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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.
| 451. |
The numbers `1,3,6,10,15,21,28."……"` are called triangular numbers. Let `t_(n)` denotes the `n^(th)` triangular number such that `t_(n)=t_(n-1)+n,AA n ge 2`. The number of positive integers lying between `t_(100)` and `t_(101)` areA. 99B. 100C. 101D. 102 |
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Answer» Correct Answer - B Given sequece `1,3,6,10,15,21,28,"…"` where `t_(n)=t_(n-1)+n,AA n ge 2` So, `t_(n)=[t_(n-2)+(n-1)]+n` `vdots " "vdots " " vdots " "` `t_(n)=t_(1)+2+3+"........"+(n-1)+n` `t_(n)=1+2+3+"......"+n` `t_(n)=(n(n+1))/(2)` `t_(100)=(100xx101)/(2)=5050` `t_(101)=(101xx102)/(2)=101xx51=5151` Number of positive integers lying between `t_(100)` and `t_(101)` `=5151-5050-1` `=101-1=100`. |
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| 452. |
The numbers `1,3,6,10,15,21,28."……"` are called triangular numbers. Let `t_(n)` denotes the bth triangular number such that `t_(n)=t_(n-1)+n,AA n ge 2`. The value of `t_(50)` isA. 1075B. 1175C. 1275D. 1375 |
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Answer» Correct Answer - C Given sequece `1,3,6,10,15,21,28,"…"` where `t_(n)=t_(n-1)+n,AA n ge 2` So, `t_(n)=[t_(n-2)+(n-1)]+n` `vdots " "vdots " " vdots " "` `t_(n)=t_(1)+2+3+"........"+(n-1)+n` `t_(n)=1+2+3+"......"+n` `t_(n)=(n(n+1))/(2)` `t_(50)=(50xx51)/(2)=25xx51=1275`. |
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| 453. |
There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If `p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series. The value of q isA. 200B. 160C. 120D. 80 |
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Answer» Correct Answer - C Let `A={A-D,A,A+D},B={a-d,a,a+d}` according to the question, `A-D+A+A+D=15` `implies 3A=15` `implies A=15" " "…….(i)"` and `a-d+a+a+d=15` `implies a=5" "………(ii)"` and `D=1+d" " "……..(iii)"` `p=(A-D)A(A+D)` `p=A(A^(2)-D^(2))" " ......(iv)"` `p=5(25-D^(2))" " "........(v)"` Similarly, `q=5(25-d^(2))` Given that, `p=7(q-p)` `8p=7q` From Eqs. (iv) and (v), we get `8xx5(25-D^(2))=7xx5(25-d^(2))` `200-8D^(2)=175-7d^(2)` `25=8D^(2)-7d^(2)` `25=8(1+d)^(2)-7d^(2)" " [" from Eq. (iii) "]` `25=8+8d^(2)+16d-7d^(2)` `17-d^(2)-16d=0` `d^(2)+16d-17=0` `(d+17)(d-1)=0` `d=-17` or `d=1` `implies d=1 " " [:.dgt0]` `impliesD=2` `q=5(25-d^(2))=5(25-1)=120`. |
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| 454. |
There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets. The value of `r^(R)+R^(r)` isA. 5392B. 368C. 32D. 4 |
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Answer» Correct Answer - C Let `A={(A)/(R),A,AR}` `B={(a)/(r ),a,ar}` According to the question, `(A)/(R )*A*AR=64` `implies A^(3)=64 " "implies A=4" " "……..(i)"` `(a)/(r )*a*ar=64" " implies a^(3)=64 " " implies a=4 " " "……(ii)"` and `R=r+2" " "……..(iii)"` `p=(A)/(R)*A*AR+AR*(A)/(R)` `=(A^(2))/(R)+A^(2)R+A^(2)=(16)/(R )+16R+16` `q=(a)/(r )*a+a*ar+ar*(a)/( r)` `=(a^(2))/(r )+a^(2)r+a(^2)=(16)/(r )=(16)/(r )+16r+16` Give that, `(p)/(q)=(3)/(2)` So, `((16+16R^(2)+16R)r)/((16+16r^(2)+16r)R)=(3)/(2)` `((1+R^(2)+R)r)/((1+r^(2)+r)R)=(3)/(2)` From Eq. (iii), `R=r+2` `implies ((1+r^(2)+4+4r+r+2)r)/((1+r+r^(2))(r+2))=(3)/(2)` `implies (r^(3)+5r^(2)+7r)/(r^(3)+3r^(2)+3r+2)=(3)/(2)` `implies r^(3)-r^(2)-5r+6=0` `implies (r-2)(r^(2)+r-3)=0` `impliesr=2" or " r=(-1pmsqrt(13))/(2)` So, `R=4` `r^(R)+R^(r )=(4)^(2)+(2)^(4)=16+16=32` |
