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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.
| 401. |
Let `a_1,a_2,a_3...` be in A.P. and `a_p,a_q,a_r` be in G.P. then value of `a_q/a_p` isA. `(q-p)/(r-p)`B. `(r-q)/(q-p)`C. `(q-p)/(r-q)`D. none of thses |
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Answer» Correct Answer - C It is given that `a_(p),a_(q),a_(r)` are in G.P. `rArr" "(a_(q))/(q_(p))=(a_(r))/(a_(q))` `rArr" "(a_(1)+(q-1)d)/(a_(1)+(p-1)d)=(a_(1)+(r-1)d)/(a_(1)+(q-1)d)` `rArr" "((q-p)d)/(a_(1)+(p-1)d)=((r-q)d)/(a_(1)+(q-1)d)" "[("Subtracting 1 from"),("both sides")]` `rArr" "(q-p)/(a_(p))=((r-q))/(a_(q))rArr(a_(p))/(a_(q))=(q-p)/(r-q)` |
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| 402. |
If `a_r>0, r in N` and `a_1.a_2,....a_(2n)` are in A.P then `(a_1+a_2)/(sqrta_1+sqrta_2)+(a_2+a_(2n-1))/(sqrta_2+sqrta_3)+.....+(a_n+a_(n+1))/(sqrt a_n+sqrta_(n+1))=`A. n-1B. `(n(a_(1)+a_(2n)))/(sqrt(a_(1))+sqrt(a_(n+1)))`C. `(n-1)/(sqrt(a_(1))+sqrt(a_(n+1)))`D. none of these |
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Answer» Correct Answer - B We have, `a_(1)+a_(2n)=a_(2)+a_(2n-1)=a_(3)+a_(2n-2)= . . . =a_(n)+a_(n+1)` Let d be the common difference of the given A.P. Then, Given expression `=(a_(1)+a_(2n)){(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+ . . .+(1)/(sqrt(a_(n))+sqrt(a_(n+1)))}` `=((a_(1)+a_(2n)))/(-d){sqrt(a_(1))-sqrt(a_(2))+sqrt(a_(2))-sqrt(a_(3))+ . . .+(1)/(sqrt(a_(n))-sqrt(a_(n+1)))}` `=((a_(1)+a_(2n)))/(d){sqrt(a_(n))-sqrt(a_(n+1))}` `=-((a_(1)+a_(2n)))/(d)xx(a_(1)-a_(n+1))/(sqrt(a_(1))+sqrt(a_(n+1)))` `=(n(a_(1)+a_(2n)))/(sqrt(a_(1))+sqrt(a_(n+1)))" "[becausea_(1)-a_(n+1)=-nd]` |
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| 403. |
Find the value of `n`so that `(a^(n+1)+b^(n+1))/(a^n+b^n)`may be the geometric mean between `aa n dbdot` |
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Answer» ` therefore (a^(n+1)+b^(n+1))/(a^(n)+b^(n))=sqrt(ab)` ` implies (b^(n+1)[((a)/(b))^(n+1)+1])/(b^(n)[((a)/(b))^(n)+1])=bsqrt((a)/(b)) implies (((a)/(b))^(n+1)+1)/(((a)/(b))^(n)+1)=((a)/(b))^((1)/(2))` Let `(a)/(b)= lambda` ` implies (lambda^(n+1)+1)/(lambda^(n)+1)=lambda^((1)/(2)) implies lambda^(n+1)+1=lambda^(n+(1)/(2))+lambda^((1)/(2)` ` implies lambda^(n+(1)/(2))(lambda^((1)/(2))-1)-(lambda^((1)/(2))-1)=0` ` implies (lambda^((1)/(2))-1)(lambda^(n+(1)/(2))-1)=0` ` implies lambda^((1)/(2))-1 ne 0 " " [ therefore a ne b]` ` therefore lambda^(n+(1)/(2))-1=0` `implies lambda^(n+(1)/(2))=1=lambda^(0)` `implies n+(1)/(2)=0 " or "n=-(1)/(2)` |
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| 404. |
Find the value of n so that `(a^(n+1)+b^(n+1))/(a^n+b^n)`may be the geometric mean between a and b. |
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Answer» As given expession is geometric mean of `a` and `b`. `(a^(n+1)+b^(n+1))/(a^n+b^n) = (ab)^(1/2)` `=>(a^(n+1)+b^(n+1)) = a^(n+1/2)b^(1/2)+b^(n+1/2)a^(1/2)` `=>a^(n+1)-a^(n+1/2)b^(1/2) = b^(n+1/2)a^(1/2)-b^(n+1)` `=>a^(n+1/2)(a^(1/2)-b^(1/2)) = b^(n+1/2)(a^(1/2)-b^(1/2))` `=>(a/b)^(n+1/2) = 1` As `a !=b`, `:. n+1/2 = 0` `n = -1/2` |
