Question
Show that the progressions given below is an AP. Find the first term, common difference and next term.
$\sqrt{20},\sqrt{45},\sqrt{80},\sqrt{125},\ ....$
$\sqrt{20},\sqrt{45},\sqrt{80},\sqrt{125},\ ....$
✓
Answer
A sequence in which each term differs from its preceding term by a constant is called an AP.
$\sqrt{45}-\sqrt{20}=3\sqrt{5}-2\sqrt{5}=\sqrt{5}$
$\sqrt{80}-\sqrt{45}=4\sqrt{5}-3\sqrt{5}=\sqrt{5}$
$\sqrt{125}-\sqrt{80}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$
clearly, the progression an AP.
The first term $=\sqrt{20}=2\sqrt{5}$
common difference $=\sqrt{5}$
The next term $\sqrt{125}+\sqrt{5}=5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$
$\sqrt{45}-\sqrt{20}=3\sqrt{5}-2\sqrt{5}=\sqrt{5}$
$\sqrt{80}-\sqrt{45}=4\sqrt{5}-3\sqrt{5}=\sqrt{5}$
$\sqrt{125}-\sqrt{80}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$
clearly, the progression an AP.
The first term $=\sqrt{20}=2\sqrt{5}$
common difference $=\sqrt{5}$
The next term $\sqrt{125}+\sqrt{5}=5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$
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|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
|
Frequency
|
16
|
p
|
30
|
32
|
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|