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| 455. |
There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If `p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series. The value of `7D+8d` isA. 37B. 22C. 67D. 52 |
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Answer» Correct Answer - B Let `A={A-D,A,A+D},B={a-d,a,a+d}` according to the question, `A-D+A+A+D=15` `implies 3A=15` `implies A=15" " "…….(i)"` and `a-d+a+a+d=15` `implies a=5" "………(ii)"` and `D=1+d" " "……..(iii)"` `p=(A-D)A(A+D)` `p=A(A^(2)-D^(2))" " ......(iv)"` `p=5(25-D^(2))" " "........(v)"` Similarly, `q=5(25-d^(2))` Given that, `p=7(q-p)` `8p=7q` From Eqs. (iv) and (v), we get `8xx5(25-D^(2))=7xx5(25-d^(2))` `200-8D^(2)=175-7d^(2)` `25=8D^(2)-7d^(2)` `25=8(1+d)^(2)-7d^(2)" " [" from Eq. (iii) "]` `25=8+8d^(2)+16d-7d^(2)` `17-d^(2)-16d=0` `d^(2)+16d-17=0` `(d+17)(d-1)=0` `d=-17` or `d=1` `implies d=1 " " [:.dgt0]` `impliesD=2` `7D+8d=14+8=22` |
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| 456. |
There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets. The value of q isA. 54B. 56C. 58D. 60 |
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Answer» Correct Answer - B Let `A={(A)/(R),A,AR}` `B={(a)/(r ),a,ar}` According to the question, `(A)/(R )*A*AR=64` `implies A^(3)=64 " "implies A=4" " "……..(i)"` `(a)/(r )*a*ar=64" " implies a^(3)=64 " " implies a=4 " " "……(ii)"` and `R=r+2" " "……..(iii)"` `p=(A)/(R)*A*AR+AR*(A)/(R)` `=(A^(2))/(R)+A^(2)R+A^(2)=(16)/(R )+16R+16` `q=(a)/(r )*a+a*ar+ar*(a)/( r)` `=(a^(2))/(r )+a^(2)r+a(^2)=(16)/(r )=(16)/(r )+16r+16` Give that, `(p)/(q)=(3)/(2)` So, `((16+16R^(2)+16R)r)/((16+16r^(2)+16r)R)=(3)/(2)` `((1+R^(2)+R)r)/((1+r^(2)+r)R)=(3)/(2)` From Eq. (iii), `R=r+2` `implies ((1+r^(2)+4+4r+r+2)r)/((1+r+r^(2))(r+2))=(3)/(2)` `implies (r^(3)+5r^(2)+7r)/(r^(3)+3r^(2)+3r+2)=(3)/(2)` `implies r^(3)-r^(2)-5r+6=0` `implies (r-2)(r^(2)+r-3)=0` `impliesr=2" or " r=(-1pmsqrt(13))/(2)` So, `R=4` `q=16((1)/(r )r+1)=16((1)/(2)+2+1)=(16)/(2)xx7=8xx7=56` |
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| 457. |
There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series. The value of p isA. 105B. 140C. 175D. 210 |
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Answer» Correct Answer - A Let `A={A-D,A,A+D},B={a-d,a,a+d}` according to the question, `A-D+A+A+D=15` `implies 3A=15` `implies A=15" " "…….(i)"` and `a-d+a+a+d=15` `implies a=5" "………(ii)"` and `D=1+d" " "……..(iii)"` `p=(A-D)A(A+D)` `p=A(A^(2)-D^(2))" " ......(iv)"` `p=5(25-D^(2))" " "........(v)"` Similarly, `q=5(25-d^(2))` Given that, `p=7(q-p)` `8p=7q` From Eqs. (iv) and (v), we get `8xx5(25-D^(2))=7xx5(25-d^(2))` `200-8D^(2)=175-7d^(2)` `25=8D^(2)-7d^(2)` `25=8(1+d)^(2)-7d^(2)" " [" from Eq. (iii) "]` `25=8+8d^(2)+16d-7d^(2)` `17-d^(2)-16d=0` `d^(2)+16d-17=0` `(d+17)(d-1)=0` `d=-17` or `d=1` `implies d=1 " " [:.dgt0]` `impliesD=2` `p=5(25-D^(2))=5(25-4)=5(21)=105`. |
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| 458. |
Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4 The sum of all numbersA. less than 1000B. lies between 1200 to 1500C. greater than 1500D. None of these |