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| 405. |
If a, b, c are in A.P., b, c, d are in G.P. and `1/c ,1/d ,1/e`are in A.P. prove that a, c, e are in G.P. |
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Answer» As a,b and c are in A.P. `a-b = b-c` `a/b - 1 = 1-c/b->(1)` As, b,c and d are in G.P. `bd = c^2->(2)` As, 1/c,1/d,1/e are in A.P. `1/d-1/e = 1/e-1/d` `1 - d/c = d/e -1->(3)` From (2)`1-c/b = d/e-1` From (1)`a/b-1 = d/e-1` `ae = bd` As, `bd = c^2,``ae = c^2`,Thus, a,c and e are in G.P. |
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| 406. |
Find the sum of the following series up to n terms : `(1^3)/1+(1^3+2^2)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+dotdotdot` |
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Answer» `r^(th)` term in the given expression can be given as, `T_r = (1^3+2^3+3^3+...)/(1+3+5+...)` `=>T_r = (((r(r+1)))/2)^2/(r^2)` `=>T_r = (r^2(r+1)^2)/(4r^2)` `=>T_r = 1/4(r+1)^2` `=>T_r = 1/4(r^2+1+2r)` `:.` Sum of given series ` = sum T_n = 1/4 (sum n^2+ sum 1 + sum 2n)` `= 1/4[(n(n+1)(n+2))/6+n+2*(n(n+1))/2]` `=n/4[((n+1)(n+2))/6+(n+2)]` |
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| 407. |
How many terms of G.P. `3,3^2,3^3,dotdotdot`are needed to give the sum 120? |
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Answer» In the given GP, `a = 3, r = 3` `S_n = 120` `:. a(r^n-1)/(r-1) = 120` `=>3(r^n-1)/(3-1) = 120` `=>3^n -1 = 80 => 3^n = 81` `=> 3^n = 3^4` `n = 4` |
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| 408. |
Find the sum of the products of the corresponding terms of the sequences 2,4, 8, 16, 32 and 128, 32, 8, 2, `1/2`. |
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Answer» let the first sequence is `a_n` let the second sequence as `b_n` we have to find `C_n = a_nb_n` `= 256(2*1/4)^(n-1)` `sum C_n = s_5= (256(1-(1/2)^5))/(1- 1/2)` `= (256(1-1/32))/(1/2)` `=(2*256*31)/32` `= 16*31 = 496` answer |
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| 409. |
Find the sum to n terms of the series, whose `n^(t h)`terms is given by : `(2n-1)^2` |
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Answer» Here, `a_n = (2n-1)^2 = 4n^2+1-4n` So, `S_n = sum_(n=1)^na_n = sum_(n=1)^n4n^2+sum_(n=1)^n1-sum_(n=1)^n4n` `S_n = 4((n(n+1)(2n+1))/6)+n-4((n(n+1))/2)` `=2/3(n(n+1)(2n+1))+n-2(n(n+1))` `=2n(n+1)((2n+1)/3-1)+n` `=2n(n+1)((2n-2)/3)+n` `=4/3n(n+1)(n-1)+n` `=n(4/3(n^2-1)+1)` `=n((4n^2-1)/3)` `:. S_n = n/3(4n^2-1)` |
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| 410. |
Find the sum to indicated number of terms in each of the geometric progressions : 0.15, 0.015, 0.0015, . . . , 20 terms. |
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Answer» In the given GP, First-term, `a = 0.15` Common ratio, `r = 0.1` So, Sum of first 20 terms, `S_20 = (a(1-r^20))/(1-r)` `S_20 = (0.15(1-(0.1)^20))/(1-0.1) = 0.15/0.9(1-(0.1)^20)` `S_20 = 1/6(1-(0.1)^20)` |
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| 411. |
Does there exist a geometric progression containing 27,8 and 12 as three of its term ? If it exists, then how many such progressions are possible ? |
