Choose the odd letters from the given alternatives.
1). PON
2). SRQ
3). XYZ
4). VUT
1) PON ⇒ P – 1 = O; O – 1 = P
2) SRQ ⇒ S – 1 = R; R – 1 = Q
3) XYZ ⇒ X + 1 = Y; Y + 1 = Z
4) VUT ⇒ V – 1 = U; U – 1 = T
Therefore, “XYZ” is odd one among given alternatives.