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Answer» Correct Answer - B Let p and `(p+1)` be removed numbers from `1,2,3,"……"n`, then Sum of the remaining numbers `=(n(n+1))/(2)-(2p+1)` From given condition, `(105)/(4)=((n(n1))/(2)-(2p+1))/((n-2))` `implies 2n^(2)-103n-8p+206=0` Since, n and p are integers, so n must be even. Let `n=2r`, we get `p=(4r^(2)+103(1-r))/(4)` Since, p is an integer, then `(1-r)` must be divisible by 4. Let `r=1+4t`, we get `n=2+8t` and `p=16t^(2)-95t+1` Now, `1le pltn` `implies 1le16t^(2)-95t+1lt8t+2` `implies t=6 " " implies n=50` and `p=7`. Sum of all numbers `=(50(50+1))/(2)=1275`. |
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| 459. |
Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4 The sum of all numbersA. are less than 10B. lies between 10 to 30C. lies between 30 to 70D. greater than 70 |
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Answer» Correct Answer - A Let p and `(p+1)` be removed numbers from `1,2,3,"……"n`, then Sum of the remaining numbers `=(n(n+1))/(2)-(2p+1)` From given condition, `(105)/(4)=((n(n1))/(2)-(2p+1))/((n-2))` `implies 2n^(2)-103n-8p+206=0` Since, n and p are integers, so n must be even. Let `n=2r`, we get `p=(4r^(2)+103(1-r))/(4)` Since, p is an integer, then `(1-r)` must be divisible by 4. Let `r=1+4t`, we get `n=2+8t` and `p=16t^(2)-95t+1` Now, `1le pltn` `implies 1le16t^(2)-95t+1lt8t+2` `implies t=6 " " implies n=50` and `p=7`. Hence, removed numbers are 7 and 8. |
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| 460. |
Two consecutive numbers from `1,2,3,"……n"` are removed. The arithmetic mean of the remaining numbers is `(105)/4)`. The value of n lies inA. `(41,51)`B. `(52,62)`C. `(63,73)`D. `(74,84)` |
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Answer» Correct Answer - A Let p and `(p+1)` be removed numbers from `1,2,3,"……"n`, then Sum of the remaining numbers `=(n(n+1))/(2)-(2p+1)` From given condition, `(105)/(4)=((n(n1))/(2)-(2p+1))/((n-2))` `implies 2n^(2)-103n-8p+206=0` Since, n and p are integers, so n must be even. Let `n=2r`, we get `p=(4r^(2)+103(1-r))/(4)` Since, p is an integer, then `(1-r)` must be divisible by 4. Let `r=1+4t`, we get `n=2+8t` and `p=16t^(2)-95t+1` Now, `1le pltn` `implies `1le16t^(2)-95t+1lt8t+2` `implies t=6 " " implies n=50` and `p=7`. Hence, the value of n lies in `(41,51)`. |
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| 461. |
If first term of a GP is a, third term is b and `(n+1)th` term is c. The `(2n+1)th` term of a GP isA. `csqrt((b)/(a))`B. `(bc)/(a)`C. `abc`D. `(c^(2))/(a)` |
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Answer» Let common ratio =r ` therefore b=ar^(2) implies r=sqrt((b)/(a))` Also, `c=ar^(n) implies r^(n)=(c)/(a)` `therefore t_(2n+1)= ar^(2n)=a(r^(n))^(2)=a((c)/(a))^(2)=(c^(2))/(a)` Hence, (d) is the correct answer. |
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| 462. |
Show that the sequence `t_(n)` defined by `t_(n)=2*3^(n)+1` is not a GP. |
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Answer» We have, `t_(n)=2*3^(n)+1` On replacing n by `(n-1)` in `t_(n)`, we get `t_(n-1)=2*3^(n-1)+1` `implies t_(n-1)=(2*3^(n)+3)/(3)` `therefore (t_(n))/(t_(n-1))=((2*3^(n)+1))/(((2*3^(n)+3))/(3))=(3(2*3^(n)+1))/((2*3^(n)+3))` Clearly, `(t_(n))/(t_(n-1))` is not independent of n and is therefore not constant. So, the given sequence is not a GP. |
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| 463. |
