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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.
| 351. |
Suppose a,b,c are in A.P and `a^2,b^2,c^2` are in G.P If `a |
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Answer» `a,b,c` are in AP. Note whenever 3 terms are given in AP assume them to be m-d,m,m+d. `a+b+c=3m=3/2` `m=1/2` given `a^2,b^2,c^2` are in G.P. so `b^4=a^2 xx c^2` `m^4=(m-d)^2 xx (m+d)^2` on solving we get, `d^2(d^2-2m^2)=0` d cannot be equal to 0 so `(d^2-2m^2)=0` `m=1/2` thus `d= +-1/sqrt(2)` using +/- value will give the same AP but opposite order so it wont affect the terms. now `a=1/2-1/sqrt(2)` |
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| 352. |
if `alpha , beta ` be roots of `x^2-3x+a=0` and `gamma , delta ` are roots of `x^2-12x+b=0` and `alpha,beta,gamma,delta`(in order) form a increasing GP then find the value of `a & b` |
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Answer» ` alpha, beta,gamma ,delta ` are in GP so assume them to be `a,ar,ar^2,ar^3`where a is the first term and r common ratio .also`alpha +beta=a+ar=3`...(1) and `gamma+delta=ar^2+ar^3=12`...(2) dividing (2) by(1) we get=>`r^2=4,r=2` from(1) we get a=1 thus by product of roots formula we get =>`a^2r=A=2` =>`a^2r^5=B=32` |
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| 353. |
If `f(n)=sum_(r=1)^(n) r^(4)`, then the value of `sum_(r=1)^(n) r(n-r)^(3)` is equal toA. `(1)/(4){n^(2)(n+1)^(3)-4f(n)}`B. `(1)/(4){n^(3)(n+1)^(2)-4f(n)}`C. `(1)/(4){n^(2)(n+1)^(2)-4f(n)}`D. none of these |
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Answer» Correct Answer - B We have, `underset(r=1)overset(n)sumr(n-r)^(3)=underset(r=1)overset(n)sum(n-r)r^(3)=underset(r=1)overset(n)sum(nr^(3)-r^(4))` `=n underset(r=1)overset(n)sumr^(3)-underset(r=1)overset(n)sumr^(4)` `=n{(n(n+1)^(2))/(2)}^(2)-f(n)` `=(n^(3)(n+1)^(2))/(4)-f(n)=(1)/(4){(n^(3)(n+1)^(2)-4f(n)}` |
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| 354. |
Let `T_r` be the rth term of an A.P. whose first term is -1/2 and common difference is 1, then `sum_(r=1)^n sqrt(1+ T_r T_(r+1) T_(r+2) T_(r+3))` |
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Answer» Here, ` a= -1/2 and d = 1` `:. T_r = a+(r-1)d = -1/2+(r-1)1 = r-3/2` `:. T_rT_(r+1)T_(r+2)T_(r+3) = (r-3/2)(r-1/2)(r+1/2)(r+3/2)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = (r-3/2)(r+3/2)(r-1/2)(r+1/2)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = (r^2-9/4)(r^2-1/4)` `=> T_rT_(r+1)T_(r+2)T_(r+3) = r^4-10/4r^2+9/16` `:. 1+T_rT_(r+1)T_(r+2)T_(r+3) = r^4-10/4r^2+25/16 = (r^2-5/4)^2` `:. sum_(r=1)^n sqrt(1+T_rT_(r+1)T_(r+2)T_(r+3) )= sum_(r=1)^n (r^2-5/4)` `= sum_(r=1)^n r^2 - sum_(r=1)^n 5/4` `= (n(n+1)(2n+1))/6-5/4n` So, option `A` is the correct answer. |
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| 355. |
If sum of x terms of a series is `S_(x)=(1)/((2x+3)(2x+1))` whose `r^(th)` term is `T_(r)`. Then, `sum_(r=1)^(n) (1)/(T_(r))` is equal toA. `(1)/(4)sum(2r+1)(2r-1)(2r+3)`B. `-(1)/(4)sum(2r+1)(2r-1)(2r+3)`C. `sum(2r+1)(2r-1)(2r+3)`D. none of these |
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Answer» Correct Answer - B we have, `S_(x)=(1)/((2x+3)(2x+1))` `:." "T_(x)=S_(x)-S_(x-1)` `rArr" "T_(x)=(1)/((2x+3)(2x+1))-(1)/((2x+1)(2x-1))` `rArr" "(-4)/((2x-1)(2x+1)(2x+3))` `:." "underset(r=1)overset(n)sum(1)/(T_(r))=-(1)/(4)underset(r=1)overset(n)sum(2r-1)(2r+1)(2r+3)` |
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| 356. |
If `sum_(r=1)^(n)T_(r)=(n)/(8)(n+1)(n+2)(n+3)," find "sum_(r=1)^(n)(1)/(T_(r))`. |
