)` [because y being the number of terms in series `implies y in N`] From Eq. (i), we have `2log_(10)x=10 implies log_(10)x=5` `:.x=10^(5)` Hence, required solution s `x=10(5),y=10`. |
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| 179. |
The pth, `(2p)th` and `(4p)th` terms of an AP, are in GP, then find the common ratio of GP. |
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Answer» Let `T_(n)=An+B` ` :. T_(p)=Ap+B,` `T_(2p)=2Ap+B,T_(4p)=4Ap+B` `:.T_(p),T_(2p)T_(4p)` are in GP. `:.(2Ap+B)^(2)=(Ap+B)(4Ap+B)` `implies ABp=0` `:.B=0,A ne 0,p ne 0` ` implies` Common ratio, `r=(T_(2p))/(T_(p))=(2Ap+0)(Ap+0)=2`. |
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| 180. |
The sides of a right angled triangle arein `A.P.`, then they are in the ratioA. `2:3:4`B. `3:4:5`C. `4:5:6`D. none of these |
| Answer» Correct Answer - B | |
| 181. |
If `l_(n)=int_(0)^((pi)/(4)) tan^(n) xdx` show that `(1)/(l_(2)+l_(4)),(1)/(l_(3)+l_(5)),(1)/(l_(4)+l_(6)),(1)/(l_(5)+l_(7)),"...."` from an AP. Find its common difference. |
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Answer» We have, `l_(n)+l_(n+2)=int_(0)^((pi)/(4)) (tan^(n)x+tan^(n+2)x) dx` `=int_(0)^((pi)/(4))tan^(n)x(1+tan^(2)x) dx` `=int_(0)^((pi)/(4))tan^(n)x*sec^(2)xdx=[(tan^(n+1)x)/(n+1)]_(0)^((pi)/(4))=(1)/(n+1)` Hence, `(1)/(l_(n)+l_(n+2))=n+1` On putting `n=2,3,4,5"...."` `:.(1)/(l_(2)+l_(4))=3,(1)/(l_(3)+l_(5))=4,(1)/(l_(4)+l_(6))=5,(1)/(l_(5)+l_(7))=6,"...."` Hence, `(1)/(l_(2)+l_(4)),(1)/(l_(3)+l_(5)),(1)/(l_(4)+l_(6)),(1)/(l_(5)+l_(7)),"...."` are in AP with common difference1. |
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| 182. |
Statement 1 The difference between the sum of the first 100 even natural numbers and the sum of the first 100 odd natural numbers is 100. Statement 2 The difference between the sum opf the first n even natural numbers and sum of the first n odd natural numbers is n.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 Sum of first 100 even natural numbers `E_(1)=2+4+"…"+200=(2(100xx101))/(2)=10100` Sum of 100 odd natural numbers `=1+3+"….."+199` `O=(100)/(2)(1+199)=10000` `:.E-O=100` So, Statement 1 is true. Statement 2 Sum of first n natural even numbers `E=2+4+"....."+2n=(2n(n+1))/(2)=n^(2)+n` Sum of first n odd natural numbers `O=1+3+"....."+(2n-1)` `=(n)/(2)[1=2n-1]=n^(2)` So, `E-O=n^(2)=n-n^(2)=n` Statement 2 is true and correct explanation for Statement 1. |
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| 183. |
Let `S_1, S_2, `be squares such that for each `ngeq1,`the length of a side of `S_n`equals the length of a diagonal of `S_(n+1)dot`If the length of a side of `S_1i s10c m ,`then for which of the following value of `n`is the area of `S_n`less than 1 sq. cm?a. 5 b. 7 c. 9 d. 10A. 7B. 8C. 5D. 6 |
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Answer» Correct Answer - B We have, Length of a side of `S_(n)` = Length of a diagonal of `S_(n+1)` `rArr" "` Length of a side of `S_(n)=sqrt(2)` Length of a side of `S_(n+1)` `rArr" "("Length of a side of "S_(n+1))/("Length of a side of "S_(n))(1)/(sqrt(2))"for all "ngtl`. `rArr" "` Sides of `S_(1),S_(2) , . . . .S_(n)` form a G.P. with common ratio `(1)/(sqrt(2))` and first term 10. `:.` Length of the side of `S_(n)=10((1)/(sqrt(2)))^(n-1)=(10)/((n-1)/(2^(2)))` `rArr" Area of "S_(n)=("side")^(2)=((10)/((n-1)/(2^(2))))^(2)=(100)/(2^(n-1))` Now, Area of `S_(n)lt1` `rArr" "(100)/(2^(n-1))lt1rArrn-1ge7rArrnge8` |
