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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.
| 251. |
If `X=1+a+a^(2)+a^(3)+"..."+infty " and " y=1+b+b^(2)+b^(3)+"..."+infty "` show that `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty=(xy)/(x+y-1), "where " 0ltalt1" and "0ltblt1`. |
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Answer» Given, `x=1+a+a^(2)+a^(3)+"..."+infty=(1)/(1-ab)` ` implies x-ax=1` ` therefore a=((x-1))/(x)" " "......(i)"` and `y=1+b+b^(2)+b^(3)+"..."+infty` Similarly, ` b=((y-1))/(y)" " "......(ii)"` Since, `0ltalt1,0ltblt1` `therefore 0ltablt1` Now, `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty=(1)/(1-ab)` `=(1)/(1-((x-1)/(x))((y-1)/(y)))" " ["from Eqs. (i) and (ii)"]` `=(xy)/(xy-xy+x+y-1)` Hence, `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty =(xy)/(x+y-1)` |
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| 252. |
Sum of `n`terms the series : `1^2-2^2+3^2-4^2+5^2-6^2+`A. `-(n(n+1))/(2)`B. `(n(n+1))/(2)`C. `-n(n+1)`D. none of these |
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Answer» Correct Answer - A `1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+ . . . . .+(n-1)^(2)-n^(2)` `=(1-2)(1+2)+(3-4)(3+4)+ . . . . . +(n-1-n)((n-1)+n)` `=-(1+2+3+4+ . . . .+n)=-(n(n+1))/(2)` |
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| 253. |
If `S_n = 1+1/2 + 1/2^2+...+1/2^(n-1) and 2-S_n < 1/100,` then the least value of `n` must be : |
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Answer» Given, `S_(n)=1+(1)/(2)+(1)/(2^(2))+"..."+(1)/(2^(n-1))=(1[*1-((1)/(2))^(n)])/((1-(1)/(2)))` ` implies S_(n)=2-(1)/(2^(n-1))` ` implies 2- S_(n)=(1)/(2^(n-1))lt(1)/(100)" "[therefore 2- S_(n)lt(1)/(100)]` `implies 2^(n-1)gt100gt2^(6)` `implies 2^(n-1)gt2^(6)` `therefore n-1gt6 implies ngt7` Hence, the least value of n is 8. |
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| 254. |
If AM of the number `5^(1+x) and 5^(1-x)` is 13 then the set of possible real values of x is -A. `5,(1)/(5)`B. {-1,1}C. {0,1}D. none of these |
| Answer» Correct Answer - B | |
| 255. |
Sum of `n`terms the series : `1^2-2^2+3^2-4^2+5^2-6^2+`A. `(n(n+1))/(2)`B. `(-n(n+1))/(2)`C. `(n(n-1))/(2)`D. `(-n(n-1))/(2)` |
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Answer» Correct Answer - A If n is odd, then required sum is given by `1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+ . . . .+{(n-2)^(2)-(n-1)^(2)}+n^(2)` `=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+ . . . .+{(n-2)-(n-1)}{(n-2)+(n-1)}+n^(2)` `=-{1+2+3+ . . . +(n-2)+(n-1)}+n^(2)` `=-(n(n-1))/(2)+n^(2)=(n(n+1))/(2)` |
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| 256. |
The coefficient of `x^(n-2)` in the polynomial `(x-1)(x-2)(x-3)...(x-n)` isA. `(1)/(24)n(n+1)(n-1)(3n+2)`B. `(1)/(24)n(n^(2)-1)(3n+2)`C. `(n(n+1)(2n+2))/(6)`D. none of these |
| Answer» Correct Answer - B | |
| 257. |
If `a,b,c` are in A.P then `a+1/(bc), b+1/(ca), c+1/(ab)` are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - A | |
| 258. |
The coefficient of `x^15` in the product of `(1-x)(1-2x)(1-2^2 x)(1- 2^3 x) (1-2^4 x) ......(1-2^15 x)`A. `2^(105)-2^(121)`B. `2^(121)-2^(105)`C. `2^(120)-2^(104)`D. none of these |