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Answer» `S_(n)=sum_(r=1)^(n)T_(r)=((n+1)(n+2)(n+3))/(8)` `T_(r)=S_(r)-S_(r-1)=(r(r+1)(r+2)(r+3))/(8)-((r-1)(r+1)(r+2))/(8)` `=((r+1)(r+2))/(2)` `(1)/(T_(r))=(2)/(r(r+1)(r+2))=((r_+2)-r)/(r(r+1)(r+2))=((1)/(r(r+1))-(1)/(r(r+1)(r+2)))` `sum_(r=1)^(n)(1)/(T_(r))=sum_(r=1)^(n)((1)/(r(r+1))-(1)/(r(r+1)(r+2)))` `=sum_(r=1)^(n){((1)/(r)-(1)/(r+1))-((1)/(r+1)-(1)/(r+2))}` `=((1)/(1)-(1)/(n+1))-((1)/(2)-(1)/(n+2))` `=(1)/(2)+(1)/(n+2)-(1)/(n+2)=(n(n+3))/(2(n+1)(n+2))`. |
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| 357. |
A person is to count 4500 currency notes. Let an denote the numberof notes he counts in the nth minute. If `a_1=""a_2="". . . . . .""=""a_(10)=""150`and `a_(10),""a_(11),"". . . . . .`are in A.P. with commondifference 2, then the time taken by him to count all notes is(1) 34 minutes(2) 125 minutes (3) 135minutes (4) 24 minutesA. 125 minutesB. 135 minutesC. 24 mintutesD. 34 minutes |
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Answer» Correct Answer - D The number of notes counted in first 10 minutes `=150xx10=1500` Suppose the person counts the remaining 3000 currency notes in n minutes. Then, 3000=Sum of terms of an AP with first term 148 and common difference -2 `rArr" "3000=(n)/(2){2xx148+(n-1)xx(-2)}` `rArr" "3000=n(149-n)` `rArr" "n^(2)-149n+3000=0` `rArr" "(n-125)(n-24)=0rArrn=125,24` Clearly, n-125 is not possible. `:.` Total time taken =(10+42)mintutes =34 minutes. |
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| 358. |
`1+2/3+6/(3^2)+10/(3^3)+14/(3^4)+` |
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Answer» `S= 1 + 2/3 + 6/3^2 + 10/3^3 + ..........`(i) `1/3 S = 1/3 + 2/3^2 + 6/3^3 + 10/3^4 + .... `(ii) `(i)- (ii)` `2s/3 = 2/3 + 2/3 + 4/3^2 + 4/3^3 + ......` `2S/3 = 4(1/3 + 1/3^2 + 1/3^3 + ......)` `2S/3 = (4 xx 1/3)/(1 -1/3)` `= (4 xx 1 xx3)/(3 xx2) ` `2S/3 = 2` `S= 3 ` option 2 is correct |
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| 359. |
Let a,b,c are respectively the sums of the first n terms, the next n terms and the next n terms of a GP. Show that a,b,c are in GP. |
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Answer» ` a =S_n =(a(r^(n) -1))/( r-1) " "...(i) ` ` b= S _ (2n ) -S_n =( a(r^(2n) -1))/( (r-1) ) -(a(r^(n) -1))/( (r-1) ) =(a(r^(n) -1))/( (r-1) )(r^(n))" "..(ii) ` ` c= S_(3n ) -S_(2n ) =( a(r^(3n) -1))/( (r-1) ) =(a( r^(2n) - 1))/( (r-1)) ` `= ( a(r^(n) -1))/( (r-1) ) (r^(2n )+ r^(n) +1-r^(n) -1)= ( a( r^(n) -1))/( (r-1) ) (r^(n))^(2) ` From Eqs.(i) (ii) and (iii) ` b^(2) ` =ac,so a,b,c are in GP. |
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| 360. |
The GM between -9 and -16, isA. 12B. -12C. -13D. None of these |
| Answer» Correct Answer - A | |
| 361. |
`1,2,4,8,16,"…."` |
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Answer» Here, `a=1` and `r=(2)/(1)=(4)/(2)=(8)/(4)=(16)/(8)="..."=2 i.e.a=1 " and "r=2` Increasing GP `(agt0,rgt1)` |
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| 362. |
`9,3,1,(1)/(3),(1)/(9),"…."` |
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Answer» Here, `a=9` and `r=(3)/(9)=(1)/(3)=((1)/(3))/(1)=((1)/(9))/((1)/(3))="..."=(1)/(3) i.e.a=9,r=(1)/(3)` Decreasing GP `(agt0,0 lt r lt 1)` |
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| 363. |
`-2,-6,-18,"…."` |
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Answer» Here, `a=-2` and `r=(-6)/(-2)=(-18)/(-6)="..."=3` `I.e. A=-8,r=(1)/(2)` Decreasing GP `(alt0,rgt1)` |
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| 364. |
`-8,-4,-2,-1,-(1)/(2),"…"` |
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Answer» Here,`a=-8` and `r=(-4)/(-8)=(-2)/(-4)=(-1)/(-2)=((-1)/(2))/(-1)="..."=(1)/(2)` `i.e. a=-8,r=(1)/(2)` Increasing GP `(alt0,0ltrlt1)` |
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| 365. |
`5,-10,20,"…."` |
