Question

If K, 2K \(-1\), 6 are in A.P., then the value of K is :

• A. \(\frac{8}{3}\)
• B. \(\frac{7}{3}\)
• C. \(\frac{5}{3}\)
• D. \(\frac{1}{3}\)

Hint:

If three terms \(p, q, r\) are in Arithmetic Progression (A.P.), then the middle term \(q\) is the arithmetic mean of \(p\) and \(r\). So, \(q = \frac{p+r}{2}\), or equivalently, \(2q = p+r\). Given terms: \(p=K\), \(q=2K-1\), \(r=6\). 1. Set up the equation: \(2(2K-1) = K + 6\). 2. Simplify: \(4K - 2 = K + 6\). 3. Solve for K: \(4K - K = 6 + 2\) \(3K = 8\) \(K = \frac{8}{3}\).

Answer 1

The Correct Option is A

Concept: If three terms, say \(p, q, r\), are in an Arithmetic Progression (A.P.), then the common difference between consecutive terms is constant. This means \(q - p = r - q\). This can be rearranged to \(2q = p + r\), which means the middle term is the arithmetic mean of the first and third terms. Step 1: Identify the three terms in A.P. The given terms are:
First term (\(p\)) = K
Second term (\(q\)) = 2K - 1
Third term (\(r\)) = 6 Step 2: Apply the property of A.P. Since the terms are in A.P., the difference between the second and first term is equal to the difference between the third and second term: \[ (2K - 1) - K = 6 - (2K - 1) \] Alternatively, using the property \(2q = p+r\): \[ 2(2K - 1) = K + 6 \] Step 3: Solve the equation for K Let's use the second formulation: \(2(2K - 1) = K + 6\). Distribute the 2 on the left side: \[ 4K - 2 = K + 6 \] Now, gather K terms on one side and constant terms on the other. Subtract K from both sides: \[ 4K - K - 2 = 6 \] \[ 3K - 2 = 6 \] Add 2 to both sides: \[ 3K = 6 + 2 \] \[ 3K = 8 \] Divide by 3: \[ K = \frac{8}{3} \] Let's verify using the first formulation: \((2K - 1) - K = 6 - (2K - 1)\) \[ K - 1 = 6 - 2K + 1 \] \[ K - 1 = 7 - 2K \] Add 2K to both sides: \[ K + 2K - 1 = 7 \] \[ 3K - 1 = 7 \] Add 1 to both sides: \[ 3K = 8 \] \[ K = \frac{8}{3} \] Both methods yield the same result. The value of K is \(\frac{8}{3}\). This matches option (1).