| Answer» Correct Answer - A | |
| 259. |
Sum of the series `1+2.2+3.2^2 +4.2^3+.....+100.2^99` isA. `99xx2^(100)`B. `99xx2^(100)+1`C. `100xx2^(100)`D. none of these |
| Answer» Correct Answer - C | |
| 260. |
In a factory, 150 workers were engaged to finish a piece of work in a certain number of days. However, if 4 workers are dropped everyday, except the first day, it will take 8 more days to finish the work. Find the number of days in which the work was to be completed. |
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Answer» series will be like `150+146+142+..........+n` work `= n/2(2+150+ (n-1)(-4))` `= (152n-2n^2)` if all are working then w`= 150(n-8)` `152n-2n^2 = 150n - 1200` `2n^2 - 2n - 1200=6` `n^2 - n-600=0` `(n-25)(n+24)=0` `n=25,-24` as no of days cant be negative so n=`25` answer |
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| 261. |
How many terms of the G.P. 3, `3/2`,`3/4``,dotdotdot`are needed to give the sum `(3069)/(512)`? |
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Answer» Here, `a = 3 and r = 1/2` `S_n = (a(1-r^n))/(1-r)` `3069/512 = (3(1-(1/2)^n))/(1-(1/2))` `1023/1024= 1-(1/2)^n` `(1/2)^n = 1/1024=(1/2)^10` So, `n = 10` is the required answer. |
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| 262. |
If `sinalpha, sin^2alpha, 1 , sin^4alpha and sin^6alpha` are in A.P., where `-pi < alpha < pi,` then `alpha` lies in the intervalA. `(-pi//2,pi//2)`B. `(-pi//3,pi//3)`C. `(-pi//6,pi//6)`D. none of these |
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Answer» Correct Answer - D It is given that `sinalpha,sin^(2)alpha,1,sin^(4)alpha,sin^(5)alpha` are in A.P. `:." "sinalpha,sin^(2)alpha,1` are also in A.P. `rArr" "2sin^(2)alpha=sinalpha+1` `2sin^(2)alpha-sinalpha-1=0` `rArr" "(2sinalpha+a)(sinalpha-1)=0` `rArr" "sinalpha=-1//2or,sinalpha=1` `rArr" "alpha=-pi//6or,alpha=pi//2" "[because-piltalphaltpi]` For `alpha=-pi//6`, the given sequence becomes `-(1)/(2),(1)/(4),1(1)/(16),-(1)/(32)`. Clearly, it is not an A.P. for `alpha=pi//2`, the given sequence becomes 1,1,1,1,1 which is an A.P. But, `alpha=pi//2` does nnot lie in any of the given intervals. |
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| 263. |
If `a,b,c` are in AP, than show that `a^(2)(b+c)+b^(2)(c+a)+c^(2)(a+b)=(2)/(9)(a+b+c)^(3)`. |
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Answer» `therefore a,b,c` are in AP. `therefore b=(a+c)/(2)i.e., 2b=a+c " " "…….(i)"` LHS`=a^(2)(b+c)+b^(2)(c+a)+c^(2)(a+b)` `=(a^(2)b+a^(2)c)+b^(2)(2b)+(c^(2)a+c^(2)b)` `=b(a^(2)+c^(2))+ac(a+c)+2b^(3)` `=b[(a+c)^(2)-2ac]+ac(2b)+2b^(3)` `=b(a+c)^(2)+2b^(3)=b(2b)^(2)+2b^(3)=6b^(3)` RHS `=(2)/(9)(a+b+c)^(3)=(2)/(9)(2b+b)^(3)` `=(2)/(9)xx27b^(3)=6b^(3)` Hence, LHS=RHS. |
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| 264. |
If ` 2 (y - a) ` is the ` H.M.` between ` y - x and y - z ` then ` x-a, y-a, z-a` are in(i) A.P(ii) G.P(iii) H.P(iv) none of theseA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 265. |
