C-13-Q109mediumsingle_mcqThe fifth term of 1+12+14+⋯1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots1+21+41+⋯ is —a18\dfrac{1}{8}81b116\dfrac{1}{16}161c132\dfrac{1}{32}321d14\dfrac{1}{4}41ব্যাখ্যাWith a=1a=1a=1 and r=12r=\tfrac12r=21, the fifth term is (12)4=116\left(\tfrac12\right)^{4}=\dfrac{1}{16}(21)4=161.