JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\{a_k\}$ and $\{b_k\}$ , $k \in \mathbb{N}$ , 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 $c_k = a_k + b_k, k \in \mathbb{N}$ . If $c_2 = 5$ and $c_3 = \frac{13}{4}$ then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: $r_1$ aur $r_2$ ki values nikalna

Hume diya hai:

$c_2 = a_2 + b_2 = 4r_1 + 4r_2 = 5 \implies r_1 + r_2 = \frac{5}{4}$
$c_3 = a_3 + b_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4} \implies r_1^2 + r_2^2 = \frac{13}{16}$

Hum jante hain $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2$ :

$(\frac{5}{4})^2 = \frac{13}{16} + 2r_1r_2 \implies \frac{25}{16} - \frac{13}{16} = 2r_1r_2$

$\frac{12}{16} = 2r_1r_2 \implies r_1r_2 = \frac{6}{16} = \frac{3}{8}$

Ab quadratic equation $t^2 - (\frac{5}{4})t + \frac{3}{8} = 0$ ko solve karne par:

$8t^2 - 10t + 3 = 0 \implies (4t-3)(2t-1) = 0$

$t = \frac{3}{4}, \frac{1}{2}$ .

Kyonki $r_1 < r_2$ , isliye $r_1 = \frac{1}{2}$ aur $r_2 = \frac{3}{4}$ .

Step 2: Infinite Sum ( $\sum c_k$ ) nikalna

\sum_{k=1}^{\infty} c_k = \sum a_k + \sum b_k = \frac{a_1}{1-r_1} + \frac{b_1}{1-r_2}

\frac{4}{1 - 1/2} + \frac{4}{1 - 3/4} = \frac{4}{1/2} + \frac{4}{1/4} = 8 + 16 = 24

Step 3: $(12a_6 + 8b_4)$ ki value nikalna

$a_6 = a_1 r_1^5 = 4(\frac{1}{2})^5 = 4 \cdot \frac{1}{32} = \frac{1}{8}$
$b_4 = b_1 r_2^3 = 4(\frac{3}{4})^3 = 4 \cdot \frac{27}{64} = \frac{27}{16}$

Ab calculation:

$12(\frac{1}{8}) + 8(\frac{27}{16}) = \frac{3}{2} + \frac{27}{2} = \frac{30}{2} = 15$ .

Step 4: Final Answer

\text{Result} = 24 - 15 = 9

Answer:

The value is 9.

Choose the correct answer:

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