Question

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P. Find the value of G.

Answer 1

Given,

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.

∴ `(2x + 2)/x = (3x + 3)/(2x + 2)`

⇒ (2x + 2)² = x(3x + 3)

⇒ (2x)² + 2² + 2 × 2x × 2 = 3x² + 3x

⇒ 4x² + 4 + 8x = 3x² + 3x

⇒ 4x² − 3x² + 8x − 3x + 4 = 0

⇒ x² + 5x + 4 = 0

⇒ x² + 4x + x + 4 = 0

⇒ x(x + 4) + 1(x + 4) = 0

⇒ (x + 1)(x + 4) = 0

⇒ x + 1 = 0 or x + 4 = 0

⇒ x = −1 or x = −4

Substituting value of x = −1, in terms we get:

Terms: −1, 2(−1) + 2, 3(−1) + 3, G

= −1, −2 + 2, −3 + 3, G

= −1, 0, 0, G

This is not possible as in this case common ratio is different.

Substituting value of x = -4, in terms we get:

Terms: −4, 2(−4) + 2, 3(−4) + 3, G

= −4, −8 + 2, −12 + 3, G

= −4, −6, −9, G

Here, common difference = `(−6)/(−4) = 3/2`

G = `−9 × 3/2 = −27/2`

Hence, G = `−27/2`