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Answer» Here, `a=5` and `r=(-10)/(5)=(20)/(-10)="..."=-2 i.e., a=5, r=-2` Neither increasing nor decreasing `(rlt0)` |
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| 366. |
`5,5,5,5,"…."` |
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Answer» Here, `a=5` and `r=(5)/(5)=(5)/(5)=(5)/(5)="..."=1 i.e., a=5, r=1` Constant GP `(r=1)` |
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| 367. |
`1,1+I,2i,-2+2i,"…."i= sqrt(-1)` |
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Answer» Here, `a=1` and `r=(1+i)/(1)=(2i)/(1+i)=(-2+2i)/(2i)="..."` `=(1+i)=(2i(1+i))/((1+i)(1+i))=((-1+i)i)/(i)^(2)="..."` `=(1+i)=(1+i)=(1+i)="..."` `i.e.,a=1,r=1+i` Imaginary GP `(r= " imaginary ")` |
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| 368. |
The nature of the `S_(n)=3n^(2)+5n` series isA. APB. GPC. HPD. AGP |
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Answer» Correct Answer - A `:. S_(n)=3n^(2)+5n` `:. T_(n)=S_(n)-S_(n-1)` `=(3n^(2)+5n)-[3(n-1)^(2)+5(n-1)]` `=3(2n-1)+5=6n+2` The nth term is a linear function in n. Hence, sequence must be an AP. |
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| 369. |
For the `S_(n)=3n^(2)+5n` sequence, the number 5456 is theA. 153th termB. 932th termC. 707th termD. 909th term |
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Answer» Correct Answer - D Given, `T_(n)=5456` `implies 6n+2=5456 implies 6n-5454` `:. N=909` `:.` The number 5456 is the 909th term. |
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| 370. |
Which of the following sequences are unbounded?A. `(1+(1)/(n))^(n)`B. `((2n+1)/(n+2))`C. `(1+(1)/(n))^(n^(2))`D. `tan n` |
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Answer» Correct Answer - C::D (c ) `:.a_(n) =(1+(1)/(n))^(n^(2))` For `n=1,a_(1)=2`, for `n=2,a_(2)=(1+(1)/(2))^(4)=((3)/(2))^(4)=(3^(4))/(2^(4))=(81)/(16)=5.06" " [" approximate "]` `lim_(nto oo)(1+(1)/(n))^(n^(2))=e^(lim_(nto oo)(1)/(n)xx^(n^(2)))=e^(lim_(nto oo)n)=e^(oo)=oo` `:.{a_(n)}` represents unbounded sequence. (b) `:.a_(n)=tann` `a_(n)=n+(n^(3))/(3)+(2)/(15)n^(5)+"...."+oo` and we know that `-oolttan nltoo` So, `{a_(n)}` is unbounded sequence. |
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| 371. |
In a `GP` if the `(m+n)th` term is `p` and `(m-n)th` term is `q` then `mth` term isA. `p((q)/(p))^((m)/(2n))`B. `sqrt(pq)`C. `sqrt((p)/(q))`D. None of these |
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Answer» Let a be the first term and r be the common ratio, then `t_(m+n)=p implies ar^(m+n-1)=p " .....(i)"` `t_(m-n)=q implies ar^(m-n-1)=q " .....(ii)"` From Eqs. (i) and (ii), we get `ar^(m+n-1)xxar^(m-n-1)=pxxq` ` implies a^(2)r^(2m-2)=pq implies ar^(m-1)=sqrt(pq)` ` implies t_(m)=sqrt(pq)` Hence, (b) is the correct answer. |
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| 372. |
Show that the the sequence defined by ` T_(n) = 3n^(2) +2 ` is not an AP. |
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Answer» We have, `t_n=3n^(2)+2` On replacing n by `(n-1),` we get `t_(n-1)=3(n-1)^(2)+2` `implies t_(n-1)=3n^(2)-6n+5` `therefore t_n-t_(n-1)=(3n^(2)+2)-(3n^(2)-6n+5)` `" "=6n-3` Clearly, `t_n-t_(n-1)` is not independent of n and therefore it is not constant. So, the given sequence is not an AP. |
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| 373. |
Let a sequence`{a_(n)}` be defined by `a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(3n)`, thenA. `a_(2)=(11)/(12)`B. `a_(2)=(19)/(20)`C. `a_(n+1)-a_(n)=((9n+5))/((3n+1)(3n+2)(3n+3))`D. `a_(n+1)-a_(n)=(-2)/(3(n+1))` |
