Question

Given that k-5, k+7, k+55 are three consecutive terms in a geometric sequence, solve for k​

Answer 1

Answer:

$k = 9$

Step-by-step explanation:

Given:

Consecutive Terms:  k-5, k+7, k+55

Required:

Determine the value of k

To do this, we make use of the concept of common ratio.

The common ratio (r) of a geometric sequence is:

$r = \frac{T_{n}}{T_{n-1}}$

In other words:

$r = \frac{T_3}{T_2} = \frac{T_2}{T_1}$

Where 1, 2 and 3 represents the terms of the progression/sequence

So:

$r = \frac{T_3}{T_2} = \frac{T_2}{T_1}$ becomes

$\frac{k+55}{k+7} = \frac{k+7}{k-5}$

Cross Multiply:

$(k+55)(k-5) = (k+7)(k+7)$

Open Brackets

$k^2 + 55k - 5k - 275 = k^2 + 7k + 7k + 49$

$k^2 + 50k- 275 = k^2 + 14k + 49$

Collect Like Terms

$k^2 - k^2 + 50k-14k = 275 + 49$

$36k = 324$

Solve for k

$k = 324/36$

$k = 9$