Show that the sequence `t_(n)` defined by `t_(n)=(2^(2n-1))/(3)` for all values of `n in N` is a GP. Also, find its common ratio. |
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Answer» We have, `t_(n)=(2^(2n-1))/(3)` On replacing n by `n-1`, we get `t_(n-1)=(2^(2n-3))/(3) implies (t_(n))/(t_(n-1))=((2^(2n-1))/(3))/((2^(2n-3))/(3))=2^(2)=4 ` Clearly, `(t_(n))/(t_(n-1))` is independent of n and is equal to 4. So, the given sequence is a GP with common ratio 4. |
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| 464. |
If `x, 2x+ 2, 3x + 3 `are in `G.P. ,` then the fourth term is (A) `27` (B) `-27` (C)`13.5` (D)`-13.5`A. 27B. -27C. 13.5D. -13.5 |
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Answer» Correct Answer - D It is given that x,2x+2,3x+3 are in G.P. `rArr" "(2x+2)^(2)=x(3x+3)` `rArr" "4x^(2)8x+4=3x^(2)+3xrArrx^(2)+5x+4=0rArrx=-1,-4` So, the G.P. is -4,-6,-9, . . . . .. . . . . . . Hence, Fourth term `=-9xx1.5=-13.5.` |
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| 465. |
The minimum value of `4^(x)+4^(1-x),x in RR` isA. 2B. 4C. 1D. 0 |
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Answer» Correct Answer - B WE know that `AMgeGM` `implies (4^(x)+4^(1-x))/(2) gesqrt(4^(x).4^(1-x))` `implies 4^(x)+4^(1-x)ge2sqrt(4)` `implies 4^(x)+4^(1-x)ge2.2` `implies 4^(x)+4^(1-x)ge4` |
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| 466. |
Let the harmonic mean and geometric mean of two positive numbers be inthe ratio 4:5. Then the two numbers are in ratio............ (1992, 2M)A. `1:1`B. `2:1`C. `3:1`D. `4:1` |
| Answer» Correct Answer - A | |
| 467. |
If second third and sixth terms of an A.P. are consecutive terms o aG.P. write the common ratio of the G.P.A. 1B. -1C. 3D. -3 |
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Answer» Correct Answer - C Let a be the first term and d be the common difference of the A.P. It is given that a+d,a+2d,a+5d are in GP `rArr" "(a+2d)^(2)=(a+d)(a+5d)` `rArr" "4d^(2)+4ad=6ad+5d^(2)rArrd=-2a` So, the terms are -a,-3a,-9a which are in G.P. with common ratio 3. |
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| 468. |
The sum of first `2n` terms of an `AP` is `alpha`. and the sum of next `n` terms is `beta,` its common difference isA. `(alpha-2beta)/(3n^(2))`B. `(2beta-alpha)/(3n^(2))`C. `(alpha-2beta)/(3n)`D. `(2beta-alpha)/(3n)` |
| Answer» Correct Answer - B | |
| 469. |
The sum of the integers from 1 to `100` which are not divisible by 3 or `5` isA. 2489B. 4735C. 2632D. 2317 |
| Answer» Correct Answer - A | |
| 470. |
The fourth, seventh and tenth terms of a G.P. are p,q,r respectively, thenA. `p^(2)=q^(2)+r^(2)`B. `q^(2)=pr`C. `p^(2)=qr`D. pqr+pq+1=0 |
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Answer» Correct Answer - B Let A be the first term and R be the common ratio of the G.P. Then, `p=AR^(3),q=AR^(6)andr=AR^(9)` Clearly, `(AR^(6))^(2)=(AR^(3))(AR^(9))rArrq^(2)=prrArrprrArrp,q,r` are in G.P. ALITER If the terms of a G.P. are chosen at regular intervals, they also form a G.P. |
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| 471. |
The fourth ;seventh and last terms of a GP are 10; 80 and 2560 respectively . Find the first term and the no. of terms in the GP.A. `(4)/(5),12`B. `(4)/(5),10`C. `(5)/(4),12`D. `(5)/(4),10` |
| Answer» Correct Answer - B | |
