2017 H2 Maths Paper 2, Question 2

Let $d$ be the common difference. Using $S_{n} = \frac{n}{2}\left( 2a + (n - 1)d \right)$ with $a = 3$, $n = 13$:

Since $r \neq 1$, the sum of the first 13 terms of the geometric progression is

Substituting $r = 1$: $1^{13} - 52(1) + 51 = 1 - 52 + 51 = 0$, so $r = 1$ does satisfy this equation algebraically. However, the equation was derived by multiplying the sum formula (valid only for $r \neq 1$) through by $(r - 1)$, which introduces $r = 1$ as an extraneous root of the cleared equation. Directly, if $r = 1$ every term of the progression equals $3$, so $S_{13} = 13(3) = 39 \neq 156$. So $r = 1$ cannot be the actual common ratio, even though it is a root of $r^{13} - 52r + 51 = 0$.

By GC, the real roots of $r^{13} - 52r + 51 = 0$ are $r = 1$ (rejected, as shown) and

Since the common ratio is positive, $r = 1.21002397\ldots$ (retaining the unrounded GC value from part (ii)). The $n$th term of the geometric progression is $3r^{n - 1}$; the $n$th term of the arithmetic progression is $3 + (n - 1)\left( \frac{3}{2} \right)$. The given condition is

By GC, tabulating both sides for consecutive integer $n$:

At $n = 41$: $3r^{40} = 6150\ \left( 3\ \text{s.f.} \right)$, while $100 \times \left( \text{41st term of A.P.} \right) = 6300$ — inequality not satisfied.

At $n = 42$: $3r^{41} = 7440\ \left( 3\ \text{s.f.} \right)$, while $100 \times \left( \text{42nd term of A.P.} \right) = 6450$ — inequality satisfied.