Question
Show that the progressions given below is an AP. Find the first term, common difference and next term.
$-1,\frac{-5}{6},\frac{-2}{3},\frac{-1}{2},....$
$-1,\frac{-5}{6},\frac{-2}{3},\frac{-1}{2},....$
✓
Answer
A sequence in which each term differs from its preceding term by a constant is called an AP.
$\frac{-5}{6}-(-1)=\frac{-5}{6}+1=\frac{1}{6}$
$\frac{-2}{3}=\Big(\frac{-5}{6}\Big)=\frac{1}{6}$
$\frac{-1}{2}-\Big(\frac{-2}{3}\Big)=\frac{1}{6}$
clearly, the progression an AP.
The first term = -1
common difference $=\frac{1}{6}$
The next term $=\frac{-1}{2}+\Big(\frac{1}{6}\Big)=\frac{-1}{3}$
$\frac{-5}{6}-(-1)=\frac{-5}{6}+1=\frac{1}{6}$
$\frac{-2}{3}=\Big(\frac{-5}{6}\Big)=\frac{1}{6}$
$\frac{-1}{2}-\Big(\frac{-2}{3}\Big)=\frac{1}{6}$
clearly, the progression an AP.
The first term = -1
common difference $=\frac{1}{6}$
The next term $=\frac{-1}{2}+\Big(\frac{1}{6}\Big)=\frac{-1}{3}$
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