Question

For what value of k are the numbers x, 2x + k, 3x + 6 are three consecutive terms of an AP?

Answer 1

Hint: To solve the given question, we need to know how the numbers in an AP look like and that in AP, the difference between two consecutive terms of the sequence will be constant. Suppose, a is the first element, then a, a + r, a + 2r, … forms an AP with the common difference as r. We will use the common difference property of an AP to solve this question.

Complete step-by-step answer:
We have been asked in the question that for what value of k, the numbers x, 2x + k, 3x + 6 will be the three consecutive terms of an AP.
So, given that the terms, x, 2x + k, 3x + 6 will be the three consecutive terms of an AP, then we can say that the difference between the terms x and 2x + k will be equal to the difference between the terms 3x + 6 and 2x + k. So, we can write the same as follows.
2x + k – (x) = (3x + 6) – (2x + k)
2x + k – x = 3x + 6 – 2x – k
On the simplifying the LHS and RHS, we will get,
x + k = x + 6 – k
On taking x and k from the LHS to the RHS, we get,
x + 6 – k – x – k = 0
On simplifying further, we will get,
6 – 2k = 0
On taking -2k from the LHS to RHS, we get,
6 = 2k
We can also write it as,
2k = 6
Now, on dividing both the sides of the equation by 2, we get,
k = 3
Hence, when the value of k = 3, the numbers x, 2x + k and 3x + 6 forms an AP. So, the AP will be x, 2x + 3, 3x + 6.

Note: We have used the common difference property of the AP to find the value of k in this question. If the numbers were taken from a GP, then we should have used the common ratio property to find the value of k. According to the common ratio property, the ratio of the first and the second term will be equal to the ratio of the second and the third term and so on.