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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.
| 201. |
.If `a, b, c` are in `H.P`. then the value of `(b+a)/(b-a) + (b+c)/(b-c)`A. 1B. 2C. 3D. 0 |
| Answer» Correct Answer - B | |
| 202. |
If a,b,c are in H.P, thenA. `(a-b)/(b-c)=(a)/(c)`B. `(b-c)/(c-a)=(b)/(a)`C. `(c-a)/(a-b)=(c)/(b)`D. `(a-b)/(b-c)=(c)/(a)` |
| Answer» Correct Answer - A | |
| 203. |
If a,b,c, are in A.P., b,c,d are in G.P. and c,d,e, are in H.P., then a,c,e are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 204. |
For any three positive real numbers `a , b`and `c ,9(25 a^2+b^2)+25(c^2-3a c)=15 b(3a+c)dot`Then :`a ,b`and`c`are in`AdotPdot`(2) `a ,b`and `c`are in `GdotPdot``b ,c`and `a`are in `GdotPdot`(4) `b , c`and `a`are in `AdotPdot`A. a,b and c are in GPB. b,c and a are in GPC. b,c and a are in APD. a,b and c are in AP |
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Answer» Correct Answer - C `(15a)^(2)+(3b)^(2)+(5c)^(2)-45ab-15bc-75ac=0` `implies (1)/(2){(15a-3b)^(2)+(3b+5c)^(2)+(5c-15a)^(2)}=0` `implies (15a-3b)^(2)+(3b+5c)^(2)+(5c-15a)^(2)=0` or `15a-3b=0,3b-5c=0,5c-15a=0` `:. b=5a,c=3a` `implies 5a,3a,a` are in AP i.e b,c,a are in AP. |
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| 205. |
if `(a+b)/(1-ab),b,(b+c)/(1-bc)` are in `AP` then `a,1/b,c ` are inA. A.P.B. G.P.C. H.P.D. `(a-b)/(b-c)=(c)/(a)` |
| Answer» Correct Answer - C | |
| 206. |
The sum of n terms of an A. P. is `an(n-1)`. Find the sum of the squares of these terms.A. `a^(2)n^(2)(n-1)^(2)`B. `(a^(2))/(6)n(n-1)(2n-1)`C. `(2a^(2))/(3)n(n-1)(2n-1)`D. `(2a^(2))/(3)n(n+1)(2n+1)` |
| Answer» Correct Answer - C | |
| 207. |
If the sum 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)/5`m, then m is equal to:(1) 102 (2) 101 (3) 100 (4) 99A. 100B. 99C. 102D. 101 |
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Answer» Correct Answer - D `((8)/(5))^(2)+((12)/(5))^(2)+((16)/(5))^(2)+((20)/(5))^(2)+"......."+((44)/(5))^(2)` `=(16)/(25)(2^(2)+3^(2)+4^(2)+5^(2)+"........"+11^(2))` `=(16)/(25)((11*(11+1)*(22+1))/(6)-1)` `=(16)/(25)xx505=(16)/(25)xx101=(16)/(5)m(" given ")` `:.m =101`. |
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| 208. |
If a,b,c are in H.P., then `(a)/(b+c),(b)/(c+a),(c)/(a+b)` will be inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - C | |
| 209. |
If a,b,c,d are in GP and `a^x=b^x=c^z=d^u`, then `x ,y,z,u ` are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - C | |
| 210. |
Find a G.P. for which sum of the first two terms is `- 4`and the fifth term is 4 times the third term. |
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Answer» given that ,`a+ ar=-4` eqn(1) `T_5 = 4T_3` `ar^4 = 4ar^2` `r^2=4` `r= +-2` now `a(1+r)= -4` if `r=2, a= -4/3` `-4/3 + (-8/3)+ (-16/3) ........` if `r=-2, a= -4/-1=4` `4+(-8)+16+(-32).......` 2 GPs possible answer |
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| 211. |
If three positive unequal numbers `a, b, c` are in H.P., thenA. `a^(3//2)+c^(3//2)gt2b^(1//2)`B. `a^(5)+c^(5)gt2b^(5)`C. `a^(2)+c^(2)gt2b^(3)`D. none of these |
| Answer» Correct Answer - B | |
| 212. |
Find the sum to `n`terms of the series: `3/(1^2 .2^2)+5/(2^2 .3^2)+7/(3^2 .4^2)+` |
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Answer» Correct Answer - D Let `T_(r)" be the "r^(th)` term of the given series. Then, `T_(r)=((2r+1))/(r^(2)(r+1)),r=1,2,3, . . . . . ` `rArr" "T_(r)={(1)/(r^(2))-(1)/((r+1)^(2))},r=1,2,3, . . . ` Let S be the required sum. Then, `S=underset(nto oo)limunderset(r=1)overset(n)sumT_(r)` `rArr" "S=underset(nto oo)limunderset(r=1)overset(n)sum{(1)/(r^(2))-(1)/((r+1)^(2))}=underset(nto oo)lim{1-(1)/((n+1)^(2))}=1` |
