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The AP is only (i) in these three sequences.
(i) 3, 3 + √2, 3 + 2√2, 3 + 3√2
a = 3
Because each time √2 is added
d ⇒ 3 + √2 - 3 = √2
⇒ 3 + 2√2 - 3 - √2 = √2
Three more terms
= a + 4d, a + 5d, a + 6d
= a + 4√2, a + 5√2, a + 6√2
But (ii) and (iii) do not form AP, because
(ii) Each time the power increases and this cannot be an AP,
d ≠ a² - a ≠ a³ - a²
(iii) Each time 3 is added inside power and this cannot be an AP
d ≠ √6 - √3 ≠ √9 - √6 ⇒ √6 - 3 ≠ 3 - √6
(i) 3, 3 + √2, 3 + 2√2, 3 + 3√2
a = 3
Because each time √2 is added
d ⇒ 3 + √2 - 3 = √2
⇒ 3 + 2√2 - 3 - √2 = √2
Three more terms
= a + 4d, a + 5d, a + 6d
= a + 4√2, a + 5√2, a + 6√2
But (ii) and (iii) do not form AP, because
(ii) Each time the power increases and this cannot be an AP,
d ≠ a² - a ≠ a³ - a²
(iii) Each time 3 is added inside power and this cannot be an AP
d ≠ √6 - √3 ≠ √9 - √6 ⇒ √6 - 3 ≠ 3 - √6
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