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Answer» Correct Answer - B::C `:.a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(3n)` `a_(n)=(1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."(1)/(n+2n)` `a_(n)=sum_(alpha=1)^(2n)(1)/(n+alpha)` `a_(2)=sum_(alpha=1)^(4)(1)/(2+alpha)=(1)/(3)+(1)/(4)+(1)/(5)+(1)/(6)=(20+15+12+10)/(60)` `=(57)/(60)=(19)/(20)` Now, `a_(n+1)-a_(n)=((1)/(n+2)+(1)/(n+3)+"......."+(1)/(3n+3))-((1)/(n+1)+(1)/(n+2)+"......+(1)/(3n))` `=(1)/(3n+1)+(1)/(3n+2)+(1)/(3n+3)-(1)/(n+1)` `=(1)/(3n+1)+(1)/(3n+2)-(2)/(3(n+1))` `=(9n^(2)+15n+6+9n^(2)+12n+3-18n^(2)-18n-4)/((3n+1)+(3n+2)+(3n+3))` `=(9n+5)/((3n+1)+(3n+2)+(3n+3))` |
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| 374. |
The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a : b = `(m+sqrt(m^2-n^2)):(m-sqrt(m^2-n^2))`. |
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Answer» Here, we are given,`((a+b)/2)/sqrt(ab) = m/n` `=>(a+b)/(2sqrt(ab)) = m/n->(1)` Squaring both sides, `=>((a+b)^2)/(4ab) = m^2/n^2` `=>((a+b)^2-4ab)/(4ab) = (m^2-n^2)/n^2` `=>((a-b)^2)/(4ab) = (m^2-n^2)/n^2` `=>(a-b)/(2sqrt(ab)) = sqrt(m^2-n^2)/n->(2)` Adding (1) and (2), `(2a)/(2sqrt(ab)) = (m+sqrt(m^2-n^2))/n ->(3)` Subtracting (2) from (1), `(2b)/(2sqrt(ab)) = (m-sqrt(m^2-n^2))/n ->(4)` Dividing (3) by (4), `a/b = (m+sqrt(m^2-n^2))/ (m-sqrt(m^2-n^2))` So, `a:b = (m+sqrt(m^2-n^2)) : (m-sqrt(m^2-n^2))` |
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| 375. |
If the sum of n terms of an A.P. is `3n^2+5`and its `m^(t h)`term is 164, find the value of m. |
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Answer» Here, we are given, `n/2(2a+((n-1)d) = 3n^2+5n` `(2a-d)+nd = 6n+10` `d = 6 and 2a-d=10` `d=6 and a =8` Now, `m^(th)` term `= a+(m-1)d = 164` `=>8+6(m-1) = 164` `=>6(m-1) = 156` `=> m = 27` |
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| 376. |
The geometric mean between -9 and -16 is`12`b. `-12`c. `-13`d. none of theseA. 12B. -12C. -13D. 13 |
| Answer» Correct Answer - B | |
| 377. |
The sum of n terms of an A.P. is `3n^(2)+5`. The number of term which equals 159, isA. 13B. 21C. 27D. none of these |
| Answer» Correct Answer - C | |
| 378. |
If `p^(th),q^(th) and r^(th)` terms of an A.P. are in G.P., then the common ratio of G.P. is-A. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 379. |
If the `2^(nd), 5^(th) and 9^(th)` terms of a non-constant A. P. are in G.P, then the common ratio of this G. P. isA. `(8)/(5)`B. `(4)/(3)`C. 1D. `(7)/(4)` |
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Answer» Correct Answer - B Let a be the first term and d be the common difference of the A.P. It is given that `a_(2)=a+d,a_(5)=a+4danda_(9)=a+8d` are in G.P. `:." "a_(5)^(2)=a_(2)xxa_(9)` `rArr" "(a+4d)^(2)=(a+d)(a+8d)` `rArr" "a=8d` `:." "a_(2)=9d,a_(5)=12d,a_(9)=16d` The common ratio of the G.P. `(a_(5))/(a_(2))=(12d)/(9d)=(4)/(3)`. |
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| 380. |
A pack contains `n`cards numbered from 1 to `n`. Two consecutive numbered cards are removed from the pack and the sumof the numbers on the remaining cards is 1224. If the smaller of het numberson the removed cards is `k ,`then `k-20=`____________. |
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Answer» Correct Answer - 5 Let number of removed cards be k and `(k+1)` `:. (n(n +1))/(2) - k - (k +1) = 1224` `rArr n^(2) +n-4k = 2450 rArr n^(2) + n - 2450 = 4k` `rArr (n + 50) (n -49) = 4k` `:. N gt 49` Let `n = 50` `:. 100 = 4k` `rArr k = 25` Now `k - 20 = 5` |
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| 381. |
If `(10)^(9)+2(11)^(1)(10)^(8)+3(11)^(2)(10)^(7)+"........"+(10)(11)^(9)=k(10)^(9)`, then k is equal toA. 100B. 110C. `(121)/(10)`D. `(441)/(100)` |
