Question

# If $k + 2,k,3k - 2$ are three consecutive terms of an AP then what is the value of k?${\text{A}}{\text{. 0}} \\ {\text{B}}{\text{. 6}} \\ {\text{C}}{\text{. 5}} \\ {\text{D}}{\text{. 8}} \\$

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Hint- Here, we will proceed by using the condition for three consecutive terms to be in AP which is $2b = a + c$ where b is the middle term, a is first term and c is the third term. From here we will get one equation in one variable i.e. k.

Given, $k + 2, k ,3k - 2$ are three consecutive terms of an AP.
As we know that for the three consecutive terms a, b and c to be in an AP, twice of the second term should be equal to the sum of the first and last term i.e. $2b = a + c$
$2k = \left( {k + 2} \right) + \left( {3k - 2} \right) \\ \Rightarrow 2k = 4k \\ \Rightarrow 4k - 2k = 0 \\ \Rightarrow 2k = 0 \\ \Rightarrow k = 0 \\$
Note- Arithmetic progression (AP) is a series in which the common difference between any two consecutive terms remains the same. For any three consecutive terms a, b and c to be in AP, the difference between the first two terms should be equal to the difference between last two terms i.e., $b - a = c - b \\ \Rightarrow b + b = c + a \\ \Rightarrow 2b = a + c \\$.