| 472. |
The fourth, seventh and tenth terms of a G.P. are p,q,r respectively, thenA. `p^(2)=q^(2)+r^(2)`B. `p^(2)=qr`C. `q^(2)=pr`D. `r^(2)=p^(2)+q^(2)` |
| Answer» Correct Answer - B | |
| 473. |
If mth term of an AP is 1/n and its nth term is 1/m , then show that its (mn)th term is 1 |
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Answer» If A snd B are constants, then rth term of AP is `t_(r)=Ar+B` Given, `t_(m)=(1)/(m) implies Am+B=(1)/(n) " " "…."(i)` and `t_(n)=(1)/(m) implies An+B=(1)/(m) " " "…."(ii)` From Eqs. (i) and (ii), we get `A=(1)/(mn)` and B=0 mnth term `=t_(mn)=Amn+B=(1)/(mn).mn+0=1` Hence, mn term of the given AP is 1. |
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| 474. |
In a `GP` if the `(m+n)th` term is `p` and `(m-n)th` term is `q` then `mth` term is |
| Answer» Correct Answer - A | |
| 475. |
The sum of 10 terms of the series `sqrt2 + sqrt6 + sqrt18 +...` isA. `121(sqrt(6)+sqrt(2))`B. `243(sqrt(3)+1)`C. `(121)/(sqrt(3)-1)`D. `242(sqrt(3)-1)` |
| Answer» Correct Answer - B | |
| 476. |
Let `a_(1),a_(2),a_(3),a_(4)anda_(5)` be such that `a_(1),a_(2)anda_(3)` are in A.P., `a_(2),a_(3)anda_(4)` are in G.P., and `a_(3),a_(4)anda_(5)` are in H.P. Then, `a_(1),a_(3)anda_(5)` are inA. G.P.B. A.P.C. H.P.D. none of these |
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Answer» Correct Answer - A We have `a_(1),a_(2),a_(3)` are in A.P. `rArr2a_(2)=a_(1)+a_(3)` . . . .(i) `a_(2),a_(3),a_(4)` are in G.P. `rArr a_(3)^(2)=a_(2)a_(4)` . . .(ii) `a_(3),a_(4),a_(5)` are in G.P. `rArra_(4)=(2a_(3)a_(5))/(a_(3)+a_(5))` . . .(iii) Putting `a_(2)=(a_(1)+a_(3))/(2)anda_(4)=(2a_(3)a_(5))/(a_(3)+a_(5))` in (ii), we get `a_(3)^(2)=(a_(1)+a_(3))/(2)xx(2a_(3)a_(5))/(a_(3)+a_(5))` `rArr" "a_(3)^(2)=a_(1)a_(5)rArra_(1),a_(3),a_(5)` are in G.P. |
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| 477. |
If `x_1, x_2` be the roots of the equation `x^2 -3x+ A=0` and `x_3,x_4` be those of the equation `x^2-12x + B = 0` and `x_1 , x_2, x_3, x_4` be an increasing GP. find find A and B. |
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Answer» `therefore x_(1),x_(2),x_(3),x_(4) " are in GP. "` Let `therefore x_(2)=x_(1)r,x_(3)=x_(1)r^(2),x_(4)=x_(1)r^(3)` `" " [" here, product of " x_(1),x_(2),x_(3),x_(4)" are not given "]` Given, `x_(1)+x_(2)=3,x_(1)x_(2)=A` `implies x_(1)(1+r)=3,x_(1)^(2)r=A " " ".....(i)"` and `x_(3)+x_(4)=12,x_(3)x_(4)=B` `implies x_(1)r^(2)(1+r)=12,x_(1)^(2)r^(5)=B" " ".....(ii)"` from Eqs. (i) and (ii), `r^(2)=4 implies r=2" "[" for increasing GP "]` From Eqs. (i), `x_(1)=1` Now, `A=x_(1)^(2)r=1^(2)*2=2 " " [" from Eq. (i)"]` and `B=x_(1)^(2)r^(5)=1^(2)*2^(5)=32 " " [" from Eq. (ii)"]`. |
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| 478. |
Thesum of first 9 terms of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+2+3)+. . . . .`is(1)71 (2) 96 (3) 142 (4) 192 |
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Answer» `T_n = (1^3 + 2^3 + 3^3 + ...... + n^3)/(1 + 3 + 5 + 7 + ...... + (2n-1))` `= ((n(n+1))/2)^2/(n/2[1 + (2n-1)])` `= n^2(n+1)^2/(4 xx n^2)` `T_n = (n+1)^2/4` `sum_(x=1)^9 T_n = sum_(n=1)^9 (n+1)^2/4` `= 1/4 sum_(n=1)^9 (n+1)^2` = `1/4 ( 2^2 + 3^2 + 4^2 + ...+ 10^2)` `= 1/4 ( 1^2 + 2^2 + 3^2 + .... + 10^2 - 1^2)` `= 1/4[(10(10+1)(2 xx 10 + 1))/6 - 1]` `= 1/4[(10 xx 11 xx 27)/6- 1]` `= 1/4 xx 384= 96` option 1 is correct |