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Answer» Correct Answer - A The given series can be written as `k = 1+2(11/10)+3(11/10)^2""+...+9(11/10)^8""+10(11/10)^9" ...(i)"` On multiplying both sides by `(11/10)`, then `(11k)/10=(11/10)+2(11/10)^2 +3(11/10)^3""+...+9(11/10)^9""+10(11/10)^10" ...(ii)"` Now, on subtracting Eq. (ii) frm Eq. (i), then `-k/10=(11/10)+(11/10)^2""+...+(11/10)^2-10(11/10)^10` `=(1*{(11/10)^10-1})/((11/10-1))-10(11/10)^10` `rArr" "k = -100*{(11/10)^10-1}+100(11/10)^10 = 100` |
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| 382. |
The sum of first 20 terms of the sequence `0.7, 0.77, 0.777,.......,` is :A. `(7)/(81)(179-10^(-20))`B. `(7)/(9)(99-10^(-20))`C. `(7)/(9)(99+10^(-20))`D. `(7)/(81)(179+10^(-20))` |
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Answer» Correct Answer - C Let `S=0.7+0.77+0.777+ . . . +` upto 20 terms Then, `S=(7)/(10)+(77)/(100)+(777)/(1000)+. . . . ` upto 20 terms `rArr" "S=7{(1)/(10)+(11)/(100)+(111)/(1000)+ . . . ." upto 20 terms"}` `rArr" "S=(7)/(9){(9)/(10)+(99)/(100)+(999)/(1000)+ . . . ." upto 20 terms"}` `rArr" "S=(7)/(9){(1-(1)/(10))+(1-(1)/(100))+(1-(1)/(1000))+ . . . +(1-(1)/(10^(20)))}` `rArr" "S=(7)/(9){20-((1)/(10)+(1)/(10^(2))+ . . . .+(1)/(10^(20)))}` `rArr" "S=(7)/(9){20-(1)/(10)((1-(1)/(10^(20)))/(1-(1)/(10)))}` `rArr" "S=(7)/(9){20-(1)/(9)(1-(1)/(10^(20)))}` `rArr" "S=(7)/(9){(179)/(9)+(1)/(9)((1)/(10))^(20)}=(7)/(81)(179+10^(-20))` |
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| 383. |
If `(10)^9 + 2(11)^1 (10)^8 + 3(11)^2 (10)^7+...........+10 (11)^9= k (10)^9` , then k is equal to :A. 100B. 110C. `(121)/(10)`D. `(441)/(100)` |
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Answer» Correct Answer - A Let `S=(10)^(9)+2(11)^(1)(10)^(8)+3(11)^(2)(10)^(7)+ . . . .+10(11)^(9)`. Then, `(11)/(10)S=11(10)^(8)+2(11)^(2)(10)^(7)+ . . . .+9(11)^(9)+(11)^(10)` `:." "S-(11)/(10)S=10^(9)+{(11)(10)^(8)+(11)^(2)(10)^(7)+ . . .+(11)^(9)}-(11)^(10)` `rArr-(S)/(10)=(10^(9){1-((11)/(10))^(10)})/(1-(11)/(10))-(11)^(10)` `rArr-(S)/(10)=-10^(10){1-((11)/(10))^(10)}-(11)^(10)` `rArr" "S=10^(11)-10(11)^(10)+10(11)^(10)=10^(11)=100(10^(9))` `rArr" "k(10^(9))=100(10^(9))rArrk=100` |
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| 384. |
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.?A. a,c,e are in G.P.B. a,b,e are in G.P.C. a,b,e are in G.P.D. a,c,e are in G.P. |
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Answer» Correct Answer - A It si given that : a,b,c are in A.P. `rArr2b=a+c` . . .(i) b,c,d are in G.P. `rArrc^(2)=bd` . . .(ii) `(1)/(c),(1)/(d),(1)/(e)` are in A.P. `rArr(2)/(d)=(1)/(c)+(1)/(e)rArrd=(2ce)/(c+e)` . . .(iii) From (ii) and (iii), we obtain `c^(2)=b((2ce)/(c+e))" [On eliminating d]"` `rArr" "2be=c^(2)+ce` `rArr" "(a+c)e=c^(2)+ce` `rArr" "c^(2)=ae" [Using (i)]` `rArr" "a,c,e,` are in G.P. |
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| 385. |
A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then `k-20` is equal to |
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Answer» Let two consecutive nymbers are k and `k+1` such that `1leklen-1`, then `(1+2+3+"……."+n)-(k+k+1)=1224` `implies (n(n+1))/(2)-(2k+1)=1224" or " k=(n^(2)+n-2450)/(4)` Now, `1le (n^(2)+n-2450)/(4) le n-1 implies 49ltnlt51` `:. n =50 implies k=25` Hence, `k-20=25-20=5`. |
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| 386. |
Let `S_(n)=sum_(k=1)^(4n) (-1)^((k(k+1))/(2))k^(2)`. Then, `S_(n)` can take the value (s)A. 1056 and 1332B. 1056 and 1088C. 1120 and 1332D. 1332 and 1432 |