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| 184. |
If `A^(x)=G^(y)=H^(z)`, where `A,G,H` are AM,GM and HM between two given quantities, then prove that `x,y,z` are in HP. |
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Answer» Let `A^(x)=G^(y)=H^(z)=k` Then, `A=k^((1)/(x)),G=k^((1)/(y)),H=k^((1)/(z))` ` therefore G^(2)=AH implies (k^((1)/(y)))^(2)=k^((1)/(x))*k^((1)/(z))` ` implies k^((2)/(y))=k^((1)/(x)+(1)/(z))implies (2)/(y)=(1)/(x)+(1)/(z)implies (1)/(x),(1)/(y),(1)/(z)` are in AP. hence, `x,y,z`are in HP. |
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| 185. |
Let `a_1, a_2,...,a_(10)`be in A.P. and `h_1, h_2,..., h_(10)`be in H.P. If `a_1=h_1=2` and `a_(10)=h_(10)=3,` then `a_4h_7`isA. 2B. 3C. 5D. 6 |
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Answer» Correct Answer - D Let d be the common difference of the A.P. Then, `a_(10)=3rArra_(1)+9d=3rArr2+9d=3rArrd=(1)/(9)`. `:." "a_(4)=a_(1)+3d=2+(1)/(3)=(7)/(3)` Let D be the common difference of the A.P. `(1)/(h_(1)),(1)/(h_(2)), . . . .,(1)/(h_(10))`. Then, `h_(10)=3` `rArr" "(1)/(h_(10))=(1)/(3)` `rArr" "(1)/(h_(1))+9D=(1)/(3)rArr(1)/(2)+9D=(1)/(3)rArr9D=-(1)/(6)rArrD=-(1)/(54)` `:." "(1)/(h_(7))=(1)/(h_(1))+6D=(1)/(2)-(1)/(9)=(7)/(18)rArrh_(7)=(18)/(7)` Hence, `a_(4)h_(7)=(7)/(3)xx(18)/(7)=6`. |
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| 186. |
If `I_(n)=int_(0)^(pi)(1-sin2nx)/(1-cos2x)dx` then `I_(1),I_(2),I_(3),"….."` are inA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - A `:. I_(n)=int_(0)^(pi)(1-sin2nx)/(1-cos2x)dx " " implies I_(n)=int_(0)^(pi)(1-sin2nx)/(2sin^(2)x)dx` `implies I_(n+1)+I_(n-1)-2 I_(n) [ 1-sin2 (n+1)x+1-sin 2(n-1)x-2]` `=(1)/(2)int_(0)^(pi)(+sin2nx " "]/(sin^(2)x)dx` `=(1)/(2)int_(0)^(pi)([sin2nx-sin 2 (n+1)x]+[sin2nx-sin 2 (n-1)x])/(sin^(2)x)dx` `=(1)/(2)int_(0)^(pi)(-2cos(2n+1)xsin(x)+2cos(2n-1)xsinx)/(sin^(2)x)dx` `=int_(0)^(pi)(sinx[cos(2n-1)x-cos(2n+1)x])/(sin^(2)x)dx` `=int_(0)^(pi)(2sin2nxsinx)/(sinx)dx=2int_(0)^(pi)sin2nxdx=(2)/(2n)[-cos2nx]_(0)^(pi)` `=-(1)/(n)(1-1)=0` `:. I_(n+1)+I_(n-1)=2I_(n)` `implies I_(n-1)+I_(n),I_(n+1)` are in AP. `:.I_(1),I_(2),I_(3)"...."` are in AP. |
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| 187. |
Statement 1 If a and b be two positive numbers, where `agtb` and `4xxGM=5xxHM` for the numbers. Then, `a=4b`. Statement 2 `(AM)(HM)=(GM)^(2)` is true for positive numbers.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 - C `:.A=(a+b)/(2),G=sqrt(ab) " and " H=(2ab)/(a+b)` Given, `4G=5H" " "……(i)"` and `G^(2)=AH` `:. H=(G^(2)))/(A)" " "……(ii)"` From Eqs. (i) and (ii), we get `4G=(5G^(2))/(A) implies 4A=5G` `implies 2(a+b)=5sqrt(ab)` `implies4(a^(2)+b^(2)+2ab)=25ab` `implies 4a^(2)-17ab+4b^(2)=0` `implies (a-4b)(4a-b)=0` `a=4b,4a-b ne 0" " [:.agtb]` `:.` Statement 1 is true. Statement 2 is true only for two numbers, if numbers more than two, then this formula`(AM)(HM)=(GM)^(2)` is true, if numbers are in GP. Statement 2 is false for positive numbers. |
