Question
Find:
$18^{th}$ term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
$18^{th}$ term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
✓
Answer
$\text{A.P.}=\sqrt{2},3\sqrt{2},5\sqrt{2},\ .....$
Here,
First term $\text{(a)}=\sqrt{2}$
and Common difference $=3\sqrt{2}-\sqrt{2}$
$=2\sqrt{2}\text{ or }5\sqrt{2}-3\sqrt{2}=2\sqrt{2}$
$\text{Now }\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\therefore\ \text{a}_{18}=\sqrt{2}+(18-1)\times2\sqrt{2}$
$=\sqrt{2}+17\times2\sqrt{2}$
$=\sqrt{2}+34\sqrt{2}=35\sqrt{2}$
Here,
First term $\text{(a)}=\sqrt{2}$
and Common difference $=3\sqrt{2}-\sqrt{2}$
$=2\sqrt{2}\text{ or }5\sqrt{2}-3\sqrt{2}=2\sqrt{2}$
$\text{Now }\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\therefore\ \text{a}_{18}=\sqrt{2}+(18-1)\times2\sqrt{2}$
$=\sqrt{2}+17\times2\sqrt{2}$
$=\sqrt{2}+34\sqrt{2}=35\sqrt{2}$
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