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Answer» Correct Answer - A We have `S_(n)=underset(k=1)overset(4n)sum(-1)^((k(k+1))/(2))k^(2)` `rArr" "S_(n)=-1^(2)-2^(2)+3^(2)+4^(2)-5^(2)-6^(2)+7^(2)+8^(2). . . .` upto 4n terms `rArr" "S_(n)=(3^(2)-1^(2))+(4^(2)-2^(2))+(7^(2)-5^(2))+(8^(2)-6^(2))+ . . .. ` up to 2n terms `rArr" "S_(n)=2(4+6+12+14+20+22 . . . .` up to 2n terms `rArr" "S_(n)=2{underset("n-terms")(4+12+20+ . . .)+underset("n-terms")(6+14+22+ . . . )}` `rArr" "S_(n)=2{:[(n)/(2){8+(n-1)xx8}+(n)/(2){12+(n-1)xx8}]:}` `rArr" "S_(n)=8n^(2)+8n^(2)+4n` `rArr" "S_(n)=4n(4n+1)` `rArr" "S_(n)`= Product of a multiple of 4 and its successor. Clearly, `1056=32xx33=(4xx8)((4xx8)+1)and1332=326xx37=(4xx9)+1)` are products of multiple of 4 and its successor. Hence, `S_(n)` can take values 1056 and 1332. |
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| 387. |
A G.P. consists of 2n terms. If the sum of the terms occupying the odd places is `S_(1)`, and that of the terms in the even places is `S_(2)`, then `S_(2)/S_(1)`, isA. independent of aB. independent of rC. independent of a and rD. dependent on r |
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Answer» Correct Answer - D We have `a_(1)=a+ar^(2)+ar^(4)+ . . .+ar^(2n-2)` `and,S_(2)=ar+ar^(3)+ar^(5)+ . . .ar^(2n-1)` Clearly, `S_(2)/S_(1)=r`, which depends on r. |
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| 388. |
The ratio of the sum of `n`terms of two A.P. ` s`is `(7n+1):(4n+27)`. Find the ratio of their nth terms.A. (14+6) : (8n-23)B. (14n-6) : (8n+23)C. 7n-1 : 4n-27D. (8n+23) : (14n-6) |
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Answer» Correct Answer - B The ratio of then `n^(th)` terms if obtained by replacing n by (2n-1) in the given ratio. Hence, required ratio = (7(2n-1)+1):(4(2n-1)+27)=(14n-6):(8n+23) |
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| 389. |
Let `S_(n)=sum_(k=1)^(4n)(-1)^((k(k+1))/(2))*k^(2)`, then `S_(n)` can take valueA. 1056B. 1088C. 1120D. 1332 |
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Answer» Correct Answer - A::D `S_(n)=-1^(2)-2^(2)+3^(2)+4^(2)-5^(2)-6^(2)+7^(2)+8^(2)-"........"+(4n-1)^(2)+(4n)^(2)` `=(3^(2)-1^(2))+(4^(2)-2^(2))+(7^(2)-5^(2))+(8^(2)-6^(2))+"........"+[{(4n-1)^(2)-(4n-3)^(2)}+{(4n)^(2)-(4n-2)^(2)}]` `=4[2+3+6+7+10+11+"......"+(4n-2)+(4n-1)]` `=8{(1+3+5+"......"+(2n-1)}+4{3+7+11+"........."+(4n-1)}` `=16n^(2)+4n=4n(4n+1),n inN` Satisfied by a and d where n=8,9, respectively. |
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| 390. |
If `sum_(r=1)^(n) r(sqrt(10))/(3)sum_(r=1)^(n)r^(2),sum_(r=1)^(n) r^(3)` are in G.P., then the value of n, isA. 2B. 3C. 4D. non-existent |
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Answer» Correct Answer - C We have, `(10)/(9)(underset(r=1)overset(n)sumr^(2))^(2)=(underset(r=1)overset(n)sumr)(underset(r=1)overset(n)sumr^(3))` `rArr" "(10)/(9){(n(n+1)(2n+1))/(6)}^(2)=(n(n+1))/(2)xx{(n(n+1))/(2)}^(2)` `rArr" "(10)/(9)((2n+1)/(3))^(2)=(n(n+1))/(2)` Clearly, n=4 satisfies this equation. |
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| 391. |
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6: 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. isA. 5B. 6C. 8D. 9 |
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Answer» Correct Answer - C Let the term and common difference of the A.P. be a and respectively. It is given that `(S_(7))/(S_(11))=(6)/(11)rArr((7)/(2){2a+6d})/((11)/(2){2a+10d})=(6)/(11)rArr(7(a+3d))/((a+5d))=6rArra=9d` It is also given that `rArr" "130lta_(7)lt140` `rArr" "130lt9d+6dlt140rArr130lt15dlt140rArrd=9`. |
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| 392. |
In a sequence of `(4n+1)`terms, the first `(2n+1)`terms are n A.P. whose common difference is 2, and the last `(2n+1)`terms are in G.P. whose commonratio is 0.5 if the middle terms of the A.P. and LG.P. are equal ,then themiddle terms of the sequence is`(n .2 n+1)/(2^(2n)-1)`b. `(n .2 n+1)/(2^n-1)`c. `n .2^n`d. none of theseA. `(n*2^(n+1))/(2^(n)-1)`B. `(n*2^(n+1))/(2^(2n)-1)`C. `n*2^(n)`D. none of these |