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| 213. |
If `log_(2)(5.2^(x)+1),log_(4)(2^(1-x)+1)` and 1 are in A.P,then x equalsA. `log_(2)5`B. `1-log_(2)5`C. `log_(5)2`D. none of these |
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Answer» Correct Answer - B The given number are in A.P. `:." "2log_(4)(2^(1-x)+1)=log_(2)(5.2^(x)+1)+1` `rArr" "2log_(2)2((2)/(2^(x))+1)=log_(2)(5.2^(x)+1)+log_(2)2` `rArr" "(2)/(2)log_(2)((2)/(2^(x))+1)=log_(2)(5.2^(x)+1)2` `rArr" "log_(2)((2)/(2^(x))+1)=log_(2)(10.2^(x)+2)` `rArr" "(2)/(2^(x))+1=10.2^(x)+2` `rArr" "(2)/(y)+1=10y+2," where"2^(x)=y` `:." "10y^(2)+y-2=0` `rArr" "(5y-2)(2y+1)=0` `rArr" "y=2//5" "[becausey=2^(x)gt0]` `rArr" "2^(x)=2//5rArrx=log_(2)(2//5)=log_(2)2-log_(2)5=1-log_(2)5` |
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| 214. |
The numbers `3^(2sin2alpha-1),14and3^(4-2sin2alpha)` form first three terms of A.P., its fifth term isA. `-25`B. `-12`C. 40D. 53 |
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Answer» Correct Answer - D Since,`3^(2sin2alpha-1),14and3^(4-2sin2alpha)` `:." "2xx143^(2sin2alpha-1)+3^(4-2sin2alpha)` `rArr" "28=(a)/(3)+(3^(4))/(a)`, where `a=3^(2sin2alpha)` `rArr" "a^(2)-84a+243=0` `rArr" "(a-81)(a-3)=0` `rArr" "a=81,a=3` `rArr" "3^(2sin2alpha)=3^(4)or,3^(2sin2alpha)=3` `rArr" "2sin2alpha=1" "[because2sin2alpha!=4]` `rArr" "sin2alpha=(1)/(2)rArr2alpha=(pi)/(6)` Thus, the first three terms of the A.P. are 1,14,27. Hence, its fifth term is 53. |
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| 215. |
`underset("n-digits")((666 . . . .6)^(2))+underset("n-digits")((888 . . . .8))` is equal toA. `(4)/(9)(10^(n)-1)`B. `(4)/(9)(10^(2n)-1)`C. `(4)/(9)(10^(n)-1)^(2)`D. none of these |
| Answer» Correct Answer - B | |
| 216. |
If `1+(1+2)/2+(1+2+3)/3+.....` to n terms is S. Then , S is equal toA. `(n(n+3))/(4)`B. `(n(n+2))/(4)`C. `(n(n+1)(n+2))/(6)`D. `n^(2)` |
| Answer» Correct Answer - D | |
| 217. |
Find the sum to n terms, whose nth term is `tan[alpha+(n-1)beta]tan (alpha+nbeta)`. |
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Answer» `T_(n)=tan[alpha+(n-1)beta]tan (alpha+nbeta)` `tanbeta=tan[(alpha+nbeta)-{alpha+(n-1)beta}]` `tanbeta=(tan(alpha+nbeta)-tan(alpha+(n-1)beta))/(1+tan(alpha+nbeta)tan(alpha+(n-1)beta))` `:.1+T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]` `T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]-1` For `n=1` `T_(1)=cotbeta[tan(alpha+beta)-tanalpha]-1` For `n=2` `T_(2)=cotbeta[tan(alpha+2beta)-tan(alpha+beta)]-1` For `n=3`, `T_(3)=cotbeta[tan(alpha+3beta)-tan(alpha+2beta)]-1` ` " " vdots " " vdots " " vdots " "` For `n=n`, `T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]-1` `S_(n)=T_(1)+T_(2)+T_(3)+"......."+T_(n)=cotbeta[tan(alpha+nbeta)-tanalpha]-n` `=(sin(alpha+nbeta)/(cos(alpha+nbeta))+(sinalpha)/(cosalpha))/(tan beta)-n=(sin(alpha+nbeta-alpha)/(cosalpha cos(alpha+nbeta)))/(tan beta)-n` `=((sinnbeta)/(cos(alpha+nbeta)cos alpha)-n tan beta)/(tan beta)`. |
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| 218. |
Find the sum to infinite terms of the series `tan^(-1)((1)/(3))+tan^(-1)((2)/(9))+"........."+tan^(-1)((2^(n-1))/(1+2^(2n-1)))+"......"`. |
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Answer» `T_(n)=tan^(-1)((2^(n-1))/(1+2^(2n-1)))=tan^(-1)((2^(n-1))/(1+2^(n)*2^(n-1)))` `=tan^(-1)((2^(n)-2^(n-1))/(1+2^(n)*2^(n-1)))=tan^(-1)2^(n)-tan^(-1)2^(n-1)` `S_(n)=T_(1)+T_(2)+"......."+T_(n)` `=(tan^(-1)2^(1)-tan^(-1)2^(0))+(tan^(-1)2^(2)-tan^(-1)2^(1))+"........"+(tan^(-1)2^(n)-tan^(-1)2^(n-1))` `=(tan^(-1)2^(n)-tan^(-1)1)` `S_(n)=tan^(-1)2^(n)-(pi)/(4)` `S=lim_(n to oo)S_(n)=lim_(n to oo)(tan^(-1)2^(n)-(pi)/(4))=(pi)/(2)-(pi)/(4)=(pi)/(4)`. |
