C-13-Q108mediumsingle_mcqWhich term of the series 128+64+32+⋯128 + 64 + 32 + \cdots128+64+32+⋯ is 12\dfrac{1}{2}21?a999thb888thc101010thd7thব্যাখ্যাSet 128(12)n−1=12128\left(\tfrac12\right)^{n-1}=\tfrac12128(21)n−1=, so , giving , hence . So it is the th term.