Let a_k and b_k , k ∈ mathbbN, be two… - JEE Main 2023 Mathematics PYQ

JEE Main 2023 — Mathematics Algebra

2023integerhard

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in \mathbb{N}$ , be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $\mathrm{a}_1=\mathrm{b}_1=4$ and $\mathrm{r}_1<\mathrm{r}_2$ . Let $\mathrm{c}_{\mathrm{k}}=\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}, \mathrm{k} \in \mathbb{N}$ . If $\mathrm{c}_2=5$ and $\mathrm{c}_3=\frac{13}{4}$ then $\displaystyle \sum_{\mathrm{k}=1}^{\infty} \mathrm{c}_{\mathrm{k}}-\left(12 \mathrm{a}_6+8 \mathrm{~b}_4\right)$ is equal to

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given:

${c}_{k}={a}_{k}+{b}_{k}$ and ${a}_{1}={b}_{1}=4$

Also,

${a}_{2}=4{r}_{1}$ and ${a}_{3}=4{r}_{1}^{2}$

${b}_{2}=4{r}_{2}$ and ${b}_{3}=4{r}_{2}^{2}$

Now,

${c}_{2}={a}_{2}+{b}_{2}=5$

$\Rightarrow 4{r}_{1}+4{r}_{2}=5$

$\Rightarrow {r}_{1}+{r}_{2}=\frac{5}{4}$

And,

${c}_{3}={a}_{3}+{b}_{3}=\frac{13}{4}$

$\Rightarrow {r}_{1}^{2}+{r}_{2}^{2}=\frac{13}{16}$

$\Rightarrow {({r}_{1}+{r}_{2})}^{2}-2{r}_{1}{r}_{2}=\frac{13}{16}$

$\Rightarrow \frac{25}{16}-2{r}_{1}{r}_{2}=\frac{13}{16}$

$\Rightarrow 2{r}_{1}{r}_{2}=\frac{12}{16}$

$\Rightarrow {r}_{1}{r}_{2}=\frac{3}{8}$

$\Rightarrow 8{r}_{1}(\frac{5}{4}-{r}_{1})3$

$\Rightarrow 10{r}_{1}-8{r}_{1}^{2}=3$

$\Rightarrow 8{r}_{1}^{2}-10{r}_{1}+3=0$

$\Rightarrow {r}_{1}=\frac{10\pm \sqrt{100-96}}{16}$

$\Rightarrow {r}_{1}=\frac{3}{4},\frac{1}{2}$

$\Rightarrow {r}_{2}=\frac{1}{2},\frac{3}{4}$

Now,

$\sum _{k=1}^{\infty }{c}_{k}-(12{a}_{6}+8{b}_{4})$

$=({c}_{1}+{c}_{2}+{c}_{3}+....)+[12\times {4\times (\frac{1}{{2}^{5}})}+8{4\times {(\frac{3}{4})}^{3}}]$

$=({a}_{1}+{a}_{2}+{a}_{3}+....)+({b}_{1}+{b}_{2}+{b}_{3}+....)+[(\frac{3}{2})+(\frac{27}{2})]$

$=\frac{4}{1-{r}_{1}}+\frac{4}{1-{r}_{2}}-15$

$=24-15=9$

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