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| 479. |
If `a_1,a_2,...a_n` are in H.P then the expression `a_1a_2+a_2a_3+....+a_(n-1)a_n` 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 Since `a_(1),a_(2),a_(3), . . . ,a_(n)` are in H.P. `:." "(1)/(a_(1)),(1)/(a_(1)),(1)/(a_(3)), . . .,(1)/(a_(n))` are in A.P. Let d be the common difference of the A.P. Then, `(1)/(a_(2))-(1)/(a_(1))=d,(1)/(a_(3))-(1)/(a_(2))=d, . . . ,(1)/(a_(n))-(1)/(a_(n-1))=d` `rArr" "a_(1)-a_(2)=d(a_(1)a_(2)),a_(2)-a_(3)=d(a_(2)a_(3))`, `a_(3)-a_(4)=d(a_(3)a_(4)), . . . a_(n-1)-a_(n)=d(a_(n-1)a_(n))` `rArr" "(a_(1)-a_(2))+(a_(2)-a_(3))+ . . . +(a_(n-1)-a_(n))` `=d(a_(1)a_(2)+a_(2)a_(3)+ . . . +a_(n-1)a_(n))` `rArr" "a_(1)-a_(n)=d(a_(1)a_(2)+a_(2)a_(3)+ . . . +a_(n-1)a_(n))` . . . (i) Now, `(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)), . . . ,(1)/(a_(n))` are in A.P. with common difference d. `rArr" "(1)/(a_(n))=(1)/(a_(1))+(n-1)d` `rArr" "(1)/(a_(n))-(1)/(a_(1))(n-1)d` `rArr" "(a_(1)-a_(n))/(a_(1)a_(n))=(n-1)drArra_(1)-a_(n)=(n-1)d(a_(1)a_(n))` . . . (ii) From (i) and (ii), we get `(n-1)d(a_(1)a_(n))=d(a_(1)a_(2)+a_(2)a_(3)+ . . . .+a_(n-1)a_(n))` `rArr" "(n-1)a_(1)a_(n)=a_(1)a_(2)+a_(2)a_(3)+ . . .+a_(n-1)a_(n)` |
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| 480. |
If `1,log_9(3^(1-x)+2), log_3(4*3^x-1)` are in A.P then x equals toA. `log_(4)3`B. `log_(3)4`C. `1-log_(3)4`D. `log_(3)0.25` |
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Answer» Correct Answer - C Since 1, `log_(9)(3^(1-x)+2)andlog_(3)(4.3^(x)-1)` are in A.P `:." "2log_(9)(3^(1-x)+2)=1+log_(3)(4.3^(x)-1)` `rArr" "2log_(3)2(3^(1-x)+2)=log_(3)3+log_(3)(4.3^(x)-1)` `rArr" "(2)/(2)log_(3)(3^(1-x)+2)=log_(3){(4.3^(x)-1)}` `rArr" "(3^(1-x)+2)=3.(4.3^(x)-1)` `rArr" "(3)/(y)+2=12y-3,"where "y=3^(x)` `rArr" "12y^(2)-5y-3=0` `rArr" "(3y+1)(4y-3)=0` `rArr" "y=33//4" "[becausey=3^(x)gt0]` `rArr" "3^(x)=(3)/(4)` `rArr" "x=log_(3)(3//4)=log_(3)3-log_(3)4rArrx=1-log_(3)4` |
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| 481. |
If x,y,z be three positive prime numbers. The progression in which `sqrt(x),sqrt(y),sqrt(z)` can be three terms (not necessarily consecutive) isA. APB. GPC. HPD. None of these |
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Answer» Correct Answer - D `:.a,b,c` are positive prime numbers. Let `sqrt(a),sqrt(b),sqrt(c)` are 3 terms of AP. [not necessarily consecutive ] Then, `sqrt(a)=A+(p-1)D " " "....(i)"` `sqrt(b)=A+(q-1)D " " "....(ii)"` `sqrt(c)=A+(r-1)D " " "....(iii)"` [A and D be the first term and common difference of AP] `sqrt(a)-sqrt(b)=(p-q)D " " "....(iv)"` `sqrt(b)-sqrt(c )=(q-r)D " " "....(v)"` `sqrt(c)-sqrt(a)=(r-p)D " " "....(vi)"` On dividing Eq.(v), we get `(sqrt(a)-sqrt(b))/(sqrt(b)-sqrt(c))=(p-q)/(q-r)" " "....(vii)"` Since,p,q,r are natural numbers and a,b,c are positive prime numbers, so Eq. (vii) does not hold. So, `sqrt(a),sqrt(b),sqrt(c)` cannot be the 3 terms of AP. [not necessarily consecutive ] Similarly, we can show that `sqrt(a),sqrt(b),sqrt(c)` cannot be any 3 terms of GP andHP. [ not necessarily, consecutive]. |
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| 482. |
If x,y,z be three positive prime numbers. The progression in which `sqrt(x),sqrt(y),sqrt(z)` can be three terms (not necessarily consecutive) isA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - D | |
