document.write( "Question 24917: The number of terms in an Arithmetic Progress is even; the sum of the odd terms is 24, of the even term 30 and the last term exceeds the first term by 21/2. Fine the number of terms \n" ); document.write( "
Algebra.Com's Answer #13434 by venugopalramana(3286)\"\" \"About 
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The number of terms in an Arithmetic Progress is even; the sum of the odd terms is 24, of the even term 30 and the last term exceeds the first term by 21/2. Fine the number of terms
\n" ); document.write( "LET THE NUMBER OF TERMS =2N ; T1=A, LET COMMON DIFFERENCE=D.HENCE T2=T1+D=A+D
\n" ); document.write( "T1+T3+T5+......+T(2N-1)=24..HERE CD=2D..NUMBER OF TERMS =N…AND SUM=24=(N/2){2A+(N-1)2D}...........I
\n" ); document.write( "T2+T4+T6+......T(2N)=30.....HERE CD=2D…NUMBER OF TERMS =N…AND SUM=30=(N/2){2A+2D+(N-1)2D}........II
\n" ); document.write( "EQN.II - EQN.I GIVES
\n" ); document.write( "6=(N/2)(2D)=ND………………………III
\n" ); document.write( "WE ARE GIVEN THAT
\n" ); document.write( "T(2N)-T1=21/2...OR T(2N)=(21/2)+T1=(21/2)+A
\n" ); document.write( "BUT T(2N)=A+(2N-1)D...
\n" ); document.write( "HENCE A+(2N-1)D=A+21/2...OR...2ND=D+21/2.........IV
\n" ); document.write( "SUBSTITUTING FROM EQN.III…FOR ND , WE GET
\n" ); document.write( "2*6=D+21/2…OR….D=12-10.5=1.5
\n" ); document.write( "HENCE
\n" ); document.write( "6=ND=1.5*N…OR….N=4
\n" ); document.write( "HENCE NUMBER OF TERMS =2N = 2*4=8\r
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