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Question

The number of terms of an A.P. is even ; sum of all terms at odd places and even places are 24 and 30 respectively . Last term exceeds the first term by 10.5. Then the number of terms in the series is

A
8
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B
12
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C
16
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D
20
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Solution

The correct option is B 12
Given that π,T2,T3..T2n are in APT1+T3+T5++T2n1=24T2+T4+T6++T2n=30T2n=T1+10.5 let T1=a and d= common difference.
a+(2n1)d=a+10.5(2n1)d=10.5T1+T3+T5++Tn1=24
a+(a+2d)+(a+4d)++(a+(2n2)d)=24na+[0+2d+4d++(2n2)d]=24na+2d[1+2+3++(n1)]=24
na+2d[(m1)(n+1)2]=24na+d(n1)(n+0)=24n(a+n1)d=24(1)



T2+T4+T6++T2n=30(a+d)+(a+3d)+(a+5d)++(a+(2n1)d)=30na+d(1+3+5++2n1)=30na+dn2=30



n(a+dn)=30 Fuom (1) and (2)na+dn2dn=2430dn2=24dn2=6n2=6(2n1)10.5n=12

Hence B is the correct option

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