If the first, second and the last terms of an A.P. are a,b,c respectively, then the sum of the A.P. isA. `((a+b)(a+c-2b))/(2(b-a))`B. `((b+c)(a+b-2c))/(2(b-a))`C. `((a+c)(b+c-2a))/(2(b-a))`D. none of these |
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Answer» Correct Answer - C Let there be n terms in the A.P. Then, `c=a+(n-1)(b-a)" "[becaused=b-a]` `rArrn=(b+c-2a)/(b-a)` `:." Sum of n terms"=(n)/(2)(a+c)=((b+c-2a)(a+c))/(2(b-a))` |
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| 266. |
Let `a_1, a_2, a_3, ....a_n,.........`be in A.P. If `a_3 + a_7 + a_11 + a_15 = 72,` then the sum of itsfirst 17 terms is equal to :A. 153B. 306C. 612D. 204 |
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Answer» Correct Answer - B We have, `a_(3)+a_(7)+a_(11)+a_(15)=72` `rArr" "(a_(3)+a_(15))+(a_(7)+a_(11))=72` `rArr" "2(a_(3)+a_(15))=72` `rArr" "a_(3)+a_(15)=36rArra_(1)+a_(17)=36` `:." "S_(17)=(17)/(2)(a_(1)+a_(17))=(17)/(2)xx36=606` |
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| 267. |
If `a_1,a_2,...`are in A.P. and `a_1+a_5+a_10+a_15+a_20+a_24=225` then `a_1+a_2+a_3+....a_23+a_24` isA. 909B. 75C. 750D. 900 |
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Answer» Correct Answer - D We have, `a_(1)+a_(5)+a_(15)+a_(20)+a_(24)=225` `rArr" "(a_(1)+a_(24))+(a_(5)+a_(20))+(a_(10)+a_(24))=225` `rArr" "3(a_(1)+a_(24))=225rArra_(1)+a_(24)=75` `:." "a_(1)+a_(2)+a_(3)+ . . .+a_(24)=(24)/(2)(a_(1)+a_(24))=12xx75=900` |
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| 268. |
Let `S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"........+"oo`, where `0ltxlt1`. If `S(x)=(sqrt(2)+1)/(2)`, then the value of `(x+1)^(2)` is |
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Answer» `S(x)=1+x-x^(2)-x^(3)+x^(4)+x^(5)-x^(6)-x^(7)+"....."+oo` where `x in(0,1)` `S(x)=(1+x)-x^(2)(1+x)+x^(4)+(1+x)-x^(6)(1+x)+"....."+oo` `impliesS(x)=(1+x)[1-x^(2)+x^(4)-x^(6)+"....."+oo]` `impliesS(x)=(1+x)((1)/(1+x^(2)))" " [:.S_(oo)=(a)/(1-r) " for "GP]` According to the question, `S(x)=(sqrt(2)+1)/(2)` So, `(1+x)/(1+x^(2))=(sqrt(2)+1)/(2)` `implies=2+2x=(sqrt(2)+1)x^(2)+sqrt(2)+1` `implies(sqrt(2)+1)x^(2)-2x-2+sqrt(2)+1=0` `implies(sqrt(2)+1)x^(2)-2x+sqrt(2)-1=0` `implies(sqrt(2)+1)x^(2)-2x+(1)/(sqrt(2)+1)=0` `implies[(sqrt(2)+1)x]^(2)-2(sqrt(2)+1)x+1=0` `implies[(sqrt(2)+1)x-1]^(2)=0` `implies x=(1)/(sqrt(2)+1)" " ["repeated "]` So, ` x=sqrt(2)-1` `:. (x+1)^(2)=2`. |
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| 269. |
If n is a positive integer satisfying the equation `2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140` then the value of n is |
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Answer» `:.2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140` `2+6(2^(2)+3^(2)+"......."+n^(2))-4*(2+3+"....."+n)=140` `implies 2+6((n(n+1)(2n-1))/(6)-1)-4((n(n+1))/(2)-1)=140` `implies 2+n(n+1)(2n+1)-6-2n(n+1)+4=140` `implies n(n+1)(2n+1)-2n(n+1)-140=0` `implies 2n^(3)3n^(2)+n-2n^(2)-2n-140=0` `implies 2n^(3)+n^(2)-n-140=0` `implies (n-4)+(2n^(2)+9n+35)=0` `implies n=4 " or " 2n^(2)+9n+35=0` `implies 2n^(2)+9n+35=0` `implies n=(-9pmsqrt(81-280))/(4)` `:. n=(9pmsqrt(-199))/(4) " " [" complex values "]` Only positive integer value of n is 4. |