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| 188. |
Statement 1 If a,b,c are three positive numbers in GP, then `((a+b+c)/(3))((3abc)/(ab+bc+ca))=(abc)^((2)/(3))`. Statement 2 `(AM)(HM)=(GM)^(2)` is true for positive numbers.A. Statement 1 is true, Statement 2 is true, Statement 2 is a corrct explanation for Statement 1.B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation 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 - C If a,b, be two real positive and unequal numbers, then `AM=(a+b)/(2),GM=sqrt(ab)` and`HM(2ab)/(a+b)` `:.(AM)(HM)=(GM)^(2)` This result will be true for n numbers, if they are in GP. Hence, Statement 1 is true, Statement 2 is false. |
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| 189. |
Let `a_1, a_2, ,a_(10)`be in A.P. and `h_1, h_2, h_(10)`be in H.P. If `a_1=h_1=2a n da_(10)=h_(10)=3,t h e na_4h_7`is`2`b. `3`c. `5`d. `6`A. 2B. 3C. 5D. 6 |
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Answer» Correct Answer - D Since, `a_(1), a_(2), a_(3), ..., a_(10)` are in AP Now, `a_(10) = a_(1) + 9d` `rArr 3 = 2 + 9d` `rArr d = 1//9 and a_(4) = a_(1) + 3d` `rArr a_(4) = 2 +3 (1//9) = 2 + 1//3 = 7//3` Also, `h_(1), h_(2), h_(3),...,h_(10)` are in HP `rArr (1)/(h_(1)), (1)/(h_(2)), (1)/(h_(3)) ,..., (1)/(h_(10))`are in AP Given, `h_(1) - 2, h_(10) -3` `:. (1)/(h_(10)) = (1)/(h_(1)) + 9d_(1) rArr (1)/(3) = (1)/(2) + 9d_(1)` `rArr - (1)/(6) = 9d_(1)` `rArr d_(1) = -(1)/(54) and (1)/(h_(7)) = (1)/(h_(1)) + 6d_(1)` `rArr (1)/(h_(7)) = (1)/(2) + (6xx1)/(-54)` `rArr (1)/(h_(7)) = (1)/(2) - (1)/(9) rArr h_(7) = (18)/(7)` `:. a_(4) h_(7) = (7)/(3) xx (18)/(7) = 6` |
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| 190. |
The AM of teo given positive numbers is 2. If the larger number is increased by 1, the GM of the numbers becomes equal to the AM to the given numbers. Then, the HM of the given numbers isA. `(3)/(2)`B. `(2)/(3)`C. `(1)/(2)`D. 2 |
| Answer» Correct Answer - B | |
| 191. |
If `H_1. H_2...., H_n` are n harmonic means between a and b`(!=a)`, then the value of `(H_1+a)/(H_1-a)+(H_n+b)/(H_n-b)`= |
| Answer» Correct Answer - C | |
| 192. |
If `H_1. H_2...., H_n` are n harmonic means between a and b`(!=a)`, then the value of `(H_1+a)/(H_1-a)+(H_n+b)/(H_n-b)`=A. nB. `n+1`C. 2nD. `2n-2` |
| Answer» Correct Answer - B | |
| 193. |
If m is the A.M. of two distinct real numbers `l and n(l, n>1) and G_1, G_2 and G_3`, are three geometric means between `I and n`, then `G_1^4+2G_2^4+G_3^4` equals-A. `4lmn^(2)`B. `4l^(2)m^(2)n^(2)`C. `4l^(2)mn`D. `4lm^(2)n` |
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Answer» Correct Answer - D We have, `m=(l+n)/(2)` Let be the common ratio of the G.P. l, `G_(1),G_(2),G_(3),n`. Then `n=lr^(4)rArrr=((n)/(l))^(1//4)` `:." "G_(!)=lr=l^(3//4)n^(1//4),G_(2)=lr^(2)=l^(2//4)n^(2//4),G_(3)=lr^(3)=l^(1//4)n^(3//4)`. `rArr" "G_(1).^(4)+2G_(2).^(4)+G_(3).^(4)=l^(3)n+2l^(2)n^(2)+l n^(3)= ln(l+n)^(2)=l n(2m)^(2)=4lm^(2)n`. |
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| 194. |
