Sum to Infinity of a Geometric Progression

The sum to infinity $S_\infty = \dfrac{a}{1 - r}$ exists only when $|r| < 1$. This question runs the formula in reverse, given $a$ and $S_\infty$, find $r$, then use it. Checking the convergence condition at the end is part of a complete answer.

A geometric progression has first term $20$ and a sum to infinity of $25$. (i) Find the common ratio $r$. [3] (ii) Find the third term of the progression. [2]

The working

(i) Substitute into $S_\infty = \dfrac{a}{1 - r}$ with $a = 20$, $S_\infty = 25$: $25 = \frac{20}{1 - r} \quad \text{(M1)}$

Solve for $r$: $25(1 - r) = 20 \;\Rightarrow\; 1 - r = \frac{20}{25} = 0.8 \;\Rightarrow\; r = 0.2 \quad \text{(M1, A1)}$

Since $|0.2| < 1$, the sum to infinity is valid, good.

(ii) The third term is $ar^2$: $u_3 = 20 \times (0.2)^2 = 20 \times 0.04 = 0.8 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Rearranging $S_\infty = \frac{a}{1-r}$ is the method: multiply up by $(1 - r)$ before isolating $r$. Trying to invert it in one step invites slips.
• The resulting $r = 0.2$ must satisfy $|r| < 1$; if it did not, the setup would be wrong. This check is worth stating.
• The third term uses $r^2$ (power $n - 1 = 2$), not $r^3$. Off-by-one on the index is the usual error.

Common mistakes

• Writing $1 - r = 25/20$ (dividing the wrong way).
• Using $ar^3$ for the third term.
• Forgetting the convergence check, or, worse, applying $S_\infty$ to a series with $|r| \ge 1$.

Full method: Sum to Infinity notes. Topic home: Series pillar.