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Answer» Correct Answer - A Let `a_(1),a_(2),a_(3), . . . ,a_(2n),a_(2n+1), . . . ,a_(4n),a_(4n+1)` be the given sequence. Then, `a_(2n+1)=a_(1)+4nanda_(4n+a)=a_(2n+1)xx((1)/(2))^(2n)` `rArr" "a_(2n+1)=a_(1)+4nanda_(4n+1)=((a_(1)+2nd))/(2^(2n))` Also, Middle term of first (2n+1) terms = Middle term of the last (2n+1) terms `rArr" "a_(n+1)=a_(3n+1)` `rArr" "a_(1)+2n=a_(2n+1)((1)/(2))^(n)` `rArr" "a_(1)+2n=(a_(1)+4n)xx(1)/(2^(n))` `rArr" "a_(1){1-(1)/(2^(n))}=(4n)/(2^(n))-2n` `rArr" "a_(1)(2^(n)-1)=4n-2nxx2^(n)rArra_(1)=(4n-2nxx2^(n))/(2^(n)-1)` `:.` Middle term of the sequence `=a_(2n+1)=a_(1)+4n` `=(4n-2nxx2^(n))/(2^(n)-1)+4n` `=(2nxx2^(n))/(2^(n)-1)=(n*2^(n+1))/(2^(n)-1)` |
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| 393. |
The number of terms common between the series 1+ 2 + 4 + 8..... to 100 terms and 1 + 4 + 7 + 10 +... to 100 terms isA. 6B. 4C. 5D. none of these |
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Answer» Correct Answer - C Soppose `m^(th)` term of first sequence be same as the `n^(th)` term of the second sequences. Then, `2^(m-1)=1+(n-1)xx3rArr2^(m-1)=3n-2rArr2^(m-2)+1=(3n)/(2)` . . . (i) We have, `1lem,nle100` `:." "(3n)/(2)le150` `rArr" "2^(m-2)+1le150` `rArr" "2^(m-2)le149rArrm-2le7rArrnle9rArrm=1,2, . . . ,9`. From (i), we have `n=(2)/(3)(2^(m-2)+1)` We observe that `n inN` for m = 1,3,5,7,9. Hence, there are five common terms in the sequences. |
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| 394. |
if `a,a_1,a_2,a_3,.........,a_(2n),b` are in `A.P.` and `a,g_1,g_2,............g_(2n) ,b` are in `G.P.` and `h` is `H.M.` of `a,b` then `(a_1+a_(2n))/(g_1*g_(2n))+(a_2+a_(2n-1))/(g_2*g_(2n-1))+............+(a_n+a_(n+1))/(g_n*g_(n+1))` is equalA. `(2n)/(h)`B. 2nhC. nhD. `(n)/(h)` |
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Answer» Correct Answer - A We have, `a_(1)+2_(2n)=a_(2)+a_(2n-1)+ . . . =a_(n)+a_(n+1)=a+b` and, `g_(1)g_(2n)=g_(2)g_(2n-1)= . . .=g_(n)g_(n+1)=ab` `:." "(a_(1)+a_(2n))/(g_(1)g_(2n))+(a_(2)+a_(2n-1))/(g_(2)g_(2n-1))+ . . . .+(a_(n)+a_(n+1))/(g_(n)g_(n+1))` `=n((a_(1)+a_(2n))/(g_(1)g_(2n)))=((a+b)/(ab))=(2n)/(h)" "{:[becauseh=(2ab)/(a+b)]:}` |
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| 395. |
If `(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))`, then `a_(1),a_(2),a_(3),a_(4)` are inA. APB. GPC. HPD. none of these |
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Answer» Correct Answer - C We have, `(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))` `rArr" "(a_(2)+a_(3))/(a_(2)a_(3))-(a_(1)+a_(4))/(a_(1)a_(4))` `rArr" "(1)/(a_(3))+(1)/(a_(2))=(1)/(a_(4))+(1)/(a_(1))rArr(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(4))-(1)/(a_(3))` Again, `(a_(2)a_(3))/(a_(1)a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))` `rArr" "(a_(1)-a_(4))/(a_(1)a_(4))=3((a_(2)-a_(3))/(a_(2)a_(3)))` `rArr" "(1)/(a_(4))-(1)/(a_(1))=3((1)/(a_(3))-(1)/(a_(2)))` `rArr" "((1)/(a_(4))-(1)/(a_(3)))+((1)/(a_(3))-(1)/(a_(2)))+((1)/(a_(2))-(1)/(a_(1)))=3((1)/(a_(3))-(1)/(a_(2)))` `rArr" "((1)/(a_(2))-(1)/(a_(1)))+((1)/(a_(3))-(1)/(a_(2)))+((1)/(a_(2))-(1)/(a_(1)))=3((1)/(a_(3))-(1)/(a_(2)))` [Using (i)] `rArr" "(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a^(2))` . . . (ii) From (i) and (ii), we obtain `(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a_(2))=(1)/(a_(4))-(1)/(a_(3))` `rArr" "(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)),(1)/(a_(4))` are in A.P. `rArra_(1),a_(2),a_(3),a_(4)` are in H.P. |
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| 396. |
let `a_1,a_2,a_3,...........,`be an `AP` such that `(a_1+a_2+a_3+...........+a_p)/(a_1+a_2+a_3+...........+a_q)=p^3/q^3`,`(p!=q)` then find `a_6/a_21=?`A. `(41)/(11)`B. `(7)/(2)`C. `(2)/(7)`D. `(11)/(41)` |
