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The number of terms of an A.P. is even, the sum of odd terms is 24, of the even terms is 3, and the last term exceeds the first by 10 1/2 find the number of terms and the series.
A. 8
B. 4
C. 6
D. 10

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Correct Answer - A
Let the series have 2n terms and the series is a,a+d,a+2d,…,a+(2n-1)d.
According to the given conditions, we have
`[a+(a+2d)+(a+4d)+…+(a+(2n-2)d)]=24`
`rArrn/2[2a+(n-1)2d]=24`
or `n[a+(n-1)d]=24 (1)`
Also,
`[(a+d)+(a+3d)+…+(a+(2n-1)d)]=30`
or `n[(a+d)+(n-1)d]=30 (2)`
Also, the last term exceeds the first by 21/2. Therefore,
`a+(2n-1)d-a=21/2
or (2n-1)d=21/2 (3)`
Now, subtracting (1) from (2),
nd=6
Dividing (3) by (4), we get
`(2n-1)/n=21/12`
`rArrn=4`

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