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Answer» Correct Answer - yes, infinite Let 27, 8, 12 be three terms of a GP `rArr t_(m) = 27, t_(n) = 8 and t_(p) = 12` `AR^(m-1) = 27, AR^(n-1) = 8` and `AR^(p -1) = 12` `:. R = ((27)/(8))^(1//(m-n)) and R = ((8)/(12))^(1//(n -p))` `rArr ((27)/(8))^(1//(m-n)) = ((2)/(3))^(1//(n -p))` `rArr 3^(3//(m-n)).3^(1//(n -p)) = 2^(1//(n-p)).2^(3//(m-n))` `rArr (3^((3)/(m-n) + (1)/(n-p)))/(2^((1)/(n-p) + (3)/(m-n))) = 1` `:. (3)/(m-n) + (1)/(n-p) = 0 and (1)/(n-p) + (3)/(m -n) = 0` `rArr 3(n -p) = n - m and 2n = 3p - m` Hence, there exists infinite GP for which 27, 8 and 12 as three of its terms. |
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| 412. |
If the sum of the first `2n`terms of the A.P. 2, 5, 8, ..., is equal to the sum of the first `n`terms of A.P. 57, 59, 61, ..., then `n`equals`10`b. `12`c. `11`d. `13`A. 10B. 12C. 11D. 13 |
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Answer» Correct Answer - C According to given condition, `rArr (2n)/(2) [2 xx 2 + (2n -1) xx 3] = (n)/(2) [2 xx 57 + (n -1) xx 2]` `rArr (4 + 6n - 3) = (1)/(2) (114 + 2n -2)` `rArr 6n += 57 + n -1 rArr 5n = 55` `:. N = 11` |
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| 413. |
If the sum of the first `2n`terms of the A.P. 2, 5, 8, ..., is equal to the sum of the first `n`terms of A.P. 57, 59, 61, ..., then `n`equalsA. 10B. 12C. 11D. 13 |
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Answer» Correct Answer - C We have, `(2n)/(2){2xx2+(2n-1)xx3}=(n)/(2){2xx57+(n-1)xx2}` `rArr" "n(6n+1)=n(n+56)` `rArr" "5n^(2)-55n=0rArrn=11" "[becausen!=0]` |
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| 414. |
If `0ltthetaltpi` , then the minmum value of `sin^(3) theta+cosec^(3) theta +2` is |
| Answer» Correct Answer - C | |
| 415. |
The sum of n terms of the following series `1+(1+x)+(1+x+x^2)+....` will beA. `(1-x^(n))/(1-x)`B. `(x(1-x^(n)))/(1-x)`C. `(n(1-x)-x(1-x^(n)))/((1-x^(2)))`D. `(1+x^(n))/(1-x)` |
| Answer» Correct Answer - C | |
| 416. |
In an arithmetic sequence `a_(1),a_(2),a_(3), . . . . .,a_(n)`, `Delta=|{:(a_(m),a_(n),a_(p)),(m,n,p),(1,1,1):}|` equalsA. 1B. -1C. 0D. mnp |
| Answer» Correct Answer - C | |
| 417. |
For a sequence, if `a_(1)=2` and `(a_(n+1))/(a_(n))=(1)/(3).` Then, `sum_(r=1)^(20) a_(r)` isA. `(20)/(2){4+19xx3}`B. `3(1-(1)/(3^(20)))`C. `2(1-3^(20))`D. none of these |
| Answer» Correct Answer - B | |
| 418. |
Let a,b,c, `in Riff(x)=ax^(2)+bx+c` is such that a+b+c=3 and `f(x+y)=f(x)+f(y)+xy," for all "x,y inR`, then `sum_(n=1)^(10) f(n)` is equal toA. 330B. 165C. 190D. 225 |
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Answer» Correct Answer - A We have, `f(x)=ax^(2)+bx+candf(x+y)=f(x)+f(y)+xy`. `:." "f(x+y)=f(x)+f(x)+f(y)+xy" for all "x,y inR` `rArr" "a(x+y)^(2)+b(x+y)+cax^(2)+bx+c+ay^(2)+c+xy" for all "x,y in R`. `rArr" "2axy=cxy" for all "x,yinR` `rArr" "(2a-1)xy-c=0rArra=(1)/(2),c=0` But a+b+c=3 `rArr" "(1)/(2)+b+0=3rArrb=(5)/(2)` `:." "f(x)=ax^(2)+bx+c` `rArr" "f(x)=(1)/(2)x^(2)+(5)/(2)x` Hence, `underset(n=1)overset(10)sumf(n)=(1)/(2)underset(n=1)overset(10)sumn^(2)+(5)/(2)underset(n=1)overset(10)sumn=(1)/(2)xx(10(10+1)(20+1))/(6)+(5)/(2)xx(10(10+1))/(2)=330` |
