(i) a, a2, a3, a4 …
(ii) √2, √8, √18, √32 …
Solution(i):
Given here,
${{a}_{2}}-{{a}_{1}}={{a}^{2}}-a=a(a-1)$
${{a}_{3}}-{{a}_{2}}={{a}^{3}}-{{a}^{2}}={{a}^{2}}(a-1)$
${{a}_{4}}-{{a}_{3}}={{a}^{4}}-{{a}^{3}}={{a}^{3}}(a-1)$
Since, an+1 – an or the common difference varies from time to time.
Therefore, the given series does not form a A.P.
Solution(ii):
Given here,
${{a}_{2}}-{{a}_{1}}=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
${{a}_{3}}-{{a}_{2}}=\sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2}$
${{a}_{4}}-{{a}_{3}}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$
Since, an+1 – an or the common difference remains the same every time.
Therefore, d = √2 so the given series forms a A.P.
The following three terms are;
${{a}_{5}}=\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}$
${{a}_{6}}=5\sqrt{2}+\sqrt{2}=6\sqrt{2}=\sqrt{72}$
${{a}_{7}}=6\sqrt{2}+\sqrt{2}=7\sqrt{2}+\sqrt{2}=\sqrt{98}$