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`