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| 419. |
All the terms of an AP are natural numbers and the sum of the first 20 terms is greater than 1072 and lss than 1162.If the sixth term is 32, thenA. first term is 7B. first term is 12C. common difference is 4D. common difference is 5 |
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Answer» Correct Answer - A::D Let the AP start with n and common differencem, then according to question, `n+5d=32` `n=32-5d " " "…..(i)"` and `1072ltn+(n+d)+"……."+(n+19d)lt1162` `1072lt20n+(19xx20)/(2)dlt1162` `1072lt640-100d+190dlt1162` `432lt90dlt522` `4.8ltdlt5.8` Let d is natutal number, so `d=5` `:.n=32-5xx5=7` First term is 7 and common difference is 5. |
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| 420. |
If the numbers `a , b , c , d , e`form an A.P. , then find the value of `a-4b+6c-4d+edot`A. 1B. 2C. 0D. none of these |
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Answer» Correct Answer - C Let D be the common difference of the A.P. Then, a-4b+6c-4d+e =a-4(a+D)+6(a+2D)-4(a+4D)=0 |
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| 421. |
If the set of natural numbers is partitioned into subsets `S_1={1},S_2={2,3},S_3={4,5,6}`and so on then find the sum of the terms in `S_(50)dot` |
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Answer» The number of terms in the groups are `1,2,3,"...."` `therefore` The number of terms in the 50th group =50 `therefore` The last term of 1st group = 1 The last term of 2nd group `=3=1+2` The last term of 3rd group `=6=1+2+3` `vdots " "vdots " "vdots " "vdots " "vdots " "vdots` The last term of 49th group `=+2+3+"......" + 49` `therefore` First term of 50th group `=1+(1+2+3+"..."+ 49)` `=1+(49)/(2)(1+49)=1226` `therefore S_(50_=(50)/(2){2xx1226+(50-1)xx1}` `=25xx2501=62525` |
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| 422. |
Let the sum of n, 2n, 3n terms of an A.P. be `S_1,S_2`and `S_3`, respectively, show that `S_3=3(S_2-S_1)`.A. `S_(3)=S_(1)+S_(2)`B. `S_(3)=2(S_(1)+S_(2))`C. `S_(3)=3(S_(2)-S_(1))`D. none of these |
| Answer» Correct Answer - C | |
| 423. |
In a certain AP, 5 times the 5th term is equal to 8 times the 8th term, its 13th term is |
| Answer» Correct Answer - A | |
| 424. |
If `t_n=1/4(n+2)(n+3)` for `n=1, 2 ,3,....` then `1/t_1+1/t_2+1/t_3+....+1/(t_(2003))=`A. `(4040)/(6063)`B. `(4040)/(6069)`C. `(8080)/(6065)`D. `(8080)/(6069)` |
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Answer» Correct Answer - D We have, `t_(n)=(1)/(4)(n+2)(n+3)` `rArr" "(1)/(t_(n))=(4)/((n+2)(n+3))` `rArr" "(1)/(t_(n))=4((1)/(n+2)-(1)/(n+3))` `rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4underset(n=1)overset(2020)sum((1)/(n+2)-(1)/(n+3))` `rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4{((1)/(3)-(1)/(4))+((1)/(4)-(1)/(5))+ . . . . +((1)/(2022)-(1)/(2023))}` `rArr" "underset(n=1)overset(2020)sum(1)/(t_(n))=4((1)/(3)-(1)/(2023))=(2020xx4)/(3xx2023)=(8080)/(6069)` |
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| 425. |
