JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the sum and product of four positive consecutive terms of a G.P. are $126$ and $1296$ , respectively, then the sum of common ratios of all such GPs is:

Choose the correct answer:

Explanation

Mathematical Steps

Maana terms: $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ (Common ratio $= r^2$ )

1. Product Equation:

\left( \frac{a}{r^3} \right) \left( \frac{a}{r} \right) (ar) (ar^3) = 1296

a^4 = 1296 \implies a = 6

2. Sum Equation:

\frac{6}{r^3} + \frac{6}{r} + 6r + 6r^3 = 126

Divide by $6$ :

\left( r^3 + \frac{1}{r^3} \right) + \left( r + \frac{1}{r} \right) = 21

Maana $x = r + \frac{1}{r}$ :

(x^3 - 3x) + x = 21

x^3 - 2x - 21 = 0

$x = 3$ iska ek root hai ( $27 - 6 - 21 = 0$ ).

3. Finding Common Ratio ( $R = r^2$ ):

r + \frac{1}{r} = 3

Multiply by $r$ :

r^2 - 3r + 1 = 0

Roots are $r = \frac{3 \pm \sqrt{5}}{2}$ .

Common ratio $R = r^2$ :

R_1 = \left( \frac{3 + \sqrt{5}}{2} \right)^2 = \frac{9 + 5 + 6\sqrt{5}}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}

R_2 = \left( \frac{3 - \sqrt{5}}{2} \right)^2 = \frac{9 + 5 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2}

4. Sum of all Common Ratios:

\text{Sum} = R_1 + R_2 = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2}

\text{Sum} = \frac{14}{2} = 7

Correct Option: (1) 7

Choose the correct answer:

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