Let {a_k} and {b_k}, k in N,be two G.P.s with | vedclass

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(A) Given that $c_k = a_k + b_k$.Since $a_k$ and $b_k$ are $G$.$P$.s with $a_1 = b_1 = 4$,we have $a_k = 4r_1^{k-1}$ and $b_k = 4r_2^{k-1}$.Given $c_2

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$9$

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(A) Given that $c_k = a_k + b_k$.Since $a_k$ and $b_k$ are $G$.$P$.s with $a_1 = b_1 = 4$,we have $a_k = 4r_1^{k-1}$ and $b_k = 4r_2^{k-1}$.Given $c_2 = a_2 + b_2 = 4r_1 + 4r_2 = 5 \Rightarrow r_1 + r_2 = 5/4$.Given $c_3 = a_3 + b_3 = 4r_1^2 + 4r_2^2 = 13/4 \Rightarrow r_1^2 + r_2^2 = 13/16$.Using $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2$,we get $(5/4)^2 = 13/16 + 2r_1r_2$ $\Rightarrow 25/16 - 13/16 = 2r_1r_2$ $\Rightarrow 12/16 = 2r_1r_2$ $\Rightarrow r_1r_2 = 3/8$.The quadratic equation $t^2 - (5/4)t + 3/8 = 0$ gives $8t^2 - 10t + 3 = 0 \Rightarrow (4t-3)(2t-1) = 0$.Since $r_1 Now,$\sum_{k=1}^{\infty} c_k = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k = \frac{4}{1-1/2} + \frac{4}{1-3/4} = 8 + 16 = 24$.Also,$12a_6 + 8b_4 = 12(4(1/2)^5) + 8(4(3/4)^3) = 48/32 + 32(27/64) = 3/2 + 27/2 = 30/2 = 15$.Thus,the required value is $24 - 15 = 9$.

Let $\{a_k\}$ and $\{b_k\}, k \in N$,be two $G$.$P$.s with common ratios $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+b_k, k \in N$. If $c_2=5$ and $c_3=13/4$,then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to

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