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| 270. |
`S=sum_(i=1)^nsum_(j=1)^isum_(k=1)^j1` |
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Answer» `sum_(i=1)^(n)sum_(j=1)^(i)sum_(k=1)^(j)1=sum_(i=1)^(n)(i(i+1))/(2)` `=(1)/(2)[sum_(i=1)^(n)i^(2)+sum_(i=1)^(i)i]=(1)/(2)[sumn^(2)+sumn]` `=(1)/(2)[(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]` `=(n(n+1))/(12)[2n+1+3]=(n(n+1)(n+2))/(6)` |
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| 271. |
Find the sum of all possible products of the first `n`natural numbers taken two by two.A. `(1)/(24)n(n+1)(n-1)(3n+2)`B. `(n(n+1)(2n+1))/(6)`C. `(n(n+1)(n-1)(2n+3))/(24)`D. none of these |
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Answer» Correct Answer - A We have, `(a_(1)+a_(2)+ . . . . .+a_(n))^(2)=underset(i=1)overset(n)suma_(i)^(2)+2underset(iltj)suma_(i)a_(j)` Let S be the required sum. Putting `a_(1)=1,a_(2)=2, . . . . ,a_(n)=n`, we get `{(n(n+1))/(2)}^(2)=(n(n+1)(2n+1))/(6)+2S` `rArr" "S=(1)/(24)n(n+1)(n-1)(3n+2)` |
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| 272. |
The value of `sum_(i=1)^(n) sum_(j=1)^(i) sum_(k=1)^(j) 1` isA. `sumn`B. `sumn^(2)`C. `sumn^(3)`D. none of these |
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Answer» Correct Answer - D We have, `underset(i=1)overset(n)sumunderset(j=1)overset(i)sumunderset(k=1)overset(j)sum1=underset(i=1)overset(n)sumunderset(j=1)overset(i)sumj` `=underset(i=1)overset(n)sum(1+2+. . . .i)` `underset(i=1)overset(n)sum(i(i+1))/(2)=(1)/(2){underset(i=1)overset(n)sumi^(2)+underset(i=1)overset(n)sumi}` `=(1)/(2){(n(n+1)(2n+1))/(6)+(n(+1))/(2)}` `=(n(n+1))/(4){(2n+1)/(3)+1}=(n(n+1)(n+2))/(6)` |
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| 273. |
Find the 50th term of the series `2+3+6+11+18+…`.A. `49^(2)-1`B. `49^(2)`C. `50^(2)+1`D. `49^(2)+2` |
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Answer» Correct Answer - D Let `T_(n)" be the "n^(th)` term of the given series. We observe that the successive differences of the terms form an A.P. So, let its `n^(th)` term be given by `T_(n)=an^(2)+bn+c` Putting n=1,2,3, we get a+b+c=2,4a+2b+c=3 and, 9a+3b+c=6 Solving these equations, we get a=1, b=-2, c=3 `:." "T_(n)=n^(2)-2n+3=(n-1)^(2)+2` `rArr" "T_(50)=49^(2)+2` D is the answer of this question
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| 274. |
Find the first term of a HP whpse secpmd ter, os `(5)/(4)` and the third term is `(1)/(2)`. |
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Answer» Let a be the first term. Then, `a,(5)/(4),(1)/(2)` are in HP. Then, `(a-(5)/(4))/((5)/(4)-(1)/(2))=(a)/((1)/(2)) " " [" from above note "]` ` implies (4a-5)/(5-2)=2a` ` implies 4a-5=6a " or " 2a=-5` ` therefore a=-(5)/(2)` |
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| 275. |
If a, b, c are in HP, then `(a-b)/(b-c)` is equal to |
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Answer» Since, `a,b,c` are in HP, therefore `(1)/(a),(1)/(b),(1)/(c)` are in AP `i.e.(1)/(b)-(1)/(a)=(1)/(c)-(1)/(b)` or `(a-b)/(ab)=(b-c)/(bc)" or "(a-b)/(b-c)=(a)/(c)` |
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| 276. |
