Find \(a, b, c\) such that \(a,\;7,\;b,\;23,\;c\) are in AP.
\(a=-1,\; b=15,\; c=31\)
Step 1: Recall what an Arithmetic Progression (AP) means.
In an AP, the difference between two consecutive terms is always the same.
Another useful fact: Any middle term in an AP is the average of the terms just before and just after it.
Step 2: Use this idea for \(b\).
Here, \(b\) lies between \(7\) and \(23\).
So, \(b = \dfrac{7 + 23}{2} = \dfrac{30}{2} = 15\).
Step 3: Now find \(a\).
Here, \(7\) lies between \(a\) and \(b\).
So, \(7 = \dfrac{a + b}{2}\).
Substitute \(b = 15\): \(7 = \dfrac{a + 15}{2}\).
Multiply both sides by 2: \(14 = a + 15\).
So, \(a = 14 - 15 = -1\).
Step 4: Finally, find \(c\).
Here, \(23\) lies between \(b\) and \(c\).
So, \(23 = \dfrac{b + c}{2}\).
Substitute \(b = 15\): \(23 = \dfrac{15 + c}{2}\).
Multiply both sides by 2: \(46 = 15 + c\).
So, \(c = 46 - 15 = 31\).
Final Answer: \(a = -1,\; b = 15,\; c = 31\).