Find the sum to `n`terms of the series: `1/(1. 3)+1/(3. 5)+1/(5. 7)+`A. `(1)/(2n+1)`B. `(2n)/(2n+1)`C. `(n)/(2n+1)`D. `(2n)/(n+1)` |
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Answer» Correct Answer - C Let `T_(r)" be the "r^(th)` term of the given series. Then, `T_(r)=(1)/((2r-1)(2r+1)),r=1,2,3, . . .n` `rArr" "T_(r)=(1)/(2)((1)/(2r-1)-(1)/(2r+1))` Let S be the required sum. Then, `S=underset(r-1)overset(n)sumT_(r)` `S=(1)/(2){:[((1)/(1)-(1)/(3))+((1)/(3)-(1)/(5))+((1)/(5)-(1)/(7))+ . . . +((1)/(2n-1)-(1)/(2n+1))]:}` `rArr" "S=(1)/(2){1-(1)/(2n+1)}=(n)/(2n+1)` |
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| 426. |
The sum of the series `1/1.2-1/2.3+1/3.4-1/4.5+` ... isA. `(1)/(n+1)`B. `1-(1)/(n+1)`C. `(1)/(n+1)-1`D. `1+(1)/(n+1)` |
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Answer» Correct Answer - B Let `T_(r)" be the "r^(th)` term of the given series. Then, `T_(r)=(1)/(r(r+1)),r=1,2, . . .,n` `rArr" "T_(r)=(1)/(r)-(1)/(r+1),r=1,2, . . .,n` Let S be the required sum. Then, `S=underset(r=1)overset(n)sumT_(r)=underset(r=1)overset(n)sum((1)/(r)-(1)/(r+1))` `rArr" "S=(1-(1)/(2))+((1)/(2)-(1)/(3))+((1)/(3)-(1)/(4))+ . . . . +((1)/(n)-(1)/(n+1))` `rArr" "S=1-(1)/(n+1)` |
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| 427. |
If the fifth term of a G.P. is 2, then write the product of its 9terms.A. 256B. 512C. 1024D. none of these |
| Answer» Correct Answer - B | |
| 428. |
If the sum of an infinite G.P. be 3 and the sum of the squares of its term is also 3, then its first term and common ratio areA. `3//2,1//2`B. `1//2,3//2`C. `1,1//2`D. none of these |
| Answer» Correct Answer - A | |
| 429. |
If the surm of the first ten terms of the series,`(1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+........`, is `16/5m` ,then m is equal toA. 102B. 101C. 100D. 99 |
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Answer» Correct Answer - B Let S be the sum of the following series. `(1(3)/(5))^(2)+(2(2)/(5))^(2)+(3(1)/(5))^(2)+4^(2)+ . . . . .` upto 10 terms i.e. `S=((8)/(5))^(2)+((12)/(5))^(2)+((16)/(5))^(2)+((20)/(5))^(2)+((24)/(5))^(2)+ . . . . . .` upto 10 terms `rArr" "S=underset(r=1)overset(10)sum{(8+4(r-1))/(5)}^(2)` `rArr" "S=(16)/(25)underset(r=1)overset(10)sum(r+1)^(2)` `rArr" "S=(16)/(25){underset(r=0)overset(10)sum(r+1)^(2)-1^(2)}` `rArr" "(16)/(25)m=(16)/(25){(11xx12xx23)/(6)-1}" "{:[becauseS=(16)/(5)m"(given)"]:}` `rArr" "m=(1)/(5)(22xx23-1)rArrm=101` |
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| 430. |
`1^3-2^3+3^3-4^3+........+9^3` is equal toA. 425B. `-425`C. 475D. `-475` |
| Answer» Correct Answer - A | |
| 431. |
In the sum of first n terms of an A.P. is `cn^2`, then the sum of squares of these n terms isA. `(n(4n^(2)-1))/(6)c^(2)`B. `(n(4n^(2)+1))/(3)c^(2)`C. `(n(4n^(2)-1))/(3)c^(2)`D. `(n(4n^(2)+1))/(6)c^(2)` |
