Let a,d and 2n be the first term, common difference and total number of terms of an A.P. respectively i.e. a +(a+d)+(a+2d)+...+(a+(2n−1)d) No. of even terms =n, No. of odd terms =n Sum of odd terms : So=
n
2
[2a+(n−1)(2d)]=24 ⇒n[a+(n−1)d]=24 . . . (i) Sum of even terms: Se=
n
2
[2(a+d)+(n−1)2d]=30 ⇒n[a+d+(n−1)d]=30 . . . (ii) Subtracting equation (i) from (ii), we get nd=6 . . . (iii) Also, given that last term exceeds the first term by