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| 219. |
Evaluate `S=sum_(n=0)^n(2^n)/((a^(2^n)+1)` (where `a>1)`. |
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Answer» `S=sum_(n=0)^(oo)(2^(n))/(a^(2n)+1)(agt1)` `S_(n)=sum_(n=0)^(n)(2^(n))/(a^(2n)+1)` `=(1)/(a+1)+(2)/(a^(2)+1)+(4)/(a^(4)+1)+(8)/(a^(8)+1)+"........"+(2^(n))/(a^(2^n)+1)` `=(1)/(1+a)+(2)/(1+a^(2))+(4)/(1+a^(4))+(8)/(1+a^(8))+"........"+(2^(n))/(1+a^(2^n))` `=(-(1)/(1-a)+(1)/(1-a))(1)/(1+a)+(2)/(1+a^(2))+(4)/(1+a^(4))+(8)/(1+a^(8))+"........"+(2^(n))/(1+a^(2^n))` `=(1)/(a-1)+((1)/(1-a)+(1)/(1+a))+(2)/(1+a^(2))+(4)/(1+a^(4))+"........"+(2^(n))/(1+a^(2^n))` `=(1)/(a-1)+((2)/(1-a^(2))+(2)/(1+a^(2)))+(4)/(1+a^(4))+"........"+(2^(n))/(1+a^(2^n))` `" " vdots " "vdots " " vdots " "` `S_(n)=(1)/(a-1)+(2^(n+1))/(1-a^(2^n+1))` `S=lim_(n to oo)S_(n)=lim_(n to oo)((1)/(a-1)+(2^(n+1))/(1-a^(2^n+1)))` `=lim_(n to oo)((1)/(a-1)+((2^(n+1))/(a^(2^n+1)))/((1)/(a^(2^n+1))-1))=(1)/(a-1)+(0)/(0-1)=(1)/(a-1)`. |
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| 220. |
Show that,`(1+5^(-1))(1+5^(-2))(1+5^(-4))(1+5^(-8))"....."(1+5^(-2n))=(5)/(4)(1-5^(-2(n+1)))`. |
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Answer» `LHS=(1+5^(-1))(1+5^(-2))(1+5^(-4))(1+5^(-8))"....."(1+5^(-2n))` `=(1+(1)/(5))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))` `=((1-(1)/(5)))/((1-(1)/(5)))(1+(1)/(5))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))` `=(5)/(4)[(1+(1)/(5^(2)))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))]` `" " vdots " " vdots " "vdots" "` `=(5)/(4)(1-(1)/(5^(2^(n+1))))=(5)/(4)(1-5^(2^(n+1)))=RHS`. |
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| 221. |
If `theta_(1),theta_(2),theta_(3),".......",theta_(n)` are in AP whose common difference is d, then show that `sind{sectheta_(1) sectheta_(2)+ sectheta_(2)sectheta_(3)+"........"+sectheta_(n-1)sectheta_(n)}=tantheta_(n)-tantheta_(1)`. |
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Answer» `theta_(1),theta_(2),theta_(3),".......,"theta_(n)` are in AP. So, `theta_(2)-theta_(1)=theta_(3)-theta_(2)="......"=theta_(n)-theta_(n-1)=d` `:.LHS =sind[sectheta_(1) sectheta_(2) sectheta_(2)sectheta_(3)+"....."+sectheta_(n-1)sectheta_(n)]` `=sind[(1)/(costheta_(1)costheta_(2))+(1)/(costheta_(2)costheta_(3))+"......."+(1)/(costheta_(n-1)costheta_(n))]` `=(sind)/(costheta_(1)costheta_(2))+(sind)/(costheta_(2)costheta_(3))+"......."+(sind)/(costheta_(n-1)costheta_(n))` `=(sin(theta_(2)-theta_(1)))/(costheta_(1)costheta_(2))+(sin(theta_(3)-theta_(2)))/(costheta_(2)costheta_(3))+"......."+(sin(theta_(n)-theta_(n-1)))/(costheta_(n-1)costheta_(n))` `=(tantheta_(2)-tantheta_(1))+(tantheta_(3)-tantheta_(2))+"......."+(tantheta_(n)-tantheta_(n-1))` `=tantheta_(n)-tantheta_(1)=RHS`. |
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| 222. |
Find out the largest term of the sequence `(1)/(503),(4)/(524),(9)/(581),(16)/(692),"...."`. |
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Answer» General term cam be written as `T_(n)=(n^(2))/(500+3n^(3))` Let `U_(n)=(1)/(T_(n))=(500)/(n^(2))+3n` Then, `(dU_(n))/(dn)=(1000)/(n)+3` and `(d^(2)U_(n))/(dn^(2))=(3000)/(n^(4))` For maxima or minima of `U_(n)`, we have `(dU_(n))/(dn)=0implies n^(3)=(1000)/(3)` `implies n=((1000)/(3))^((1)/(3)) (" not an integer ) and " 6lt((1000)/(3))^((1)/(3))lt7` But n is an integer, `therefore` for the maxima or minima of `U_(n)` we will take n as the nearest integer to `((1000)/(3))^((1)/(3))` Since, `((1000)/(3))^((1)/(3))` is more close to 7 then to 6. Thus, we take `n=7`. Further `(d^(2)U^(n))/(dn^(2))=+ve` , then `U_(n)` will be minimum and therefore, `T_(n)` will be maximum for `n=7`. Hence, `T_(7)` is largest term. So, largest term in the given sequence is `(49)/(1529)`. |