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Answer» Correct Answer - C We have, `S_(n)=cn^(2)` `:.` First term = c and common difference d=2c So, given A.P. is c,3c,5c,7c, . . . . . . Let S be the sum of first n terms of A.P. then, `S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)` `rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)` `rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)` ALITER We have, `S_(n)=cn^(2)` `:." "a_(n)=S_(n)-S_(n-1)=cn^(2)-c(n-1)^(2)=(2n-1)c` Let S be the sum of n terms of the sequence. Then, `S=underset(r=1)overset(n)suma_(r)^(2)` `rArr" "S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)` `rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)` `rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)` |
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| 432. |
The sum of infinite number of terms in G.P. is 20 and the sum of theirsquares is 100. Then find the common ratio of G.P.A. 5B. `3//5`C. `8//5`D. `1//5` |
| Answer» Correct Answer - B | |
| 433. |
If the sum of first n terms of an AP is `cn^(2)`, then the sum of squares of these n terms isA. `(n(4n^(2)-1)c^(2))/(6)`B. `(n(4n^(2)+1)c^(2))/(3)`C. `(n(4n^(2)-1)c^(2))/(3)`D. `(n(4n^(2)+1)c^(2))/(6)` |
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Answer» Correct Answer - C `:.S_(n)=cn^(2)` `:. T_(n)=S_(n)-S_(n-1)=c(2n-1)` `sumt_(n)^(2)=c^(2)sum(2n-1)^(2)` `=c^(2)sum(4n^(2)-4n+1)=c^(2){4sumn^(2)-4sumn+sum1}` `=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}` `=c^(2)n{(2)/(3)(2n^(2)3n+1)-2n-2+1}` `=(c^(2)n)/(3)(4n^(2)-1)=(n(4n^(2)-1)c^(2))/(3)`. |
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| 434. |
The coefficient of `x^(49)`in the product `(x-1)(x-3)(x+99)i s``-99^2`b. `1`c. `-2500`d. none of theseA. `-99^(2)`B. 1C. `-2500`D. none of these |
| Answer» Correct Answer - C | |
| 435. |
If `a,b,c` are in A.P and `a^2, b^2, c^2` are in H.P thenA. a=b=cB. 2 b=3 a+cC. `b^(2)=sqrt((ac//8))`D. none of these |
| Answer» Correct Answer - A | |
| 436. |
`0. 423 `isequivalent to the fraction`(94)/(99)`(b) `(49)/(99)`(c) `(491)/(990)`(d) `(419)/(990)`A. `(419)/(999)`B. `(419)/(990)`C. `(423)/(1000)`D. `(409)/(999)` |
| Answer» Correct Answer - B | |
| 437. |
Find the sum upto n terms of the series `a+aa+aaa+aaaa+"......",AA a in N " and "1 le a le 9.` |
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Answer» Let ` S = a+aa+aaa+aaaa+"......" " upto n terms " ` ` = a(1+11+111+1111+"......" "upto n terms") ` ` = (a)/(9)(9+99+999+9999+"......" "upto n terms") ` ` = (a)/(9){(10^(1)-1)+(10^(2)-1)+(10^(3)-1)+(10^(4)-1)+"......" "upto n terms"} ` ` = (a)/(9){10+10^(2)+10^(3)+"......" "upto n terms")-(1+1+1+"......" "upto n terms")}` ` = (a)/(9){(10(10^(n)-1))/(10-1)-n}=(a)/(9){(10)/(9)(10^(n)-1)-n}` |
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| 438. |
If N, the set of natural numbers is partitionaed into groups `S_(1)={1},S_(2)={2,3},S_(3)={4,5,6,7},S_(4)={8,9,10,11,12,13,14,15},"....",` find the sum of the numbers n `C_(50).` |
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Answer» The numbers of terms in the groups are `1,2,2^(2),2^(3),"…."` `therefore` The number of terms in the 50th group `=2^(50-1)=2^(49)` `therefore` The first term of 1st group `=1=2^(0)=2^(1-1)` The first term of 2nd group `=2=2^(1)=2^(2-1)` The first term of 3rd group `=4=2^(2)=2^(3-1)` `vdots" "vdots " " vdots " "vdots " " vdots "` The first term of 50th group `2^(50-1)=2^(49)` ` therefore S_(50)=(2^(49))/(2){2xx2^(49)+(2^(49)-1)xx1}` ` =2^(48)(2^(50)+2^(49)-1)` ` =2^(48)[(2^(49)(2+1)-1]=2^(48)(3*2^(49)-1)` |