If `x, y` and `z` are `pth, gth` and `rth` terms respectively of an `A.P` and also of a `G.P.` then `x^(y-z)*y^(z-x)*z^(x-y) ` is equal toA. xyzB. 0C. 1D. -1 |
| Answer» Correct Answer - C | |
| 277. |
Thefirst two terms of a geometric progression add up to 12. The sum of the thirdand the fourth terms is 48. If the terms of the geometric progression arealternately positive and negative, then the first term is |
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Answer» `a,ar,ar^2,ar^3` `a+ar=12---(1)` `ar^2+ar^3=48 ` `r^2(a+ar)=48---(2)` `(1)/(2)=1/r^2=1/4` `r^2=4` `r=+2,-2` `r=2` `a+ar=12` `a(1-2)=12` `a=-12` option `2`is correct |
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| 278. |
If `x=111"...."(20 digits),y=333"...."(10digits) and z=222".....2"(10 digits), then (x-y^(2))/(z)` equals.A. `(1)/(2)`B. 1C. 2D. 4 |
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Answer» Correct Answer - B `:.x=(1)/(9)(999"...."9)=(1)/(9)(10^(20)-1)` `y=(1)/(3)(999"...."9)=(1)/(3)(10^(10)-1)` and `z=(2)/(9)(999"...."9)=(2)/(9)(10^(10)-1)` `:.(x-y^(2))/(z)=((1)/(9)(10^(20)-1)-(1)/(9)(10^(10)-1)^(2))/((2)/(9)(10^(10)-1))` `=(10^(10)+1-(10^(10)-1))/(2)=1` |
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| 279. |
Consider the sequence `1,2,2,3,3,3,"……",` where n occurs n times that occuts as 201th rerms isA. 61B. 62C. 63D. 64 |
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Answer» Correct Answer - C The last 4 occurs as `1+2+3+4+=10th` term. The last n occurs as `((n(n+1))/(2))^(th)` term, the last 62 occurs as `((63xx63)/(2))^(th)=1953rd` term and the last 63 occurs as `((64xx64)/(2))^(th)=2016th` term. `:. 63` occurs from 1954th term to 2016th term. Hence, `(2011)^(th)` term os 63. |
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| 280. |
If a, b, c are in A.P. and x, y, z are in G.P., then prove that : `x^(b-c).y^(c-a).z^(a-b)=1`A. -1B. 0C. 1D. none of these |
| Answer» Correct Answer - C | |
| 281. |
If a,b,c are unequal numbers in A.P. such that a,b-c,c-a are in G.P. thenA. `(a)/(1)=(b)/(2)=(c)/(3)`B. `(a)/(3)=(b)/(5)=(c)/(7)`C. `(a)/(1)=(b)/(3)=(c)/(5)`D. none of these |
| Answer» Correct Answer - C | |
| 282. |
A man borrows Rs. 8190 without interset and repays the loan in 12 monthly instalments. If each instalment is double the preceding one then the first and last instalments are (in rupees)A. 5 and 1200B. 2 and 4096C. 3 and 7200D. none of these |
| Answer» Correct Answer - B | |
| 283. |
Sum of three numbers in G.P. be 14. If one is added to first and second and 1 is subtracted from the third, the new numbers are in A.P. The smallest of them isA. 2B. 4C. 6D. 8 |
| Answer» Correct Answer - A | |
| 284. |
If x,1,z are in A.P. and x,2,z are in G.P., then x,4,z are inA. APB. G.PC. H.P.D. none of these |
| Answer» Correct Answer - C | |
| 285. |
If `(3+5+7+ . . . .+"n terms")/(5+8+11+ . . .. +"10 terms")=7`, then the value of n, isA. 35B. 36C. 37D. 40 |
| Answer» Correct Answer - A | |
| 286. |
If `a,b,c,d` be four disinct positive quantities in HP, then (a) `a+dgtb+c` (b)`adgt bc` |
