Formula for a geometric sequence read from a graph

Read the points: $g_1=8$, $g_2=4$, $g_3=2$, $g_4=1$, $g_5=\frac{1}{2}$. Each term is half the one before, so the common ratio is $r=\frac{1}{2}$.

The standard form starts from the first term: $g_n=g_1{r}^{n-1}=8{\left(\frac{1}{2}\right)}^{n-1}$. None of the choices is written that way, so test each by plugging in values of $n$.

Choice A, $4{\left(\frac{1}{2}\right)}^{n-2}$: - $n=1$: $4{\left(\frac{1}{2}\right)}^{-1}=4\cdot 2=8$ - $n=2$: $4{\left(\frac{1}{2}\right)}^0=4$ - $n=3$: $4{\left(\frac{1}{2}\right)}^1=2$

It matches. That makes sense: any term can anchor the formula, and here it is anchored at $g_2=4$, so the exponent counts steps from $n=2$. Algebraically $4{\left(\frac{1}{2}\right)}^{n-2}=4\cdot 2\cdot {\left(\frac{1}{2}\right)}^{n-1}=8{\left(\frac{1}{2}\right)}^{n-1}$.

The answer is A.

Each choice

• A (correct). Anchored at the second term, $g_2=4$, with ratio $\frac{1}{2}$, it gives 8, 4, 2, 1 for $n=1$ to 4.
• B. The first term is right but the ratio is 2, so this sequence grows: 8, 16, 32. The graph falls by half each step. Halving means multiplying by $\frac{1}{2}$, not by 2.
• C. With exponent $n$ instead of $n-1$, the first term comes out as $8\cdot \frac{1}{2}=4$. The whole sequence is shifted one step, so every value is half of what the graph shows.
• D. This uses the right ratio but the wrong anchor. At $n=1$ it gives $16{\left(\frac{1}{2}\right)}^0=16$, not 8. The value 16 would fit an exponent of $n$, not $n-1$.