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| 439. |
The sum of all `2` digited odd numbers isA. 2475B. 2530C. 4905D. 5049 |
| Answer» Correct Answer - A | |
| 440. |
Let `sum_(r=1)^(n)r^(4)=f(n)," then " sum_(r=1)^(n) (2r-1)^(4)` is equal toA. f(2n)-16f(n)B. f(2n)-7f(n)C. f(2n-1)-8f(n)D. none of these |
| Answer» Correct Answer - A | |
| 441. |
If the sum of the series 2, 5, 8, 11, ... is 60100, then find the valueof `ndot`A. 100B. 200C. 150D. 250 |
| Answer» Correct Answer - B | |
| 442. |
Find the sum upto n terms of the series `0.b+0.b b+0.b b b +0.b b b b+"......",AA " b" in N " and "1 le b le 9.` |
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Answer» Let `S=0.b+0.b b+0.b b b +0.b b b b+"......""upto n terms"` `=b(0.1+0.11+0.111 +0.1111+"......""upto n terms")` `=(b)/(9)(0.9+0.99+0.999+0.9999+"......""upto n terms")` `=(b)/(9){(1-0.1)+(1-0.01)+(1-0.001)+(1-0.0001)+"......""upto n terms")}` `=(b)/(9){(1+1+1+1)+"......""upto n terms")-(0.1+0.01+0.001+0.0001+"......""upto n terms")}` `=(b)/(9){n-((1)/(10)+(1)/(10^(2))+(1)/(10^(3))+(1)/(10^(4))+"......""upto n terms")}` `=(b)/(9){n-(((1)/(10)(1-((1)/(10))^(n)))/(1-(1)/(10))}=(b)/(9){n-(1)/(9)[1-((1)/(10))^(n)]}` |
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| 443. |
The pollution in a normal atmosphere is less than `0.01%`. Due to leakage of a gas from a factory, the pollution is increased to 20%. If every day 80% of pollution is neutralised, in how many days the atmosphere will be normal? |
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Answer» Let the pollution on 1st day =20 The pollution on 2nd day `=20xx20%=20(0.20)` The pollution on 3rd day `=20(0.20)^(2)` `" " vdots " "vdots " "vdots " "vdots ` Let in n days the atmosphere will be normal ` therefore " " 20(0*20)^(n-1)lt0.01` ` implies " " ((2)/(10))^(n-1)lt(1)/(2000)` Taking logarithm on base 10, we get `(n-1)(log2 -log10)ltlog1- log2000` `(n-1)(0*3010-1)lt0-(0*3010+3)` ` implies n-1gt(3*3010)/(0*6990)` ` implies ngt5*722` Hence, the atmosphere will be normal in 6 days. |
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| 444. |
An insect starts from a point and travels in a straight path 1 mm in the first second and half of the distance covered in the previous second in the succeeding second. In how much time would it reach a point 3 mm away from its starting point. |
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Answer» Distance covered by the insect in the 1st second =1 mm Distance coverd by it in the 2nd second `=1xx(1)/(2)=(1)/(2) " mm "` Distance covered by it in the 3rd second `=(1)/(2)xx(1)/(2)=(1)/(4) " mm "` `" "vdots " "vdots " "vdots " "vdots` The distance covered by the insect in 1st second, 2nd second, 3rd second,`"...."` are respectively `1,(1)/(2),(1)/(4),"....",` which are in GP with `a=1,r=(1)/(2)`. Let time taken by the insect in covering 3 mm be n seconds. `therefore 1+(1)/(2)+(1)/(4)+"...."+n " terms "=3` `implies (1*[1-((1)/(2))^(n)])/(1-(1)/(2))=3` `implies 1-((1)/(2))^(n)=(3)/(2)` `implies ((1)/(2))^(n)=-(1)/(2)` `implies 2^(n)=-2` which is impossible because `2^(n)gt0` `therefore` Our supposition is wrong. `therefore` There is no `n in N`, for which the insect could never 3 mm in n seconds. Hence, it will never to able to cover 3 mm. |
