Geometric Progression from Two Given Terms

A GP mirrors an AP but with multiplication: terms are $ar^{n-1}$, so you divide the term equations to cancel $a$ and isolate a power of $r$. Once $r$ and $a$ are known, the sum formula finishes it.

In a geometric progression, the 2nd term is $6$ and the 5th term is $48$. (i) Find the common ratio $r$ and the first term $a$. [4] (ii) Find the sum of the first $8$ terms. [2]

The working

(i) Write each term with $u_n = ar^{n-1}$: $u_2 = ar = 6 \qquad u_5 = ar^4 = 48 \quad \text{(M1 for both)}$

Divide $u_5$ by $u_2$ (the $a$ cancels): $\frac{ar^4}{ar} = \frac{48}{6} \;\Rightarrow\; r^3 = 8 \;\Rightarrow\; r = 2 \quad \text{(A1)}$

Back-substitute into $ar = 6$: $2a = 6 \;\Rightarrow\; a = 3 \quad \text{(A1)}$

(ii) Use $S_n = \dfrac{a\left(r^n - 1\right)}{r - 1}$ with $n = 8$, $a = 3$, $r = 2$: $S_8 = \frac{3\left(2^8 - 1\right)}{2 - 1} = \frac{3(256 - 1)}{1} = 3 \times 255 = 765 \quad \text{(M1, A1)}$

Where the marks are won and lost

• Divide, do not subtract. $\frac{ar^4}{ar} = r^3$ is the move that isolates $r$. Subtracting leaves $ar^4 - ar$, which does not simplify usefully.
• $r^3 = 8$ gives $r = 2$ (real cube root). Since the term equations are consistent with a positive ratio, $r = 2$.
• In the sum, $2^8 = 256$ (not $2 \times 8$). Powers of $2$ are worth knowing cold for the non-calculator paper.

Common mistakes

• Subtracting the equations (AP method) on a GP.
• Miscomputing $2^8$ as $16$ or $64$.
• Using $\frac{a(1 - r^n)}{1 - r}$ with a sign slip, it is equivalent, but mixing the two forms mid-calculation causes errors.

Full method: Geometric Progressions notes. Topic home: Series pillar.