Sequences And Series Question 7
Question 7 - 29 January - Shift 2
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 $c_3=\frac{13}{4}$ then $\sum _{k=1}^{\infty} c_k-(12 a_6+8 b_4)$ is equal to ________
Show Answer
Answer: 9
Solution:
Formula: Sum of an infinite G.P. (III)
$\begin{gathered} a_1=b_1=4 ; \quad c_2=a_1 r_1+b_1 r_2 ; \quad c_2=4 r_1+r_2=5 \\ c_3=4 r_1^2+r_2^2=\frac{13}{4} \ r_1+r_2^2-2 r_1 r_2=\frac{13}{16} \Rightarrow \frac{25}{16}-\frac{13}{16}=2 r_1 r_2 \\ \Rightarrow 2 r_1 r_2=\frac{12}{16}=\frac{3}{4} \Rightarrow r_1=\frac{1}{2}, r_2=\frac{3}{4} \\ \sum_{k=1}^{\infty} C_k=41+\frac{1}{2}+\frac{1}{2}+\ldots+41+\frac{3}{4}+\frac{3^2}{4}+\ldots \\ =8+4 \times 4=24 \\ \text { Ans }=24-15=9 \end{gathered}$
