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Question: If \(k+2,4k-6,3k-2\) are the three consecutive terms of an AP, then the value of k is \[\begin{ali...

If k+2,4k−6,3k−2k+2,4k-6,3k-2 are the three consecutive terms of an AP, then the value of k is

& A.2 \\\ & B.3 \\\ & C.4 \\\ & D.5 \\\ \end{aligned}$$
Explanation

Solution

In this question, we are given the three consecutive terms of an arithmetic progression in terms of k and we need to find the value of k. For this, we will use the arithmetic mean property. According to the arithmetic mean property, the sum of the first term and the third term of an arithmetic progression is equal to the twice of the second term of the arithmetic progression.

Complete step by step answer:
Here we are given the three consecutive terms of an arithmetic progression as k+2,4k−6,3k−2k+2,4k-6,3k-2. Since these are in order, we can suppose the first term as k+2, the second term as 4k-6 and the third term as 3k-2. Now, we know by arithmetic mean property that, the sum of the first term and the third term of an arithmetic progression is equal to twice of the second term of an arithmetic progression. So we can say for a given arithmetic progression that, (k+2)+(3k−2)\left( k+2 \right)+\left( 3k-2 \right) is equal to the twice of (4k−6)\left( 4k-6 \right).
In mathematical terms we can say that,
k+2+3k−2=2(4k−6)⇒k+2+3k−2=8k−12k+2+3k-2=2\left( 4k-6 \right)\Rightarrow k+2+3k-2=8k-12.
Taking variables on one side and constant on the other side we get,
k+3k−8k=−12−2+2k+3k-8k=-12-2+2.
Simplifying it we get,
4k−8k=−14+2⇒−4k=−124k-8k=-14+2\Rightarrow -4k=-12.
Cancelling negative sign from both sides we get,
4k=124k=12.
Dividing both sides by 4, we get:
k=124=3k=\dfrac{12}{4}=3.

So, the correct answer is “Option B”.

Note: Students can solve this sum using the following way. We know that, in an arithmetic progression, the difference between the second term and the first term is equal to the difference between the third term and the second term. So we get,
(4k−6)−(k+2)=(3k−2)−(4k−6) ⇒4k−6−k−2=3k−2−4k−6 \begin{aligned} & \left( 4k-6 \right)-\left( k+2 \right)=\left( 3k-2 \right)-\left( 4k-6 \right) \\\ & \Rightarrow 4k-6-k-2=3k-2-4k-6 \\\ \end{aligned}
Taking variables on one side and constants on the other side we get,
4k−k−3k+4k=−2+6+6+2 ⇒4k=12 ⇒k=3 \begin{aligned} & 4k-k-3k+4k=-2+6+6+2 \\\ & \Rightarrow 4k=12 \\\ & \Rightarrow k=3 \\\ \end{aligned}
Which is the same answer. Students should take care of the signs while solving the equation.