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Answer» `therefore a,b,c,d` are in HP. (a) Applying AMgtHM For first three members, `(a+c)/(2)gtb` `implies a+cgt2b" " "….(ix)"` For last three members, `(b+d)/(2)gtc` `implies b+dgt2c " " "….(x)"` From Eqs. (ix) and (x), we get `a+c+b+dgt2b+2c` or `a+dgtb+c` (b) Applying GMgt HM For first three members, `sqrt(ac)gtb` `implies acgtb^(2) " " "....(xi)"` For last three members, `sqrt(bd)gtc` `implies bdgtc^(2) " " "....(xii)"` From Eqs. (xi) and (xii), we get `(ac)(bd)gtb^(2)c^(2)` or `adgtbc` |
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| 287. |
The value of `2.bar(357)`, isA. `(2355)/(1001)`B. `(2355)/(999)`C. `(2355)/(1111)`D. `(2354)/(1111)` |
| Answer» Correct Answer - B | |
| 288. |
If a,b,c are in A.P. as well as in G.P. thenA. `a=b!=c`B. `a!=b=c`C. `a!=b!=c`D. a=b=c |
| Answer» Correct Answer - D | |
| 289. |
The sum to infinity of the series `1+(4)/(5)+(7)/(5^(2))+(10)/(5^(3))+ . . . ,` is |
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Answer» (ii) ` therefore S_(infty)=1+4((1)/(5))+7((1)/(5))^(2)+10((1)/(5))^(3)+"...." "upto "infty ".....(iii)"` Multiplying both sides of Eq. (i) by `(1)/(5)`, we get `(1)/(5) S_(infty)=((1)/(5))+4((1)/(5))^(2)+7((1)/(5))^(3)+"...." "upto "infty ".....(iv)"` Subtracting Eq. (iv) from Eq. (iii), we get `(1-(1)/(5)) S_(infty)=1+3[((1)/(5))+((1)/(5))^(2)+((1)/(5))^(3)+"...." "upto "infty "]` `=1+3(((1)/(5))/(1-(1)/(5)))=1+(3)/(4)` `=(4)/(5)S_(infty)=(7)/(4)=` `therefore S_(infty)=(35)/(16)` |
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| 290. |
If `(1)/(b+c),(1)/(c+a),(1)/(a+b)` are in A.P., thenA. a,b,c are in A.P.B. `a^(2),b^(2),c^(2)` are in A.P.C. `(1)/(a),(1)/(b),(1)/( c)` are in A.P.D. none of these |
| Answer» Correct Answer - B | |
| 291. |
Find the sum of `n`terms of the series `1+4/5+7/(5^2)+10/5^3+dot` |
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Answer» The given series can be written as `1+4((1)/(5))+7((1)/(5))^(2)+10((1)/(5))^(3)+"....."` The series is an Arithmetico-Geometric series, since each term is formed by multiplng corresponding terms of the series `1,4,7,"….."` which are in AP and `1,(1)/(5),(1)/(5)^(2),"...."` which are in GP. ` therefore T_(n)=[" nth term of "1,4,7,"..."][" n th term of " 1,(1)/(5),)((1)/(5))^(2)+"....."]` `=[1+(n-1)3]xx1*((1)/(5))^(n-1)=(3n-2)((1)/(5))^(n-1)"...(i)"` ` therefore T_(n-1)=(3n-5)((1)/(5))^(n-2)` (i) Let sum of n terms of the series is denoted by `S_(n)` Then, ` S_(n)=1+4((1)/(5))+7((1)/(5))^(2)+"...."+(3n-5)((1)/(5))^(n-2)+(3n-2)((1)/(5))^(n-1)" ......(i)"` Multiplying both the sides of Eq. (i) by `(1)/(5)`, we get ` (1)/(5)S_(n)=(1)/(5)+4((1)/(5))^(2)+7((1)/(5))^(3)+"...."+(3n-5)((1)/(5))^(n-1)+(3n-2)((1)/(5))^(n)"......(ii)"` Subtracting Eq. (ii) from Eq.(i), we get ` (1-(1)/(5))S_(n)=1+3[(1)/(5)+((1)/(5))^(2)+((1)/(5))^(3)+"...."+((1)/(5))^(n-1)]-(3n-2)((1)/(5))^(n)` or ` (4)/(5)S_(n)=1+3[((1)/(5))+((1)/(5))^(2)+((1)/(5))^(3)+"...."+(n-1)" terms "]-(3n-2)((1)/(5))^(n)` ` =1+3[[(1)/(5)[1-((1)/(5))^(n-1)])/(1-(1)/(5))]-(3n-2)((1)/(5))^(n)` ` =1+(3)/(4)[1-((1)/(5))^(n-1)]-(3n-2)((1)/(5))^(n)` ` therefore S_(n) =(5)/(4)+(15)/(16)[1-((1)/(5))^(n-1)]-((3n-2))/(4)((1)/(5))^(n-1)` `=(35)/(16)-((12n+7))/(16)((1)/(5))^(n-1)` |
