Is it true that \(-1,\; -\dfrac{3}{2},\; -2,\; -\dfrac{5}{2},\; \ldots\) forms an AP because \(a_2-a_1=a_3-a_2\)? Justify.
True.
Step 1: Write the first three terms clearly.
\(a_1 = -1, \; a_2 = -\tfrac{3}{2}, \; a_3 = -2\).
Step 2: Find the difference between the 2nd and 1st term.
\(a_2 - a_1 = -\tfrac{3}{2} - (-1) = -\tfrac{3}{2} + 1\).
\(-\tfrac{3}{2} + 1 = -\tfrac{3}{2} + \tfrac{2}{2} = -\tfrac{1}{2}\).
Step 3: Find the difference between the 3rd and 2nd term.
\(a_3 - a_2 = -2 - (-\tfrac{3}{2}) = -2 + \tfrac{3}{2}\).
\(-2 = -\tfrac{4}{2}\), so \(-\tfrac{4}{2} + \tfrac{3}{2} = -\tfrac{1}{2}\).
Step 4: Both differences are equal.
\(a_2 - a_1 = a_3 - a_2 = -\tfrac{1}{2}\).
Step 5: If the difference between consecutive terms is the same, the sequence is an Arithmetic Progression (AP).
So, the given sequence is an AP with common difference \(d = -\tfrac{1}{2}\).