Question

For a G.P., its fourth term is x, seventh term is y and tenth term is z. 

Assertion (A): x, y and z are in G.P. 

Reason (R): \[y^{2}=(ar^{6})^{2}=ar^{3}\times ar^{9}=xz\].

Options

• A is true, R is false.

• A is false, R is true.

• Both A and R are true and R is the correct reason for A.

• Both A and R are true and R is the incorrect reason for A.

Answer 1

Both A and R are true and R is the correct reason for A.

Explanation:

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

By formula:

⇒ \[T_n = a \cdot r^{n \: - \: 1}\]

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

⇒ \[a_4 = x\], \[a_7 = y\] and \[a_{10} = z\]

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

⇒ \[ar^3 = x\], \[ar^6 = y\] and \[ar^9 = z\]

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

Ratio between \[y\] and \[x\]:

⇒ \[\frac{y}{x} = \frac{ar^6}{ar^3} = r^3.\]

Ratio between \[z\] and \[y\]:

⇒ \[\frac{z}{y} = \frac{ar^9}{ar^6} = r^3.\]

Since, the ratio between the consecutive terms are equal.

Thus, \[x\], \[y\] and \[z\] are in G.P.

∴ Assertion (A) is true.

⇒ \[y^2 = (ar^6)^2\]

⇒ \[y^2 = ar^{12}\]

⇒ \[y^2 = ar^3 \times ar^9\]

⇒ \[y^2 = xz.\]

∴ Reason (R) is true.