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| 223. |
IF `f(r )=1+(1)/(2)+(1)/(3)+"...."+(1)/(r )` and `f(0)=0`, find `sum _(r=1)^(n)(2r+1)f(r )`. |
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Answer» Since, `sum _(r=1)^(n)(2r+1)f(r )` `=sum _(r=1)^(n)(r^(2)+2r+1-r^2))f(r )=sum _(r=1)^(n){(r+1)^(2)-r^(2)}f(r )` `=sum _(r=1)^(n){(r+1)^(2)f(r )-(r+1)^(2)f(r+1)+(r+1)^(2)f(r+1)-r^(2)f(r )}` `=sum _(r=1)^(n)(r+1)^(2){f(r )-(r+1)}+sum _(r=1)^(n){(r+1)^(2)f(r+1)-r^(2)f(r )}` `=-sum _(r=1)^(n)((r+1)^(2))/((r+1))+sum _(r=1)^(n)(r+1)^(2)f(r+1)+(n+1)^(2) f(n+1)-sum_r^(r=1)r^(2)f(r)[:.f(r+1)-f(r)=(1)/(r+1)]` `=-sum _(r=1)^(n)(r+1)+{2^(2)f(2)+3^(2)f(3)+"...."n^(2)f(n)}+(n+1)^(2)f(n+1)={1^(2)f(1)+2^(2)f(2)+3^(2)+"...."n^(2)f(n)}` `=-sum _(r=1)^(n)r-sum _(r=1)^(n)1+(n+1)^(2)f(n+1)-1^(2)f(1)` `=(n(n+1))/(2)-n+(n+1)^(2)f(n+1)-f(1)` `=(n+1)^(2)f(n+1)-(n(n+1))/(2)-1 " "[:.f(1)=1]` `=(n+1)^(2)f(n+1)-((n^(2)+3n+2))/(2)` Hence, this is the required result. |
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| 224. |
If the equation `x^(4)-4x^(3)+ax^(2)+bx+1=0` has four positive roots, fond the values of a and b. |
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Answer» Let `x_(1),x_(2),x_(3),x_(4),` are the roots of the equation `x_(1)x_(1)x_(1)x^(4)-4x^(3)+ax^(2)+bx+1=0 " " "....(i)"` `:.x_(1)+x_(2)+x_(3)+x_(4)=4 " and "x_(1)x_(2)x_(3)x_(4)=1` `:.AM=(x_(1)+x_(2)+x_(3)+x_(4))/(4)=(4)/(4)=1` and `GM=(x_(1)x_(2)x_(3)x_(4))^((1)/(4))=1` `i.e., AM=GM` which is true only when `x_(1)=x_(2)=x_(3)=x_(4)=1` Hence, given equation has all roots identical, equal to i.e., equation have form `(x-1)^(4)=0` `implies x^(4)-4x^(3)+6x^(2)-4x+1=0" " "....(ii)"` On comparing Eqs. (i) and (ii), we get `a=6,b=-4`. |
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| 225. |
An A.P., and a H.P. have the same first and last terms and the same odd number of terms. The middle terms of the three series are inA. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 226. |
If `a,b,c` are in G.P and `a + x, b +x, c + x` are in H.P, then the value of x is `(a,b,c` are distinct numbers)A. cB. bC. aD. none of these |
| Answer» Correct Answer - B | |
| 227. |
How many terms of the series `20+19(1)/(3)+18(2)/(3)+"...."` must be taken to make 300? Explain the double answer. |
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Answer» Here, given series is an AP with first term `a=20` and the common difference, `d=-(2)/(3).` Let the sum of n terms of the series be 300. Then, `S_(n)=(n)/(2){2a+(n-1)d}` `implies 300=(n)/(2){2xx20+(n-1)(-(2)/(3))}` ` implies 300=(n)/(3){60=n+1}` `implies n^(2)-61n+900=0` `implies (n-25)(n=36)=0` `implies n=25 " or " n=36` `therefore` Sum of 25 terms = Sum of terms = 300 Explaination of double answer Her, the common difference is negetive, therefore terms go on diminishing and `t_(31)=20+(31-1)((-2)/(3))=0 i.e.,31st` term becomes zero. all terms after 31st term are negative. These negative terms `(t_(32),t_(33),t_(34),t_(35)t_(36))` when added to positive terms `(t_(26),t_(27),t_(28),t_(29)t_(30))`, they cancel out each other i.e., sum of ter,s from 26th to 36th terms is zero. Hence, the sum of 25 terms as well as that of 36 terms is 300. |
