For a G.P., its fourth term = x, seventh term = y and tenth term = z. Assertion (A): x, y and z are in G.P. Reason (R): y2 = (ar6)² = ar³ × ar⁹ = xz. 1. A is true, R is false. 2. A is false, R is true. 3. Both A and R are true and R is correct reason for A. 4. Both A and R are true and R is incorrect reason for A.

Mathematics

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Let first term of the G.P. be a and common ratio be r.

By formula :

⇒ T_n = a.r^{n - 1}

Given, fourth term = x, seventh term = y and tenth term = z

⇒ a₄ = x, a₇ = y and a₁₀ = z

⇒ ar^{4 - 1} = x, ar^{7 - 1} = y and ar^{10 - 1} = z

⇒ ar³ = x, ar⁶ = y and ar⁹ = z

If x, y and z are in G.P., then the ratio between the consecutive terms will be equal.

Ratio between y and x :

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3.$

Ratio between z and y :

$\Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3.$

Since, the ratio between the consecutive terms are equal.

Thus, x, y and z are in G.P.

∴ Assertion (A) is true.

⇒ y² = (ar⁶)²

⇒ y² = ar¹²

⇒ y² = ar³ × ar⁹

⇒ y² = xz.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

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