C-13-Q143mediumsingle_mcqThe number of terms in 1+3+5+⋯+1251 + 3 + 5 + \cdots + 1251+3+5+⋯+125 is —a636363b626262c646464d656565ব্যাখ্যাThese are consecutive odd numbers, so tn=1+(n−1)⋅2=125t_n = 1+(n-1)\cdot 2 = 125tn=1+(n−1)⋅2=125 gives 2n−1=1252n-1=1252n−1=125, hence n=63n=63n=63.