QJEEM23Ma2df4: JEE Main 2023 Mathematics - Class 11 JEEM23Ma2df4 Solution: Let and , be two G.P.s with common ratios and respectively such …

$\{ {a_k}\}$ be a G.P. with ${a_1} = 4,r = {r_1}$

And

$\{ {b_k}\}$ be G.P. with ${b_1} = 4,r = {r_2}$ ({r_1} < {r_2})

Now

${C_k} = {a_k} + {b_k}$

${c_1} = 4 + 4 = 8$ and ${c_2} = 5$

${a_2} + {b_2} = 5$

$\therefore$ ${r_1} + {r_2} = {5 \over 4}$

and ${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$

$\therefore$ ${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$

$\therefore$ ${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$

$\therefore$ ${r_2} = {3 \over 4},{r_1} = {1 \over 2}$

$\therefore$ ${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$

and $\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24}$

$\therefore$ $\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09}$