Sequences & Series
Finding a Term of Increasing GP
nta_pyq_2025_apr
Grade 11
Question:
Let an increasing GP have first term $a_1$ and common ratio $r$. If $a_1 a_5 = 28$ and $a_2+a_4=29$, then $a_6$ equals
Step-by-Step Solution
Key Concept: Express $a_3=\sqrt{a_1a_5}=\sqrt{28}$ as the GP's geometric mean, then write $a_2+a_4=a_3(r^{-1}+r)=29$ and verify by trying $r=\sqrt{28}$.
GM: $a_3=\sqrt{a_1a_5}=\sqrt{28}$, so $a_1r^2=\sqrt{28}$.
$a_2+a_4=a_1r+a_1r^3=a_1r(1+r^2)=\dfrac{\sqrt{28}}{r}\cdot(1+r^2)=\sqrt{28}\left(\dfrac{1}{r}+r\right)=29$.
Try $r=\sqrt{28}$: $\sqrt{28}\left(\dfrac{1}{\sqrt{28}}+\sqrt{28}\right)=1+28=29$ ✓.
So $a_1=\dfrac{\sqrt{28}}{r^2}=\dfrac{\sqrt{28}}{28}=\dfrac{1}{\sqrt{28}}$.
$a_6=a_1r^5=\dfrac{1}{\sqrt{28}}\cdot(\sqrt{28})^5=(\sqrt{28})^4=28^2=784$.
Correct Answer: 4