Q.Find the sum to infinity of the GP $8, 4, 2, 1, \ldots$

The common ratio here is $r = \tfrac12$, and since $|r| < 1$ the terms shrink toward zero, so a finite sum to infinity exists.

Given: GP $8, 4, 2, 1, \ldots$

Step 1 — Identify $a$ and $r$: $a = 8$; $r = 4/8 = \tfrac12$ (check: $2/4 = \tfrac12$, $1/2 = \tfrac12$ ✓).

Step 2 — Confirm $S_\infty$ exists: $|r| = \tfrac12 < 1$, so the sum to infinity is defined.

Step 3 — Apply the formula: $S_\infty = \dfrac{a}{1-r} = \dfrac{8}{1 - \tfrac12} = \dfrac{8}{\tfrac12}$.

Step 4 — Simplify: dividing by $\tfrac12$ is the same as multiplying by 2: $S_\infty = 8 \times 2 = 16$. …

Trying to apply $S_\infty = a/(1-r)$ to a GP where $|r| \geq 1$ — the formula only works when $|r|<1$; otherwise …