Find the 8^("th ") term in the sequence -(1)/(2),-1,-2,-4,dots

Identify pattern: Step 1: Identify the pattern in the sequence. The sequence given is -1/2, -1, -2, -4, ... Notice that each term is twice the previous term. This is a geometric sequence with the first term $ a = -\frac{1}{2} $ and the common ratio $ r = 2 $. Write formula: Step 2: Write the formula for the nth term of a geometric sequence. The nth term $ a_n $ of a geometric sequence can be found using the formula: \[ a_n = a \cdot r^{(n-1)} \] where $ a $ is the first term and $ r $ is the common ratio. Substitute values: Step 3: Substitute the values into the formula to find the 8th term. Using the formula from Step 2: \[ a_8 = -\frac{1}{2} \cdot 2^{(8-1)} \] \[ a_8 = -\frac{1}{2} \cdot 2^7 \] \[ a_8 = -\frac{1}{2} \cdot 128 \] \[ a_8 = -64 \]

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Q. Find the

8^{\text{th}}

term in the sequence

\newline

-\frac{1}{2},-1,-2,-4,\dots

\newline

Identify pattern: Step $1$ : Identify the pattern in the sequence. $\newline$ The sequence given is $-1$ / $2$ , $-1$ , $-2$ , $-4$ , ... $\newline$ Notice that each term is twice the previous term. This is a geometric sequence with the first term $a = -\frac{1}{2}$ and the common ratio $r = 2$ .
Write formula: Step $2$ : Write the formula for the nth term of a geometric sequence. $\newline$ The nth term $a_n$ of a geometric sequence can be found using the formula: $\newline$ $a_n = a \cdot r^{(n-1)}$ $\newline$ where $a$ is the first term and $r$ is the common ratio.
Substitute values: Step $3$ : Substitute the values into the formula to find the $8$ th term. $\newline$ Using the formula from Step $2$ : $\newline$ $a_8 = -\frac{1}{2} \cdot 2^{(8-1)}$ $\newline$ $a_8 = -\frac{1}{2} \cdot 2^7$ $\newline$ $a_8 = -\frac{1}{2} \cdot 128$ $\newline$ $a_8 = -64$