C-13-Q108mediumsingle_mcq

Which term of the series $128 + 64 + 32 + \cdots$ is $\dfrac{1}{2}$ ?

a $9$ th
b $8$ th
c $10$ th
d $7$ th

ব্যাখ্যা

Set $128\left(\tfrac12\right)^{n-1}=\tfrac12$ , so $\left(\tfrac12\right)^{n-1}=\dfrac{1}{256}=\left(\tfrac12\right)^{8}$ , giving $n-1=8$ , hence $n=9$ . So it is the $9$ th term.

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Finite Series — সব প্রশ্ন দেখুন (160টি) →

Which term of the series $128 + 64 + 32 + \cdots$ is $\dfrac{1}{2}$ ?

9

8

10

7

ব্যাখ্যা

Set $128\left(\tfrac12\right)^{n-1}=\tfrac12$ , so $\left(\tfrac12\right)^{n-1}=\dfrac{1}{256}=\left(\tfrac12\right)^{8}$ , giving $n-1=8$ , hence $n=9$ . So it is the $9$ th term.

Which term of the series 128+64+32+⋯128 + 64 + 32 + \cdots128+64+32+⋯ is 12\dfrac{1}{2}21​?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Which term of the series 128+64+32+⋯128 + 64 + 32 + \cdots128+64+32+⋯ is 12\dfrac{1}{2}21​?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

Which term of the series $128 + 64 + 32 + \cdots$ is $\dfrac{1}{2}$ ?

Which term of the series $128 + 64 + 32 + \cdots$ is $\dfrac{1}{2}$ ?