# 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}} \\

$

Last updated date: 29th Mar 2023

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Answer

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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.

“Complete step-by-step answer:”

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. $2b = a + c$

For the given terms to be in an AP, twice of the middle term which is k should be equal to the sum of the first term which is (k+2) and the second term which is (3k-2).

$

2k = \left( {k + 2} \right) + \left( {3k - 2} \right) \\

\Rightarrow 2k = 4k \\

\Rightarrow 4k - 2k = 0 \\

\Rightarrow 2k = 0 \\

\Rightarrow k = 0 \\

$

Therefore, the required value of k is 0.

Hence, option A is correct.

Also, the three consecutive terms of an AP are 2 , 0 ,-2.

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 \\

$.

“Complete step-by-step answer:”

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. $2b = a + c$

For the given terms to be in an AP, twice of the middle term which is k should be equal to the sum of the first term which is (k+2) and the second term which is (3k-2).

$

2k = \left( {k + 2} \right) + \left( {3k - 2} \right) \\

\Rightarrow 2k = 4k \\

\Rightarrow 4k - 2k = 0 \\

\Rightarrow 2k = 0 \\

\Rightarrow k = 0 \\

$

Therefore, the required value of k is 0.

Hence, option A is correct.

Also, the three consecutive terms of an AP are 2 , 0 ,-2.

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 \\

$.

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