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| 228. |
If `a , b , c ,`are in `A P ,a^2,b^2,c^2`are in HP, then prove that either `a=b=c`or `a , b ,-c/2`from a GP (2003, 4M) |
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Answer» Given, `a,b,c` are in AP. `therefore " " b=(a+c)/(2) " " "……(i)` and `a^(2),b^(2),c^(2)` in HP. `therefore " " b^(2)=(2a^(2)c^(2))/(a^(2)+c^(2)) " " "………(ii)` From Eq. (ii) `b^(2){(a+c)^(2)-2ac}=2a^(2)c^(2)` ` implies b^(2){(2b)^(2)-2ac}=2a^(2)c^(2) " " [" from Eq. (i)"]` `implies 2b^(4)-acb^(2)-a^(2)c^(2)=0` `implies (2b^(2)+ac)(b^(2)-ac)=0` `implies 2b^(2)+ac =0` or `b^(2)-ac=0` If `2b^(2)+ac=0`, than `b^(2)=-(1)/(2)ac` or `-(a)/(2),b,c` are in GP and if `b^(2)-ac =0 implies a,b,c` are in GP. But given, `a,b,c` are in AP. Which is possible only when `a=b=c`. |
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| 229. |
The maximum sum of the series `20+19 1/3+18 2/3+`is`310`b. `300`c. `0320`d. none of theseA. 310B. 300C. 320D. none of these |
| Answer» Correct Answer - A | |
| 230. |
The maximum sum of the series `20+19 1/3+18 2/3+`is`310`b. `300`c. `0320`d. none of theseA. 310B. 290C. 320D. none of these |
| Answer» Correct Answer - A | |
| 231. |
If `a,b,c` are in HP,b,c,d are in GP and `c,d,e` are in AP, than show that `e=(ab^(2))/(2a-b)^(2)`. |
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Answer» Given, `a,b,c` are in HP. `therefore b=(2ac)/(a+c)" or "c=(ab)/(2a-b) "……..(i)"` Given,` b,c,d` are in GP. `therefore " " c^(2)=bd " " "……..(ii)"` and given, `c,d,e` are in AP. `therefore " " d=(c+e)/(2)` `implies e=2d-c` `e=((2c^(2))/(b)-c) " " [" from Eq. (ii) "] "........(iii)"` From Eqs. (i) and (iii), `e=(2)/(b)((ab)/(2a-b))^(2)-((ab)/(2a-b))` `=(ab)/(2a-b)^(2){(2a-(2a-b)}` `=(ab^(2))/(2a-b)^(2)` |
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| 232. |
Given a,b,c are in A.P.,b,c,d are in G.P and c,d,e are in H.P .If a=2 and e=18 , then the sum of all possible value of c is ________. |
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Answer» Given,`a,b,c` are in AP. `therefore " " b=(a+c)/(2) " " "…….(i)"` `b,c,d` are in GP. `therefore c^(2)=bd " " "…….(ii)"` and `c,,e` are in HP. `therefore d=(2ce)/(c+e) " " "…..(iii)"` Now, substituting the values of b and d in Eq. (ii), than ` c^(2)= ((a+c)/(2))((2ce)/(c+e))` `implies c(c+e)=e(a+c)` `implies c^(2)=ae " " ".......(iv)"` Given, `a=2,e=18` From Eq. (iv), `c^(2)=(2)(18)=36` ` therefore " " c= pm 6 ` From Eq. (i), `b=(2 pm 6)/(2)=4,-2` and from Eq. (ii), `d=(c^(2))/(b)=(36)/(b)=(36)/(4) " or " (36)/(-2)` `therefore " " d=9 " or " -18` Hence, `c=6,b=4,d=9 " or " c=-6,b=-2,d=-18` |
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| 233. |
If `a,b,c` are in AP and p is the AM between a and b and q is the AM between b and c, then show that b is the AM between p and q. |
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Answer» `therefore a,b,c` are inAP. `therefore " " 2b=a+c " " "…….(i)"` `therefore` p is the AM between a and b. `therefore " " p=(a+b)/(2) " " "……..(ii)"` `therefore` q is the AM between b and c. `therefore " " q=(b+c)/(2) " " "......(iii)"` On adding Eqs. (ii) and (iii), than `p+q=(a+b)/(2)+(b+c)/(2)=(a+c+2b)/(2)=(2b+2b)/(2) [" using Eq. (i)"]` `therefore p+q=2b " or " b=(p+q)/(2)` Hence, b is the AM between p and q. |
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| 234. |
If three positive numbers `a,b` and c are in AP, GP and HP as well, than find their values. |
