Sequences & Series
GP with Varying Denominator — Finding a₁
nta_pyq_2026_jan
Grade 11
Question:
Let $a_1,\dfrac{a_2}{2},\dfrac{a_3}{2^2},\ldots,\dfrac{a_{10}}{2^9}$ be a G.P. of common ratio $\dfrac{1}{\sqrt{2}}$. If $a_1+a_2+\ldots+a_{10}=62$, then $a_1$ is equal to:
$2-\sqrt{2}$
$2(2-\sqrt{2})$
$\sqrt{2}-1$
$2(\sqrt{2}-1)$
Step-by-Step Solution
Key Concept: Consecutive ratio: $\tfrac{a_{n+1}/2^n}{a_n/2^{n-1}}=\tfrac{a_{n+1}}{2a_n}=\tfrac{1}{\sqrt{2}}\Rightarrow\tfrac{a_{n+1}}{a_n}=\sqrt{2}$. So $a_n=a_1(\sqrt{2})^{n-1}$.
$a_1=2(\sqrt{2}-1)$.
Correct Answer: 4