Sequences & Series
Sum of infinite GP with two GPs
nta_pyq_2023_jan
Grade 11
Question:
Let $\{a_k\}$ and $\{b_k\}$, $k \in \mathbb{N}$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1 = b_1 = 4$ and $r_1 < r_2$. Let $c_k = a_k + b_k$, $k \in \mathbb{N}$. If $c_2 = 5$ and $c_3 = \dfrac{13}{4}$, then $\displaystyle\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to ______.
Step-by-Step Solution
Key Concept: From $c_2 = a_2 + b_2 = 4r_1 + 4r_2 = 5$ and $c_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4}$. Solve for $r_1$ and $r_2$.
$r_1 + r_2 = \frac{5}{4}$, $r_1^2 + r_2^2 = \frac{13}{16}$, so $r_1 r_2 = \frac{3}{8}$, giving $r_1 = \frac{1}{2}$, $r_2 = \frac{3}{4}$. $\sum c_k = \frac{4}{1-1/2} + \frac{4}{1-3/4} = 8 + 16 = 24$. $12a_6 + 8b_4 = 12 \cdot \frac{4}{32} + 8 \cdot \frac{4 \cdot 27}{64} = \frac{3}{2} + \frac{27}{2} = 15$. Answer $= 24 - 15 = 9$.
Correct Answer: 9