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Answer» Since, `a,b,c` are in AP,GP and HP as well `therefore " " b=(a+c)/(2) " " "…..(i)"` ` " " b^(2)=ac " " "…..(ii)"` and `b= (2ac)/(a+c)" " "…..(iii)"` Form Eq. (i) and (ii), we have `((a+c))^(2)/(2)` or `(a+c)^(2)=4ac` or `(a+c)^(2)-4ac=0` or `(a-c)^(2)=0` `therefore " "a=c " " "........(iv)"` On putting `c=a` in Eq. (i), we get `b=(a+a)/(2)=a " " "......(v)"` From Eqs. (iv) and (v), `a=b=c`, thus the three numbes will be equal. |
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| 235. |
The sum of the products of the ten numbers `+-1,+-2,+-3,+-4,+-5` taking two at a time is: |
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Answer» we know, `(a+b)^2 = a^2 + b^2 + 2ab` `(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca)` numbers given `1,2,3,4,5,-1,-2,-3,-4,-5` `(1+2+3+4+5-1-2-3-4-5)^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 2(s)` `0 = 2(1^2 + 2^2 + 3^2 + 4^2 + 5^2) + 2(s) ` `s = -(1^2 + 2^2 + 3^2 + 4^2 + 5^2)` `s = -(5 xx 6 xx 11)/6` `= - 55` option b is correct answer |
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| 236. |
The arithmetic mean between two numbers is A and the geometric mean is G. Then these numbers are: |
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Answer» let nos be `x & y` `(x+y)/2 =A` `x+y = 2A` `sqrt(xy) = a` `xy = a^2` `x-y = sqrt((x+y)^2 - 4xy)` `= sqrt(4A^2 - 4G^2)` `x-y = 2 sqrt(A^2 - G^2 )` `2x = 2 sqrt(A^2 - G^2) + 2A` `x = A + sqrt(A^2 - G^2)` now `y= 2A-x` `= 2A-A - sqrt(A^2 - G^2)` `y= A- sqrt(A^2 - G^2)` option c is correct answer |
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| 237. |
Three numbers whose sum is 18 are in AP if 2,4, 11 are added to them respective the resulting numbers are in GP. Find the numbers. |
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Answer» Let the numbers are `a-r, a and a+r`. Then,`a+a-r+a+r = 18=> 3a = 18=> a = 6` We are given if we add 2,4 and 11 to these numbers, it will be in GP.So, `(6+4)/(6-r+2) = (6+r+11)/((6+4)` `=>10/(8-r) = (17+r)/10` `=> (17+r)(8-r) = 100` `=>r^2+9r-136+100 = 0` `=>r^2+9r-36 = 0` `=>(r+12)(r-3) = 0` `r = -12 and r = 3` When `r = -12`, then numbers are, `18, 6, -6`. When `r = 3`, then numbers are `3,6, 9`. |
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| 238. |
Prove that:`3^(1/2)xx3^(1/4)xx3^(1/8)xx...=3` |
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Answer» `L.H.S. = 3^(1/2)xx3^(1/4)xx3^(1/8)+...` `=3^(1/2+1/4+1/8+1/16...)` Now, we will solve, `(1/2+1/4+1/8+1/16...)` It forms a GP with first term, `a= 1/2` and common ratio, `r = 1/2` So, Sum of the given series will be, `S = a/(1-r) = (1/2)/(1-1/2) = 1` So, our expression becomes, `3^(1/2+1/4+1/8...) = 3^1 = 3 = R.H.S.` |
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| 239. |
Find the sum to infinity of the following Geometric Progression: `5,(20)/7 , (80)/(49) ,...` |
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Answer» `S=5,5*4/7,5*(4/7)^2+.....` `S=5/(1-(4/7))` `S=35/3` |
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| 240. |
The friends whose ages from a G.P. divide a certain sum of money in proportion to their ages. If they do that three years later, when the youngest is halfg the age of the oldest, then he will receive 105 rupees more that the he gets now and the middle friends will get 15 reupees more that he gets now, then ages of the friends are |