| 483. |
If `a (1/b+1/c), b(1/c+1/a), c(1/a+1/b)`are in A.P. prove that `a , b , c`are in A.P.A. a,b,c are in A.P.B. `(1)/(a),(1)/(b),(1)/(c)` are in A.P.C. a,b,c are in H.PD. `(1)/(a),(1)/(b),(1)/(c)` are in G.P. |
| Answer» Correct Answer - B | |
| 484. |
If the `m^(th),n^(th)andp^(th)` terms of an A.P. and G.P. be equal and be respectively x,y,z, thenA. `x^(y)y^(z)z^(x)=x^(2)y^(x)z^(y)`B. `(x-y)^(x)(y-z)^(x)=(z-x)^(z)`C. `(x-y)^(z)(y-z)^(x)=(z-x)^(y)`D. none of these |
| Answer» Correct Answer - A | |
| 485. |
If `p,q,r,s in N` and the are four consecutive terms of an A.P., then `p^(th),q^(th),r^(th)ands^(th)` terms of a G.P. are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 486. |
If `5^(th)` , `8^(th)` and `11^(th)` terms of a G.P. are p,q and s respectively thenA. p+r=2qB. q+r=2pC. p+q=2rD. pr=`q^(2)` |
| Answer» Correct Answer - D | |
| 487. |
If the `5^(th)` and `8^(th)` terms of a G.P. are 32 and 256 respectively . Then its `4^(th)` term isA. 8B. 12C. 16D. 20 |
| Answer» Correct Answer - C | |
| 488. |
Sum of all two digit numbers which when divided by 4 yield unity as remainder is.A. 1100B. 1200C. 1210D. none of these |
| Answer» Correct Answer - C | |
| 489. |
Sum of all two digit numbers which when divided by 4 yield unity as remainder is.A. 1012B. 1201C. 1212D. 1210 |
| Answer» Correct Answer - D | |
| 490. |
Sum of all two digit numbers which when divided by 4 yield unity as remainder is. |
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Answer» The two-digit numbers, which when divided by `4`, yield `1` as remainder, are, `13, 17, … 97`. This series forms an A.P. with first term, `a = 13` and common difference, `d =4`. Let `n` be the number of terms of the A.P. It is known that the nth term of an A.P. is given by,` a_n = a + (n -1) d`. `:. 97 = 13 + (n -1) (4)` `=> 4 (n –1) = 84` `=> n – 1 = 21` `=> n = 22` Now, `S_22 = n/2(2a+(n-1)d) = 22/2(2(13)+(22-1)4) = 11**110 = 1210` So, the required sum will be `1210`. |
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| 491. |
Find the sum of the following series: `5+55+555+ to n t e r m sdot`A. `(5)/(9)[(10)/(9)(1-10^(n))-1]`B. `(5)/(9)[(10)/(9)(10^(n)-1)-n]`C. `(5)/(9)[(1)/(9)(1-10^(n))-10_(n)]`D. none of these |
| Answer» Correct Answer - B | |
| 492. |
1.2+3.02+5.002+7.0002+ `cdots` to n terms =A. `n^(2)+(2)/(9)(1+(1)/(10^(n)))`B. `n^(2)-(2)/(9)(1-(1)/(10^(n)))`C. `n^(2)+(2)/(9)(1-(1)/(10^(n)))`D. none of these |
| Answer» Correct Answer - C | |
| 493. |
Find the sum of n terms of the series whose nth terms is (i) `n(n-1)(n+1)`. |
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Answer» (i) We have, `T_(n)=n(n-1)(n+1)=n^(3)-n` ` therefore " sum of n terms " S_(n)= sumT_(n)` `2=sumn^(3)+2sumn^(2)+sumn` `={(n(n+1))/(2)}^(2)-{(n(n+1))/(2)}` `=(n(n+1))/(2)-{(n(n+1))/(2)-1}` `=(n(n+1)(2)(n-1)(n+2))/(4)` |
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| 494. |
Find the sum of n terms of the series whose nth terms is (ii) `n^(2)+3^(n)`. |
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Answer» We have, `T_(n)=n^(2)+3^(n)` `" S_(n)= sumT_(n) =sumn^(2)+sum3^(n)` `=sumn^(2)+(3^(1)+3^(2)+3^(3)+"...."+3^(n))` `=(n(n+1)(2n+1))/(6)+(3(3^(n)-1))/((3-1))` `=(n(n+1)(2n+1))/(6)+(3)/(2)(3^(n-1))` |
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| 495. |
