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Sequence and Series

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Question : 84 of 86

Marks: +1, -0

For what values of

x,

-\;\frac{2}{7}, x,-\;\frac{7}{2}

are in G.P.?

±1
±3
±2
none of these

Solution:

Concept:

For three numbers to be in G.P., the square of the middle term equals the product of the first and third terms.

Explanation:

Given terms:

-\frac{2}{7}

,

x

,

-\frac{7}{2}

are in G.P.

So,

x^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right)

.

Compute product:

\left(-\frac{2}{7}\right) \times \left(-\frac{7}{2}\right) = \frac{2 \times 7}{7 \times 2} = 1

.

Thus,

x^2 = 1 \implies x = \pm 1

.

Answer:

x = \pm 1

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