8. Sequences and Series
hard

माना $\left\{a_{\mathrm{k}}\right\}$ तथा $\left\{\mathrm{b}_{\mathrm{k}}\right\}, \mathrm{k} \in \mathbb{N}$, दो G.P. है, जिनके सार्व अनुपात क्रमशः $r_1$ तथा $r_2$ है और $a_1=b_1=4$, $\mathrm{r}_1<\mathrm{r}_2$ है। माना $\mathrm{c}_{\mathrm{k}}=\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}, \mathrm{k} \in \mathbb{N}$ है। यदि $\mathrm{c}_2=5$ तथा $\mathrm{c}_3=\frac{13}{4}$ है तो $\sum_{\mathrm{k}=1}^{\infty} \mathrm{c}_{\mathrm{k}}-\left(12 \mathrm{a}_6+8 \mathrm{~b}_4\right)$ बराबर है________. 

A

$9$

B

$18$

C

$20$

D

$22$

(JEE MAIN-2023)

Solution

Given that

$c_k=a_k+b_k$ and

also

$a _2=4 r _1$ $\quad a _3=4 r _1{ }^2$

$b _2=4 r _2$ $\quad b _3=4 r _2{ }^2$

Now $c_2=a_2+b_2=5$ and $c_3=a_3+b_3=\frac{13}{4}$

$\Rightarrow r_1+r_2=\frac{5}{4}$ and $r_1^2+r_2^2=\frac{13}{16}$

Hence $r_1 r_2=\frac{3}{8}$ which gives $r_1=\frac{1}{2} \quad \& \quad r_2=\frac{3}{4}$

$\sum \limits_{ k =1}^{\infty} c _{ k }-\left(12 a _6+8 b _4\right)$

$=\frac{4}{1-r_1}+\frac{4}{1-r_2}-\left(\frac{48}{32}+\frac{27}{2}\right)$

$=24-15=9$

Standard 11
Mathematics

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