Question
Find:
The $18^{th}$ term of the AP $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ....$
The $18^{th}$ term of the AP $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ....$
✓
Answer
The given AP is $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98},\ ...$
⇒ The AP can be rewritten as $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\ ...$
$\text{a}=\sqrt{2}$ and $\text{d}=3\sqrt{2}-\sqrt{2}=2\sqrt{2}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{a}_{18}=\sqrt{2}+17(2\sqrt{2})$
$\Rightarrow\text{a}_{18}=\sqrt{2}+34\sqrt{2}$
$\Rightarrow\text{a}_{18}=\sqrt{2450}$
So, the $18^{th}$ term is $\sqrt{2450}.$
⇒ The AP can be rewritten as $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\ ...$
$\text{a}=\sqrt{2}$ and $\text{d}=3\sqrt{2}-\sqrt{2}=2\sqrt{2}$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\text{a}_{18}=\sqrt{2}+17(2\sqrt{2})$
$\Rightarrow\text{a}_{18}=\sqrt{2}+34\sqrt{2}$
$\Rightarrow\text{a}_{18}=\sqrt{2450}$
So, the $18^{th}$ term is $\sqrt{2450}.$
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