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| 292. |
If `p^(t h), q^(t h)a n d r^(t h)`terms of a G.P. are `x , y , z`respectively then write the value of `x^(q-r)y^(r-p)z^(p-q)dot` |
| Answer» Correct Answer - B | |
| 293. |
If `x ,y ,a n dz`are in G.P. and `x+3+,y+3,a n dz+3`are in H.P., then`y=2`b. `y=3`c. `y=1`d. `y=0`A. y=2B. y=3C. y=1D. y=0 |
| Answer» Correct Answer - B | |
| 294. |
If a,b,c are in AP, then `(a)/(bc),(1)/(c ), (2)/(d)` are inA. A.P.B. G.P.C. H.P.D. AGP |
| Answer» Correct Answer - D | |
| 295. |
If a,b,c are in AP, then `(a)/(bc),(1)/(c ), (1)/(b)` are inA. APB. GPC. HPD. None of these |
| Answer» Correct Answer - C | |
| 296. |
If a, b, c are in AP and a, b, d are in GP, show that `a, (a-b)` and `(d-c)` are in GP.A. APB. GPC. HPD. None of these |
| Answer» Correct Answer - C | |
| 297. |
If `a,b,c,d` be four distinct positive quantities in AP, then (a) `bcgtad` (b) ` c^(-1)d^(-1)+a^(-1)b^(-1)gt2(b^(-1)d^(-1)+a^(-1)c^(-1)-a^(-1)d^(-1))` |
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Answer» `therefore a,b,c,d` are in AP. (a) Applying AMgtGM For first three members, `bgtsqrt(ac)` `implies b^(2)gt ac " " "….(i)"` and for last three members, `cgtsqrt(bd)` `implies c^(2)gt bd " " "……(ii)"` From Eqs. (i) and (ii), we get `b^(2)c^(2)gt(ac)(bd)` Hence, `bcgtad` (b) Applying AMgtHM For first three members, `bgt(2ac)/(a+c)` `implies ab+bcgt2ac " " ".....(iii)"` For last three members, `cgt(2bd)/(b+d)` `bc+cdgt2bd " " "...(iv)"` From Eqs. (iii) and (iv), we get `ab+bc+bc+cdgt2ac+2bd` or `ab+cdgt2(ac+bd-bc)` Dividing in each term by `abcd`, we get ` c^(-1)d^(-1)+a^(-1)b^(-1)gt2(b^(-1)d^(-1)+a^(-1)c^(-1)-a^(-1)d^(-1))` |
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| 298. |
Statement -1: If a,b,c are distinct real numbers in H.P, then `a^(n)+c^(n)gt2b^(n)" for all "ninN`. Statement -2: `AMgtGMgtHM` |
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Answer» `therefore GgtH` `therefore sqrt(ac)gtb` `implies (ac)^((n)/(2))gtb^(n) " or " a^((n)/(2))c^((n)/(2))gtb^(n) " " "....(i)"` Also, `(a^((n)/(2))-c^((n)/(2)))^(2)gt0 implies a^(n)+c^(n)-2a^((n)/(2))c^((n)/(2))gt0` `implies a^(n)+c^(n)gt2a^((n)/(2))c^((n)/(2))gt2b^(n)" " [" from Eq.(i) "]"....(i)"` `therefore a^(n)+c^(n)gt2b^(n)` |
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| 299. |
In a G.P. of positive terms if any terms is equal to the sum of nexttow terms, find the common ratio of the G.P.A. `-1`B. `-3`C. `-3`D. `-1//2` |
| Answer» Correct Answer - C | |
| 300. |
Statement -1: If a,b,c are distinct real numbers in H.P, then `a^(n)+c^(n)gt2b^(n)" for all "ninN`. Statement -2: `AMgtGMgtHM`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 If a,b,c are in H.P., then b is the HM of a and c. But, the GM of a and c is `sqrt(ac)`. . . .(i) Using `AMgtGM`, we have `(a^(n)+c^(n))/(2)gtsqrt(a^(n)c^(n))` `rArr" "(a^(n)+c^(n))/(2)gt(ac)^(n//2)` `rArr" "(a^(n)+c^(n))/(2)gtb^(n)` `rArr" "a^(n)+c^(n)gt2b^(n)` Hence, statement -1 and 2 are true statement -2 is a correct explanation for statement -1. |
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