8ProgressionsLEVEL 2 66.Thegivensequencecanbewrittenas 5220131092023, , , , The reciprocal of the numbers in the sequence are820132018202320, , , , Thisisan A.P.nthtermofthisA.P.isan= 8 5 120+ - ( ) n= 3 520+nThus,nthtermofthegivensequenceis 205 3 n +.67.H.M.betweenaandbis 2 a ba b += 2128 1 10128 1 10( / ) ( / )/ / += 11968.a1 1b c+,b 1 1c a+,c1 1a b+areinA.P.a1 1b c++1,b 1 1c a++1,c1 1a b++1are in A.P.abc ca aba b c+ + ,bbc ca ababc+ +,cbc ca ababc+ +areinA.P.a,b,careinA.P.69.sin d[cosec a1cosec a2+cosec a2cosec a3+L+ cosecan1cosecan]= sin ( )sin sina aa a2 11 2-+ sin ( )sin sina aa a3 22 3-+Lsin ( )sin sina aa an nn n---11=(cota1cota2)+(cota2cota3)+L+(cotan1cotan)=cota1cotan70.1+ 3253+ +L+ 2 1 nn-=(21)+212213- + -+L+21-n=2nHn71. a,b,careinH.P. 1 1 1a b c, , areinA.P. a b ca+ +2, a b cb+ +2, a b cc+ + 2 are inA.P. b c aa+ -, c a bb+ -, a b cc+ -areinA.P. ab c a + -, bc a b + -, ca b c + -areinH.P.72.Lasttermofthenthrow=1+2+3+L+n= 12n(n+1)Notethatsumofthetermsinthenthrow8.2JEE Main MathematicsLevel 2, Hints and Solutions=sumtontermsofanA.P.whosersttermin 12n(n+1)andcommondifferenceis(1).= 122121 1 1 n n n n+ + - -( ) ( ) ( ) =1212n n ( ) +73.WehaveA1=a+ 13(ba)= 23a b +andA2= a b + 23\A1+A2=a+bAlso, 11H= 132 1a b+= 132b aa b+ and 12H= 132a bab+Also, 1 11 2H H+ = 1 1a b+H1+H2= a ba b+H1 H2Therefore, A AH H1 21 2++= abHH1 2= a b a b a bab 92 22( ) ( )( )+ += 2 592 2( ) a b a ba b+ +74. 1112 n n ++++L+ 12n=11212+ + +n1121+ + +n=112131412 112- + - + +--n n+2121412+ + +n 1121+ + +n=a(2n)75.Let m be the minimum number lled in the boxes. Let a, b, c, d be the numbers in the boxes sharing asidewiththeboxcontainingm,thenm= 14(a+b+c+d)(am)+(bm)+(cm) + (dm) = 0(1)Asam0,bm0,cm0,dm0,(1)ispossibleifandonlyifam=0,bm=0,cm=0dm=0Thus,eachoftheboxcontainsm.Astherstboxcontains5,m=5.Therefore,x=576.a+8b+7c=09a+2b+3c=05a+5b+5c=0asb=6,wegeta=1,c=7Therootsofquadraticequationx2+6x7=0x=1,7Now, 1 1a b+ =117- = 67Thus, 1 10a b+=nn= 670=nn= 11 6 7 - /=7