if `m` is the AM of two distinct real number `l and n ` and `G_1,G_2 and G_3` are three geometric means between `l` and `n` , then ` (G_1^4 + G_2^4 +G_3^4 )` equals toA. `4l^(2)mn`B. `4lm^(2)n`C. `lmn^(2)`D. `l^(2) m^(2) n^(2)` |
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Answer» Correct Answer - B Given, m is the AM of l and n ltbr. `l + n = 2m` ..(i) and `G_(1), G_(2), G_(3)` are geometric means between l and n `l, G_(1), G_(2), G_(3), n` are in GP Let r be the common ratio of this GP `:. G_(1) = lr, G_(2) = lr^(2), G_(3) = lr^(3), n - lr^(4) rArr r = ((n)/(l))^((1)/(4))` Now, `G_(1)^(4) + 2G_(2)^(4) + G_(3)^(4) = (lr)^(4) + 2(lr^(2))^(4) + (lr^(3))^(4)` `= l^(4) xx r^(4) (1 + 2r^(4) + r^(6)) = l^(4) xx r^(4) (r^(4) + 1)^(2)` `= l^(4) xx (n)/(l) ((n +l)/(l))^(2) = ln xx 4m^(2) = 4l m^(2) n` |
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| 195. |
A squence of positive terms `A_(1),A_(2),A_(3),"....,"A_(n)` satisfirs the relation `A_(n+1)=(3(1+A_(n)))/((3+A_(n)))`. Least integeral value of `A_(1)` for which the sequence is decreasing can be |
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Answer» `:.A_(n+1)=(3(1+A_(n)))/((3+A_(n)))" For "n=1, A_(2)=(3(1+A_(1)))/((3+A_(1)))` For `n=2,A_(3)=(3(1+A_(2)))/((3+A_(2)))` `=(3(1+(3(1+A_(1)))/((3+A_(1)))))/(3+(3(1+A_(1)))/((3+A_(1))))=(6+4A_(1))/(4+2A_(1))=(3+2A_(1))/(2+A_(1))` `:.` Given, sequence ccan be written as `A_(1),(3(1+A_(1)))/((3+A_(1))),((3+2A_(1)))/((2+A_(1)))"...."` Given,`A_(1)gt0` and sequence is decreasing, then `A_(1)gt(3(1+A_(1)))/((3+A_(1))),((3+A_(1)))/((3+A_(1)))gt((3+2A_(1)))/((2+A_(1)))` `implies A_(1)^(2)gt3 " or "A_(1)gtsqrt(3)` `:. A_(1)=2 " " [" least integral value of " A_(1)]` |
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| 196. |
When the ninth term of an AP is divided by its second term we get 5 as the quotient, when the thirteenth term ia devided ny sixth term the quotient is 2 and the remainderis 5, then the seonnd term is |
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Answer» Let a be the first term ad d be the common difference, then`T_(9)=5T_(2)` `implies (a+8d)=5(a+d)` `:. 4a=3d" " "….(i)"` and `T_(13)=T_(6)xx2+5` `implies a+12d=2(a+5d)+5` `implies 2d=a+5" " "….(ii)"` From Eqs.(i) and (ii), we get `a=3` and `d=4` `:.T_(2)=a+d=7` |
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| 197. |
The sum of the series `(1+2)+(1+2+2^(2))+(1+2+2^(2)+2^(3))+ . . .` up to n terms, isA. `2^(n+2)-n-4`B. `2(2^(n)-1)-n`C. `2^(n+1)-n`D. `2^(n+1)-1` |
| Answer» Correct Answer - A | |
| 198. |
If a,b,c are in H.P., then the value of `((1)/(b)+(1)/(c)-(1)/(a))((1)/(c)+(1)/(a)-(1)/(b))` isA. `(2)/(bc)-(1)/(b^(2))`B. `(1)/(4)((3)/(c^(2))+(2)/(ca)-(1)/(a^(2)))`C. `((2)/(b^(2))-(2)/(ab))`D. all of these |
| Answer» Correct Answer - D | |
| 199. |
The 5th term of the series `(10)/(9),(1)/(3)sqrt((20)/(3)),(2)/(3),…` isA. `(1)/(3)`B. 1C. `(2)/(5)`D. `sqrt((2)/(3))` |
| Answer» Correct Answer - C | |
| 200. |
The sum of `i-2-3i+4`up to 100 terms, where `i=sqrt(-1)`is`50(1-i)`b. `25 i`c. `25(1+i)`d. `100(1-i)`A. 50(1-i)B. 25 iC. 25(1+i)D. 100 (1-i) |
| Answer» Correct Answer - A | |