If $k+7, 2k-2$ and $2k+6$ are three consecutive terms of an A.P., then the value of $k$ is:
(A) 15
(B) 17
(C) 5
(D) 1
Answer (1)
Set up the equation based on the constant difference between consecutive terms in an A.P.: ( 2 k − 2 ) − ( k + 7 ) = ( 2 k + 6 ) − ( 2 k − 2 ) .
Simplify the left side of the equation: ( 2 k − 2 ) − ( k + 7 ) = k − 9 .
Simplify the right side of the equation: ( 2 k + 6 ) − ( 2 k − 2 ) = 8 .
Solve for k : k − 9 = 8 ⇒ k = 17 . The value of k is 17 .
Explanation
Understanding the Problem We are given three consecutive terms of an arithmetic progression (A.P.): k + 7 , 2 k − 2 , and 2 k + 6 . Our goal is to find the value of k . In an arithmetic progression, the difference between consecutive terms is constant. This means that the difference between the second term and the first term is equal to the difference between the third term and the second term.
Setting up the Equation We can set up the equation: ( 2 k − 2 ) − ( k + 7 ) = ( 2 k + 6 ) − ( 2 k − 2 ) . Now, we simplify the equation.
Simplifying the Left Side First, let's simplify the left side of the equation: ( 2 k − 2 ) − ( k + 7 ) = 2 k − 2 − k − 7 = k − 9
Simplifying the Right Side Next, let's simplify the right side of the equation: ( 2 k + 6 ) − ( 2 k − 2 ) = 2 k + 6 − 2 k + 2 = 8
Solving for k Now we have the simplified equation: k − 9 = 8 To solve for k , we add 9 to both sides of the equation: k = 8 + 9 = 17
Final Answer Therefore, the value of k is 17.
Examples
Arithmetic progressions are useful in many real-world scenarios, such as predicting the cost of items over time with a constant rate of increase, calculating the number of seats in a stadium where each row has a fixed number of additional seats, or determining the amount of savings with regular, consistent deposits. For instance, if you save $100 in the first month, $150 in the second month, and $200 in the third month, this forms an arithmetic progression. Understanding A.P. helps you predict your savings in future months assuming the same pattern continues.