Find the sum of the sequence 7, 77, 777, 7777, . . . to n terms. |
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Answer» `S=7+77+777+7777..........n terms` `S=7(1+11+111.......n terms)` `S=7/9(9+99+999.......n terms)` `S=7/9((10-1)+(100-1)+(1000-1).......n terms)` `S=7/9(10+100+1000......n terms.-(1+1+1+1....n terms)` `S=7/9( (10((10)^n-1))/(10-1)-n)` `S=7/9( (10((10)^n-1))/(9)-n)` |
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| 496. |
Find the sum of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+`up to `n`terms. |
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Answer» Let `T_(n)` be the nth term of the given series. Then, `T_(n)=(1^(3)+2^(3)+3^(3)+"..."n^(3))/((1+3+5+"...."+(2n-1)))={(n(n+1))/(2)}^(2)/((n)/(2)(1+2n-1))` `((n+1)^(2))/(4)=(1)/(4)(n^(2)+2n+1)` Let `S_(n)` denotes the sum of n terms of the given series. Then, `S_(n)=sumT_(n)=(1)/(4)sum(n^(2)+2n+1)` `=(1)/(4)(sumn^(2)+sum2n+sum1)` `=(1)/(4){n(n+1)(2n+1))/(6)+(2n(n+1))/(2)+n}` `=(n)/(24){2n^(2)+3n+1+6n+6+6}` Hence, `S_(n)=(n(2n^(2)+9n+13))/(24)` |
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| 497. |
Find the sum of the series `1*2*3+2*3*4+3*4*5+"...." " upto n terms "` . |
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Answer» Here, `T_(n)={" nth term of " 1,2,3,"....""}` `xx={" nth term of " 2,3,4,"...."}xx={" nth term of " 2,3,4,"...."}` ` therefore T_(n)= n(n+1)(n+2)=n^(3)+3n^(2)+2n` ` therefore S_(n)= " Sum of n terms of the series "` `sum T_(n)=sumn^(3)+3sumn^(2)+2sumn` `={(n(n+1))/(2)}^(2)+3{(n(n+1)(2n+1))/(6)}+2{(n(n+1))/(2)}` `=(n(n+1))/(2)+{(n(n+1))/(2)+(2n+1)+2}` `=(n(n+1))/(4)(n^(2)+n+4n+2+4)` `=(n(n+1)(n+2)(n+3))/(4)` |
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| 498. |
In Illustration 6, which one of the following statement is correct ?A. `H_(1)gtH_(2)gtH_(3)gt . . .. `B. `H_(1)ltH_(2)ltH_(3)lt . . .. `C. `H_(1)gtH_(3)gtH_(5)gt . . .. andH_(2)ltH_(4)ltH_(6)lt . . ..`D. `H_(1)ltH_(3)ltH_(5)lt . . .. andH_(2)gtH_(4)gtH_(6)gt . . ..` |
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Answer» Correct Answer - B We have, `A_(1)gtG_(1)gtH_(1)` `H_(2)` is the harmonic mean of `A_(1)andH_(1)andA_(1)gtH_(1)`. `:." "A_(1)gtH_(2)gtH_(1)" "{:[(because"Harmonic mean lies"),("between the numbers")]:}` `H_(3)` is the Harmonic mean of `A_(2)andH_(2)andA_(2)gtH_(2)` `:." "A_(2)gtH_(3)gtH_(2)" "{:[(because"Harmonic mean lies"),("between the numbers")]:}` Similarly, we have `A_(3)gtH_(4)gtH_(3),A_(4)gtH_(5)gtH_(4) . . . .` Hence, `H_(1)ltH_(2)ltH_(3),A_(4) . . .. .` |
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| 499. |
`60^(2)-59^(2)+58^(2)-57^(2)+ cdots +2^(2)-1^(2)` =A. 1830B. 3180C. 1380D. none of these |
| Answer» Correct Answer - A | |
| 500. |
Find sum to n terms of the series `1+(2+3)+(4+5+6)+"...."`. |
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Answer» Now, number of terms in first bracket is `1`,in thr second bracket is 2, in the third bracket is 3, etc. Therefore, the number of terms in the nth bracket will be n. Let the sum of the given sries of n terms =S `therefore ` Number of terms in `S=1+2+3+"..."+n=(n(n+1))/(2)` Also, the first term of S is 1 and common difference is also 1. `therefore S= {(n(n+1))/(2)}/(2)[2*1+((n(n+1))/(2)-1)*1]` `= (n(n+1))/(8)(4+n^(2)+n-2)` `= (n(n+1)(n^(2)+n+2))/(8)` |
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