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Answer» Let their ages be `a,ar,ar^(2)`, After 3 yr, their ages will be `a+3,ar+3,ar^(2)+3`. Given, `2(a+3)=ar^(2)+3" " "……..(i)"` Let x rupee be the sum of the money divided. And Let `y=a+ar+ar^(2)" " "………(ii)"` Then, `y+9=a+3+(ar+3)+(ar^(2)+3)` We have, `(x(a+3))/((y+9))=(xa)/(y)+105` `impliesx((a+3)/(y+9)-(a)/(y))=105" " ".........(iii)"` Also, `(x(ar+3))/((y+9))=(xar)/(y)+15` `impliesx[(ar+3)/(y+9)-(ar)/(y)]=15" " ".........(iv)"` On dividing Eq. (iii)" by Eq. "(iv), we get `implies(y(a+3)-a(y+9))/(y(ar+3)-ar(y+9))=7 implies (y-3a)/(y-3ar)=7` ` implies 6y=21ar-3a implies y=(a(7r-1))/(2)` From Eq. (ii), `(a(7r-1))/(2)=a+ar+ar^(2)` `implies 5ar=3a+2ar^(2)" " ".........(v)"` From Eqs. (i) and (ii), `a=12,r=(3)/(2)` Let ages of these friends are `12,12xx(3)/(2),12xx((3)/(2))^(2)i.e.12,18,27`. |
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| 241. |
If the first and the nth terms of a G.P., are `aa n db ,`respectively, and if `P`is hte product of the first `n`terms prove that `P^2=(a b)^ndot`A. abB. `(ab)^(n)`C. `(ab)^(n//2)`D. `(ab)^(2n)` |
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Answer» Correct Answer - B Let r be the common ratio of the G.P. Then, `b=n^(th)term=ar^(n-1)rArrr^(n-1)=(b)/(a)rArrr=((b)/(a))^((1)/(n-1))` and, P = Product of the first n terms `rArr" "P=a.ar.ar^(2) . . .ar^(n-1)=a^(n)r^(1+2+3+ . . .+(n-1)` `rArr" "P=a^(n)r^((n(n-1))/(2))=a^(n){((b)/(a))^((1)/(n-1))}^((n(n-1))/(2)` `rArr" "P=a^(n)((b)/(a))^(n//2)=a^(n//2)b^(n//2)=(ab)^(n//2)` `rArr" "P^(2)=[(ab)^(n//2)]^(2)=(ab)^(n)` |
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| 242. |
Statement -1: `(1^(2))/(1.3)+(2^(2))/(3.5)+(3^(2))/(5.7)+ . . . .+(n^(2))/((2n-1)(2n+1))=(n(n+1))/(2(2n+1))` Statement -2: `(1)/(1.3)+(1)/(3.5)+(1)/(5.7)+ . . . .+(1)/((2n-1)(2n+1))=(1)/(2n+1)`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 - C We, have `(1^(2))/(1.3)+(2^(2))/(3.7)+(3^(2))/(5.7)+ . . . . . . . . . . . . . +(n^(2))/((2n-1)(2n+1))` `=underset(r=1)overset(n)sum(r^(2))/((2r-1)(2r+1))` `=(1)/(4)underset(r=1)overset(n)sum(4r^(2))/((2r-1)(2r+1))` `=(1)/(4)underset(r=1)overset(n)sum((2r-1)(2r+1)+1)/((2r-1)(2r+1))` `=(1)/(4)underset(r=1)overset(n)sum{1+(1)/((2r-1)(2r+1))}` `=(1)/(4)underset(r=1)overset(n)sum1+(1)/(8)underset(r=1)overset(n)sum((1)/(2r-1)-(1)/(2r+1))` `=(n)/(4)+(1)/(8)(1-(1)/(2n+1))=(n(n+1))/(2(2n+1))` So, statement -1 is true. Statement -2 is fals, because `(1)/(1.3)+(1)/(3.5)+(1)/(5.7)+ . . . . .+(1)/((2n-1)(2n+1))` `=underset(r=1)overset(n)sum(1)/((2r-1)(2r+1))=(1)/(2)underset(r=1)overset(n)sum((1)/(2r-1)-(1)/(2r+1))` `=(1)/(2)(1-(1)/(2n+1))=(n)/(2n+1)` |
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| 243. |
If `sin theta,sqrt(2) (sintheta + 1),6 sin theta +6` are in GP, than the fifth term isA. 81B. `81sqrt(2)`C. 162D. `162sqrt(2)` |
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Answer» `[sqrt(2)(sintheta + 1)]^(2)= sin theta (6 sin theta +6)` `implies [(sintheta + 1) 2(sin theta+1)-6 sin theta]=0` We get, `sin theta=-1, (1)/(2)` ` therefore sin theta=(1)/(2)" [" sin theta =-1 " is not possible ]"` then first term` =a=sin theta=(1)/(2)` and common ratio `=r= sqrt(2)((1)/(2)+1)/((1)/(2))=3sqrt(2)` `therefore t_(5)=ar^(4)=(1)/(2)(3sqrt(2))^(4)=162` Hence, (c) is the correct answer. |
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| 244. |
find sum of `1+3x+6x^2+10x^3+15x^4 + -------oo` where ` |x| < 1 , x!= 0`A. `(1)/((1-x)^(2))`B. `(1)/(1-x)`C. `(1)/((1+x)^(2))`D. `(1)/((1-x)^(3))` |
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Answer» Correct Answer - D Clearly, 1,3,6,10, . . . . . Is not an A.P. But, successive differences of terms form on A.P. Let `S=1+3x+6x^(2)+10x^(3)+15x^(4)+ . . . .oo` `rArr" "S-xS=1+2x+3x^(2)+4x^(3)+5x^(4)+ . . . . oo` . . .(i) `rArr" "x(S-xS)=x+2x^(2)+3x^(3)+4x^(4)+ . . .` . . .(ii) Subtracting (ii) and (i), we get `S(1-x)^(2)=1+x+x^(2)+x^(3)+ . . ."to "oo` `rArr" "S(1-x)^(2)=(1)/(1-x)rArrS=(1)/((1-x)^(3))` |
