Question

Look at this series: $2,\; 1,\; \tfrac{1}{2},\; \tfrac{1}{4},\; \ldots$ What number should come next?

• A. $\tfrac{1}{3}$
• B. $\tfrac{1}{8}$
• C. $\tfrac{2}{8}$
• D. $\tfrac{1}{16}$

Hint:

When terms shrink rapidly, test for a common ratio. If $a_{k+1/a_k$ is constant, you have a GP; the next term is $a_k \times r$.

Answer 1

The Correct Option is B

Step 1 (Check for additive pattern).
Compute consecutive differences: $1-2=-1$, $\tfrac{1}{2}-1=-\tfrac{1}{2}$, $\tfrac{1}{4}-\tfrac{1}{2}=-\tfrac{1}{4}$.
The differences themselves halve each time, suggesting a multiplicative (geometric) rule rather than a constant additive rule. Step 2 (Test multiplicative ratio).
Compute ratios: $\dfrac{1}{2}=\dfrac{1}{2}$, $\dfrac{\tfrac{1}{2}}{1}=\dfrac{1}{2}$, $\dfrac{\tfrac{1}{4}}{\tfrac{1}{2}}=\dfrac{1}{2}$.
Common ratio $r=\tfrac{1}{2}$ this is a geometric progression (GP). Step 3 (Write the general term to verify).
First term $a_1=2$, common ratio $r=\tfrac{1}{2}$. Then $a_n = a_1 \cdot r^{\,n-1} = 2\left(\tfrac{1}{2}\right)^{n-1}$.
Check: $a_2=2\cdot \tfrac{1}{2}=1$, $a_3=2\cdot \tfrac{1}{4}=\tfrac{1}{2}$, $a_4=2\cdot \tfrac{1}{8}=\tfrac{1}{4}$ (matches). Step 4 (Find the next term).
$a_5 = 2 \cdot \left(\tfrac{1}{2}\right)^{4} = 2 \cdot \tfrac{1}{16} = \tfrac{1}{8}$. Step 5 (Eliminate distractors).
$\tfrac{1}{3}$ is not a power of 2; $\tfrac{2}{8}=\tfrac{1}{4}$ repeats the previous term; $\tfrac{1}{16}$ would be the 6th} term ($a_6$). \[ \boxed{\tfrac{1}{8}\ \text{(Option (b)}} \]