Which of the following series forms an AP?
(i)a–2d,a–d,a,a+d,a+2d
(ii)a,a+d,a+2d,a+3d,a+4d
(iii)a–3d,a–d,a+d,a+3d
(i),(ii) and (iii)
Consider each series (i)a–2d,a–d,a,a+d,a+2d
Difference between first two consecutive terms =a–d–(a–2d)=d
Difference between third and second consecutive terms =a–(a–d)=d
Difference between fourth and third consecutive terms =a+2d–(a+d)=d
Since a2 –a1=a3 –a2=a4 –a3
So, the sequence is in AP
Similarly, (ii) has common diffrence of d and (iii) has commonn difference of 2d and all are in A.P.