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Answer» Correct Answer - D We have, `(a_(1)+a_(2)+ . . .+a_(p))/(a_(1)+a_(2)+ . . ..+a_(q))=(p^(2))/(q^(2))rArr(a_(1)+((p-1)/(2))d)/(a_(1)+((q-1)/(2))d)=(p)/(q)` . . . .(i) We have to find the value of `(a_(6))/(a_(21))` i.e. `(a_(1)+5d)/(a_(1)+20)`. So, replacing `(p-1)/(2)` by 5 and `(q-1)/(2)=20` i.e. putting p=11 and q=41 in (i), we get `(a_(6))/(a_(12))=(11)/(41)`. |
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| 397. |
Let `sum_(r=1)^(n) r^(6)=f(n)," then "sum_(n=1)^(n) (2r-1)^(6)` is equal toA. `f(n)-64f((n+1)/(2))` n is oddB. `f(n)-64f((n-1)/(2))` n is oddC. `f(n)-64f((n)/(2))`, n is evenD. none of these |
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Answer» Correct Answer - D we have, `underset(r=1)overset(n)sum(2r-1)^(6)` `=1^(6)+3^(6)+5^(6)+ . . .+(2n-1)^(6)` `={1^(6)+2^(6)+3^(6)+ . . . +(2n)^(6)}-{2^(6)+4^(6)+6^(6)+ . . . +(2n)^(6)}` `=f(2n)-2^(6)f(n)=f(2n)-64f(n)` |
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| 398. |
If A, G & H are respectively the A.M., G.M. & H.M. of three positive numbers a, b, & c, then equation whose roots are a, b, & c is given byA. `a^(2)=AH`B. A is an integer if `altbltclt4`C. A=H iff a=b=cD. `AgtGgtH,ifa!=b!=c` |
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Answer» Correct Answer - A We know that A=H=G iff a=b=c. Also, if `a!=b!=c`, then `AgtGgtH`. Thus, options (c ) and (d) are true. If `altbltclt4`, then for a=1,b=2andc=3, we have `A=(1+2+3)/(3)=2`, which is an integer. So, option (b) may also be true. Thus, option (a) is not true. |
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| 399. |
If `a_(1),a_(2),a_(3), . . .,a_(n)` are in H.P. and `f(k)=sum_(r=1)^(n) (a_(r)-a_(k))`, then `(a_(1))/(f(1)),(a_(2))/(f(2)), . . . ,(a_(n))/(f(n))`, 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 `a_(1),a_(2),a_(3), . . . ,a_(n)` are in H.P. `rArr" "(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)), . . . ,(1)/(a_(n))` are in A.P. `rArr" "(underset(r=1)overset(n)suma_(r))/(a_(1)),(underset(r=1)overset(n)suma_(r))/(a_(2)),(underset(r=1)overset(n)suma_(r))/(a_(3)), . . . .,(underset(r=1)overset(n)suma_(r))/(a_(n))` are in A.P. `rArr" "(f(1)+a_(1))/(a_(1)),(f(2)+a_(2))/(a_(2)),(f(3)+a_(3))/(a_(3)), . . . .,(f(n)+a_(n))/(a_(n))` are in A.P. `rArr" "(f(1))/(a_(1))+1,(f(2))/(a_(2))+1,(f(3))/(a_(3))+1, . . . ,(f(n))/(a_(n))+1` are in A.P. `rArr" "(f(1))/(a_(1)),(f(2))/(a_(2)),(f(3))/(a_(3)), . . . .,(f(n))/(a_(n))` are in A.P. `rArr" "(a_(1))/(f(1)),(a_(2))/(f(2)),(a_(3))/(f(3)), . . .,(a_(n))/(f(n))` are in H.P. |
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| 400. |
If `a_1, a_2, a_3, ,a_(2n+1)`are in A.P., then`(a_(2n+1)-a_1)/(a_(2n+1)+a_1)+(a_(2n)-a_2)/(a_(2n)+a_2)++(a_(n+2)-a_n)/(a_(n+2)+a_n)`is equal to`(n(n+1))/2xx(a_2-a_1)/(a_(n+1))`b. `(n(n+1))/2`c. `(n+1)(a_2-a_1)`d. none of theseA. `(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))`B. `(n(n+1))/(2)`C. `(n+1)(a_(2)-a_(1))`D. none of these |
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Answer» Correct Answer - A In an A.P. the sum of the terms equidistant from the beginning and the end is always same is equal to the sum of first and last term. Therefore, `a_(1)+a_(2n+1)=a_(2)+a_(2n)= . . .=a_(n)+a_(n+2)` `:." "(a_(2n+1)-a_(1))/(a_(2n+1)+a_(1))=(a_(2n)-a_(2))/(a_(2n)+a_(2))+ . . .+(a_(n+2)-a_(n))/(a_(n+2)+a_(n))` `=(2nd+(2n-2)d+ . . .+2d)/(a_(2n+1)+a_(1))` `=2d(n(n+1))/(2)xx(1)/(a_(2n+1)+a_(1))` `=2d(n(n+1))/(2)xx(1)/(2a_(n+1))" "[becausea_(1)+a_(2n+1)=a_(n)+a_(n+2)=2a_(n+1)]` `=d(n(n+1))/(2)xx(1)/(a_(n+1))=(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))" "[becaused=a_(2)-a_(1)]` |
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