Q.Find the sum to infinity of the GP whose first term is 27 and common ratio is $\tfrac13$.

Since $|r| = \tfrac13 < 1$, the sum to infinity exists and can be found directly from the formula.

Given: $a = 27$, $r = \tfrac13$, find $S_\infty$.

Step 1 — Confirm $S_\infty$ exists: $|r| = \tfrac13 < 1$ ✓.

Step 2 — Apply the formula: $S_\infty = \dfrac{a}{1-r} = \dfrac{27}{1-\tfrac13} = \dfrac{27}{\tfrac23}$.

Step 3 — Simplify: dividing by $\tfrac23$ is the same as multiplying by $\tfrac32$: $S_\infty = 27 \times \tfrac32 = \dfrac{81}{2} = 40.5$. …

Computing $1 - r$ incorrectly as a whole number instead of a fraction — $1 - \tfrac13 = \tfrac23$, not $1$ or a n …