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​The eleventh term in the following sequence:​​−32+34−38+316−332+⋯- \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \frac{3}{32} + \cdots−23​
Question

The eleventh term in the following sequence:

32+3438+316332+- \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \frac{3}{32} + \cdots is​

A.

31024\frac{3}{1024}​​

B.

31024-\frac{3}{1024}​​

C.

32048\frac{3}{2048}​​

D.

32048-\frac{3}{2048}​​

Correct option is D

Given:
The sequence is:
32+3438+316332+- \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \frac{3}{32} + \cdots

Formula:
Tn=arn1T_n = a \cdot r^{n - 1} \\

Solution:

T11=32(12)10 =3211024 =32048 Final Answer:32048T_{11} = -\frac{3}{2} \cdot \left(-\frac{1}{2}\right)^{10} \\\ \\= -\frac{3}{2} \cdot \frac{1}{1024} \\\ \\= -\frac{3}{2048} \\\ \\\textbf{Final Answer:} \\{-\frac{3}{2048}}


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