Question

Check if given series is $AP$ or not? If it is an $AP$, find the common difference $d$ and write three more terms.

$3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},.$

Answer 1

$3,3\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}$

$\Rightarrow$ Difference between second and first term $=3+\sqrt{2}-3=\sqrt{2}$

$\Rightarrow$ Difference between third and second term $=(3+2\sqrt{2})-(3+\sqrt{2})$

$=3+2\sqrt{2}-3-\sqrt{2}$

$=\sqrt{2}$

$\Rightarrow$ Difference between fourth and third term $=(3+3\sqrt{2})-(3+2\sqrt{2})$

$=3+3\sqrt{2}-3-2\sqrt{2}$

$=\sqrt{2}$

$\Rightarrow$ Since, difference is same, it is an $A.P$

$\Rightarrow$ Common difference $=d=\sqrt{2}$

$\Rightarrow$ We have to find next three terms. We are given $4$ terms.

$\Rightarrow$ We have to find $5^{th},6^{th}$ and $7^{th}$ terms.

$\Rightarrow$ $5^{th}$ term $=4^{th}term+d=3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}$

$\Rightarrow$ $6^{th}$ term $=5^{th}term+d=3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}$

$\Rightarrow$ $7^{th}$ term $=6^{th}term+d=3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}$

Hence $5^{th},6^{th}$ and $7^{th}$ terms are $3+4\sqrt{2},3+5\sqrt{2},3+6\sqrt{2}$