C-13-Q107mediumsingle_mcqThe eighth term of 64+32+16+8+⋯64 + 32 + 16 + 8 + \cdots64+32+16+8+⋯ is —a12\dfrac{1}{2}21b111c14\dfrac{1}{4}41d222ব্যাখ্যাWith a=64a=64a=64 and r=12r=\tfrac12r=21, the eighth term is 64(12)7=26⋅2−7=2−1=1264\left(\tfrac12\right)^{7}=2^{6}\cdot 2^{-7}=2^{-1}=\dfrac{1}{2}64(21)7=26⋅2−7=2−1=21.