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| 245. |
The sum of the first 9 terms of the series `1^3/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 +3^3)/(1 + 3 +5)` ..... is :A. 142B. 192C. 71D. 96 |
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Answer» Correct Answer - D Let `T_(n) " be "n^(th)` term of the 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)/(4)(n+1)^(2)` `:." Required sum"=underset(r=1)overset(9)sumT_(r)=underset(r=1)overset(9)sum(1)/(4)(r-1)^(2)` `=(1)/(4)(2^(2)+3^(2)+ . . . . +10^(2))` `=(1)/(4){(1^(2)+2^(2)+ . . . +10^(2))-1^(2)}` `=(1)/(4){(10(10+1)(20+1))/(6)-1}` `=(1)/(4)(384)=96` |
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| 246. |
The 1025th termm in the sequence are `1,22,4444,88888888,"…"`isA. `2^(9)`B. `2^(10)`C. `2^(11)`D. `2^(12) |
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Answer» The number of digits in each term of the sequence are `1,2,4,8,"…"` which are in GP. Let 1025th term is `2^(n)`. Then `1+2+4+8+"..."+2^(n-1)lt1025 le1+2+4+8+"..."+2^(n)` ` implies ((2-1)(1+2+2^(2)+2^(3)+"..."+2^(n-1)))/((2-1))lt1025` ` le ((2-1)(1+2+2^(2)+2^(3)+"..."+2^(n)))/((2-1))` ` implies 2^(n)-1lt1025 le2^(n+1)-1 implies 2^(n)lt1026 le 2^(n+1)"....(i)"` or `2^(n+1)ge 1026gt 1024` `implies 2^(n+1)gt2^(10) implies n+1gt10` ` therefore ngt9 therefore =10` [which is always satisfy Eq. (i)] Hence, (b) is the correct answer. |
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| 247. |
The sum of the `n` terms of the series `1+(1+3)+(1+3+5)+...` isA. `n^(2)`B. `{(n(n+1))/(2)}^(2)`C. `(n(n+1)(2n+1))/(6)`D. none of these |
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Answer» Correct Answer - C Let `a_(n)" be "n^(th)` term of the given series. Then, `a_(n)=1+3+5+ . . . +(2n-1)=n^(2)` `:." "` Sum of n-terms of the series `=underset(r=1)overset(n)suma_(r)=underset(r=1)overset(n)sumr^(2)=(n(n+1)(2n+1))/(6)` |
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| 248. |
If a, b and c are in H.P., then the value of `((ac+ab-bc)(ab+bc-ac))/(abc)^2` isA. `((a+c)(3a-c))/(4a^(2)c^(2))`B. `(2)/(bc)+(1)/(b^(2))`C. `(2)/(bc)-(1)/(a^(2))`D. `((a-c)(3a+c))/(4a^(2)c^(2))` |
| Answer» Correct Answer - A | |
| 249. |
If `a,b,c` are real numbers such that `3(a^(2)+b^(2)+c^(2)+1)=2(a+b+c+ab+bc+ca)`, than `a,b,c`are inA. AP onlyB. GP onlyC. GP and APD. None of these |
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Answer» Given, `3(a^(2)+b^(2)+c^(2)+1)=2(a+b+c+ab+bc+ca)` `implies 2(a^(2)+b^(2)+c^(2)-ab-bc-ca)+(a^(2)+b^(2)+c^(2)-2a -2b -2c+3)` `implies {(a-b)^(2)+(b-c)^(2)+(c-a)^(2)}+{(a-1)^(2)+(b-1)^(2)+(c-1)}=0` `implies a-b=b-c=c-a=0 " and "a-1=b-1=c-1=0` `implies a=b=c=1` `implies a,b,c` are in GP and AP. Hence, ia the correct answer. |
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| 250. |
Find the value of `0.3258.` |
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Answer» Let `R=0.3258` `implies R= 0.32585858585"....." " " "......(i)"` Here, number of figures which are not recurring is 2 and number of figures which are recurring is also 2. Then, `100R= 32.585858585"....." " " "......(ii)"` and `10000R= 3258.585858585"....." " " "......(iii)"` On subtracting Eq.(ii) from Eq.(iii), we get `9900R= 3226` `therefore " " R= (3226)/(9900)` Hence, `" " R= (1613)/(4950)` |
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