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| 445. |
The sum of the series `1/(2!)-1/(3!)+1/(4!)-...`uptoinfinity is(1) `e^(-2)`(2) `e^(-1)`(3) `e^(-1//2)`(4) `e^(1//2)` |
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Answer» `l^x=1+x+x^2/(2!)+x^3/(3!)+........` `if x=-1` `l^-1=1-1+1^2/(2!)-1^3/(3!)+........` `l^-1=1^2/(2!)-1^3/(3!)+1/4....` option`2` |
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| 446. |
If `(1)/(sqrt(x-1))+(1)/(sqrt(y-1))+(1)/(sqrt(z-1))gt0andx,y,z,` are in G.P., then `(logx^(2))^(-1),(logxz)^(-1),(logz^(2))^(-1)` are inA. A.P.B. G.P.C. H.P.D. none of these |
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Answer» Correct Answer - C It is given that `(1)/(sqrt(x-1))+(1)/(sqrt(y-1))+(1)/(sqrt(z-1))gt0rArrxgt1,ygt1,zgt1` It is also given that x,y,z are in G.P. `:." "logx,logy,logz` are in A.P. ltbgt `rArr" "2logx,2logy,2logz` are in A.P. `rArr" "logx^(2),logy^(2)logz^(2)` are in A.P. `rArr" "(logx^(2))^(-1),(logy^(2))^(-1),(logz^(2))^(-1)`, are in H.P. `rArr" "(logx^(2))^(-1),(logxz)^(-1),(logz^(2))^(-1)` are in H.P. |
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| 447. |
If `sum_(r=1)^(n) a_(r)=(1)/(6)n(n+1)(m+2)` for all `nge1`, then `lim_(ntooo) sum_(r=1)^(n) (1)/(a_(r))`, isA. 2B. 3C. `3//2`D. 6 |
| Answer» Correct Answer - A | |
| 448. |
If `a_(1)=2` and `a_(n)=2a_(n-1)+5` for `ngt1`, the value of `sum_(r=2)^(5)a_(r)` isA. 130B. 160C. 190D. 220 |
| Answer» Correct Answer - A | |
| 449. |
A person is to cout 4500 currency notes. Let `a_(n)` denotes the number of notes he counts in the nth minute. If `a_(1)=a_(2)="........"=a_(10)=150" and "a_(10),a_(11),"......",` are in AP with common difference `-2`, then the time taken by him to count all notes isA. 34 minB. 125 minC. 135 minD. 24 min |
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Answer» Correct Answer - A Till 10th minute, number of counted notes =1500 `:. 3000=(n)/(2){2xx148+(n-1)xx-2}=n(148-n+1)` `implies n^(2)-149n+3000=0` `implies (n-125)(n-24)=0` `:.n=125,24` `n=125` is not possible. `:.` n=24` `:. " Total time =10+24=34 min`. |
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| 450. |
If `t_(1)=1,t_(r )-t_( r-1)=2^(r-1),r ge 2`, find `sum_(r=1)^(n)t_(r )`. |
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Answer» `t_(1)=1 " and "t_(r )-t_( r-1)=2^(r-1),r ge 2` `t_(2)-t_(1)=2` `t_(3)-t_(2)=2^(2)` `t_(4)-t_(3)=2^(3)` ` vdots " " vdots " " vdots ` `t_(n)-t_(n-1)=2^(n-1)` Adding conlumnwise, we get `t_(n)-t_(1)=2+2^(2)+"........."+2^(n-1)` `t_(n)=1+2+2^(2)+"......"+2^(n-1)` `t_(n)=(1*(2^(n-1)))/(2-1) implies t_(n)=2^(n)-1` So,`sum_(r=1)^(n)t_(r )=t_(1)+t_(2)+"........"+t_(n)=(2-1)+(2^(2)-1)+"......"+(2^(n)-1)` `=(2+2^(2)+"......"+2^(n))-n=(2*(2^(n)-1))/((2-1))-n=2^(n+1)-2-n` `=2^(n+1)-n-2`. |
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