Question
In the AP, 5,?,?, $9\frac{1}{2}$ find the missing terms?
✓
Answer
Let the common difference of the given AP be d.
a = 5
4th term $ = 9\frac{1}{2}$
$ \Rightarrow 5 + (4 - 1)d = \frac{{19}}{2}\left[ {\because {a_n} = a + (n - 1)d} \right]$
$ \Rightarrow 3d = \frac{{19}}{2} - 5$
$ \Rightarrow 3d = \frac{9}{2}$
$ \Rightarrow d = \frac{3}{2}$
Therefore,
Second term $ = 5 + \frac{3}{2} = \frac{{13}}{2} = 6\frac{1}{2}$
and, third term $ = \frac{{13}}{2} + \frac{3}{2} = 8$
Hence, the missing terms in the boxes are $6\frac{1}{2}$ and 8.
a = 5
4th term $ = 9\frac{1}{2}$
$ \Rightarrow 5 + (4 - 1)d = \frac{{19}}{2}\left[ {\because {a_n} = a + (n - 1)d} \right]$
$ \Rightarrow 3d = \frac{{19}}{2} - 5$
$ \Rightarrow 3d = \frac{9}{2}$
$ \Rightarrow d = \frac{3}{2}$
Therefore,
Second term $ = 5 + \frac{3}{2} = \frac{{13}}{2} = 6\frac{1}{2}$
and, third term $ = \frac{{13}}{2} + \frac{3}{2} = 8$
Hence, the missing terms in the boxes are $6\frac{1}{2}$ and 8.
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