SOLUTION: 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 term

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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
Answer by venugopalramana(3286) (Show Source):
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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
LET THE NUMBER OF TERMS =2N ; T1=A, LET COMMON DIFFERENCE=D.HENCE T2=T1+D=A+D
T1+T3+T5+......+T(2N-1)=24..HERE CD=2D..NUMBER OF TERMS =N�AND SUM=24=(N/2){2A+(N-1)2D}...........I
T2+T4+T6+......T(2N)=30.....HERE CD=2D�NUMBER OF TERMS =N�AND SUM=30=(N/2){2A+2D+(N-1)2D}........II
EQN.II - EQN.I GIVES
6=(N/2)(2D)=ND��III
WE ARE GIVEN THAT
T(2N)-T1=21/2...OR T(2N)=(21/2)+T1=(21/2)+A
BUT T(2N)=A+(2N-1)D...
HENCE A+(2N-1)D=A+21/2...OR...2ND=D+21/2.........IV
SUBSTITUTING FROM EQN.III�FOR ND , WE GET
2*6=D+21/2�OR�.D=12-10.5=1.5
HENCE
6=ND=1.5*N�OR�.N=4
HENCE NUMBER OF TERMS =2N = 2*4=8